[R] to calculate c(rep(1, 43), rep(2,43),...., rep(10,43))
Hello I would like to find an elegant way of calculating c(rep(1, 43), rep(2,43),, rep(10,43)) Any idea ? Thank you -- Michel ARNAUD DGDRD-Drh - TA 174/04 tel : 04.67.61.75.38 port: 06.47.43.55.31 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] outputs of a function
Hello
I have a function named FctModele13 in which
1) I calculate a dataframe named Total and
2) I used ggplot2.
I have the following problem. I cannot produce simultaneously
* the graphic by ggplot2
* the dataframe
My simplified code is the following one :
TT - FctModele13(ListePlusde50ans, PourcentSexeCsp, NbAn=10)
FctModele - function(ListePlusde50ans, PourcentSexeCsp, NbAn)
{
# calculate Total
.
.
Total - data.frame()
###
# plot by ggplot
library(ggplot2)
ggplot(.) +
..
axis.title.y = element_text(size = 8)) +
labs(title=Title)
Total
}
Any idea ?
--
Michel ARNAUD
CIRAD Montpellier
tel : 04.67.61.75.38
port: 06.47.43.55.31
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] outputs of a function
Thank you Jeff and Mark for your help
Michel
Le 04/12/2014 15:09, Jeff Newmiller a écrit :
This is a poor approach from a usability perspective... I suggest you create
two separate functions rather than one.
However, you seem to be missing a crucial point in the use of ggplot, which also applies
to lattice graphics. These functions actually don't produce output at all... they produce
grid graphics objects that produce graphic output when printed.
Interactively, this printing step is done automatically for you, but inside functions
that does not happen. So, you can wrap your ggplot expression in a print function call to
have your function produce the graphic output as a side effect.
print( ggplot(.) +
..
axis.title.y = element_text(size = 8)) +
labs(title=Title) )
One of the things that is nice about grid graphics is that you can modify the
object before you print it. For example, if you make a basic scatterplot
function for your data, you can tack on things like labels or extra lines to
aid your explanation about what is in the data just as you print it. Then you
can also print it later with different notations or none at all. Having a
separate graph-generating function that just returns the grid object for you to
print or modify as you wish can be quite useful later.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On December 4, 2014 2:09:12 AM PST, Arnaud Michel michel.arn...@cirad.fr
wrote:
Hello
I have a function named FctModele13 in which
1) I calculate a dataframe named Total and
2) I used ggplot2.
I have the following problem. I cannot produce simultaneously
* the graphic by ggplot2
* the dataframe
My simplified code is the following one :
TT - FctModele13(ListePlusde50ans, PourcentSexeCsp, NbAn=10)
FctModele - function(ListePlusde50ans, PourcentSexeCsp, NbAn)
{
# calculate Total
.
.
Total - data.frame()
###
# plot by ggplot
library(ggplot2)
ggplot(.) +
..
axis.title.y = element_text(size = 8)) +
labs(title=Title)
Total
}
Any idea ?
--
Michel ARNAUD
DGDRD-Drh - TA 174/04
tel : 04.67.61.75.38
port: 06.47.43.55.31
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] To calculate the month number between 2 dates
Hello Can one calculate the month number between two dates D1 - 01/01/2007 and D2 - 01/04/2009 ? Thank you -- Michel ARNAUD Cirad __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] To build a new Df from 2 Df
Thank you David Now, the problem is to list all the combinations which verify the condition III (ie every Rapporteur has to have more or less the same number of demandeur) Have you any idea ? Michel Le 14/10/2014 13:18, david.kaeth...@dlr.de a écrit : Hello, here's a draft of a solution. I hope it's not overly complicated. # find all possible combinations combi - expand.grid(Dem$Nom, Rap$Nom); names(combi) - c(Dem, Rap) # we need the corresponding departments and units combi$DemDep - apply(combi, 1, function(x) Dem$Departement[x[1] == Dem$Nom]) combi$DemUni - apply(combi, 1, function(x) Dem$Unite[x[1] == Dem$Nom]) combi$RapDep - apply(combi, 1, function(x) Rap$Departement[x[2] == Rap$Nom]) combi$RapUni - apply(combi, 1, function(x) Rap$Unite[x[2] == Rap$Nom]) # we exclude the combinations that we don't want dep - combi[combi$DemDep != combi$RapDep, c(Dem, Rap)] dep$id - as.numeric(dep$Rap) uni - combi[combi$DemUni != combi$RapUni, c(Dem, Rap)] uni$id - as.numeric(uni$Rap) # preliminary result resDep - reshape(dep, timevar = id, idvar = Dem, direction = wide ) resUni - reshape(uni, timevar = id, idvar = Dem, direction = wide ) In resDep and resUni you find the results for Rapporteur1 and Rapporteur2. NAs indicate where conditions did not match. For Rap1/Rap2 you can now choose any column from resDep and resUni that is not NA for that specific Demandeur. I wasn't exactly sure about your third condition, so I'll leave that to you. But with the complete possible matches, you have a more general solution. Btw, you can construct data.frames just like this: Dem - data.frame( Nom = c(John, Jim, Julie, Charles, Michel, Emma, Sandra, Elodie, Thierry, Albert, Jean, Francois, Pierre, Cyril, Damien, Jean-Michel, Vincent, Daniel, Yvan, Catherine), Departement = c(D, A, A, C, D, B, D, B, C, D, B, B, B, A, C, D, B, A, D, D), Unite = c(Unite8, Unite4, Unite4, Unite7, Unite9, Unite1, Unite6, Unite5, Unite7, Unite3, Unite2, Unite6, Unite8, Unite8, Unite3, Unite8, Unite9, Unite7, Unite9, Unite5) ) -dk -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von Arnaud Michel Gesendet: Dienstag, 14. Oktober 2014 10:46 An: r-help@r-project.org Betreff: [R] To build a new Df from 2 Df Hello I have 2 df Dem and Rap. I would want to build all the df (dfnew) by associating these two df (Dem and Rap) in the following way : For each value of Dem$Nom (dfnew$Demandeur), I associate 2 different values of Rap$Nom (dfnew$Rapporteur1 and dfnew$Rapporteur2) in such a way * for each dfnew$Demandeur, dfnew$Rapporteur1 does not have the same value for Departement as Dem$Departement * for each dfnew$Demandeur, dfnew$Rapporteur2 does not have the same value for Unite as Dem$Unite * the value of table(dfnew$Rapporteur1) and the value of table(dfnew$Rapporteur2) must be balanced and not too different (Accepted differences : 1) table(dfnew$Rapporteur1) Rapporteur01 Rapporteur02 Rapporteur03 Rapporteur04 Rapporteur05 4 4 4 4 4 Thanks for your help Michel Dem - structure(list(Nom = c(John, Jim, Julie, Charles, Michel, Emma, Sandra, Elodie, Thierry, Albert, Jean, Francois, Pierre, Cyril, Damien, Jean-Michel, Vincent, Daniel, Yvan, Catherine), Departement = c(D, A, A, C, D, B, D, B, C, D, B, B, B, A, C, D, B, A, D, D), Unite = c(Unite8, Unite4, Unite4, Unite7, Unite9, Unite1, Unite6, Unite5, Unite7, Unite3, Unite2, Unite6, Unite8, Unite8, Unite3, Unite8, Unite9, Unite7, Unite9, Unite5)), .Names = c(Nom, Departement, Unite ), row.names = c(NA, -20L), class = data.frame) Rap - structure(list(Nom = c(Rapporteur01, Rapporteur02, Rapporteur03, Rapporteur04, Rapporteur05), Departement = c(C, D, C, C, D), Unite = c(Unite10, Unite6, Unite5, Unite5, Unite4)), .Names = c(Nom, Departement, Unite), row.names = c(NA, -5L), class = data.frame) dfnew - structure(list(Demandeur = structure(c(13L, 12L, 14L, 3L, 15L, 8L, 17L, 7L, 18L, 1L, 10L, 9L, 16L, 4L, 5L, 11L, 19L, 6L, 20L, 2L), .Label = c(Albert, Catherine, Charles, Cyril, Damien, Daniel, Elodie, Emma, Francois, Jean, Jean-Michel, Jim, John, Julie, Michel, Pierre, Sandra, Thierry, Vincent, Yvan), class = factor), Rapporteur1 = structure(c(3L, 1L, 3L, 5L, 1L, 5L, 1L, 2L, 5L, 4L, 2L, 4L, 2L, 3L, 5L, 4L, 4L, 2L, 3L, 1L), .Label = c(Rapporteur01, Rapporteur02, Rapporteur03, Rapporteur04, Rapporteur05), class = factor), Rapporteur2 = structure(c(1L, 3L, 4L, 4L, 2L, 4L, 5L, 1L, 2L, 3L, 3L, 3L, 5L, 5L, 1L, 1L, 2L, 5L, 4L, 2L), .Label = c(Rapporteur01, Rapporteur02, Rapporteur03, Rapporteur04, Rapporteur05), class = factor)), .Names = c(Demandeur, Rapporteur1, Rapporteur2), row.names = c(NA, -20L), class = data.frame) -- Michel ARNAUD Cirad Montpellier [[alternative HTML version deleted
[R] To build a new Df from 2 Df
Hello I have 2 df Dem and Rap. I would want to build all the df (dfnew) by associating these two df (Dem and Rap) in the following way : For each value of Dem$Nom (dfnew$Demandeur), I associate 2 different values of Rap$Nom (dfnew$Rapporteur1 and dfnew$Rapporteur2) in such a way * for each dfnew$Demandeur, dfnew$Rapporteur1 does not have the same value for Departement as Dem$Departement * for each dfnew$Demandeur, dfnew$Rapporteur2 does not have the same value for Unite as Dem$Unite * the value of table(dfnew$Rapporteur1) and the value of table(dfnew$Rapporteur2) must be balanced and not too different (Accepted differences : 1) table(dfnew$Rapporteur1) Rapporteur01 Rapporteur02 Rapporteur03 Rapporteur04 Rapporteur05 4 4 4 4 4 Thanks for your help Michel Dem - structure(list(Nom = c(John, Jim, Julie, Charles, Michel, Emma, Sandra, Elodie, Thierry, Albert, Jean, Francois, Pierre, Cyril, Damien, Jean-Michel, Vincent, Daniel, Yvan, Catherine), Departement = c(D, A, A, C, D, B, D, B, C, D, B, B, B, A, C, D, B, A, D, D), Unite = c(Unite8, Unite4, Unite4, Unite7, Unite9, Unite1, Unite6, Unite5, Unite7, Unite3, Unite2, Unite6, Unite8, Unite8, Unite3, Unite8, Unite9, Unite7, Unite9, Unite5)), .Names = c(Nom, Departement, Unite ), row.names = c(NA, -20L), class = data.frame) Rap - structure(list(Nom = c(Rapporteur01, Rapporteur02, Rapporteur03, Rapporteur04, Rapporteur05), Departement = c(C, D, C, C, D), Unite = c(Unite10, Unite6, Unite5, Unite5, Unite4)), .Names = c(Nom, Departement, Unite), row.names = c(NA, -5L), class = data.frame) dfnew - structure(list(Demandeur = structure(c(13L, 12L, 14L, 3L, 15L, 8L, 17L, 7L, 18L, 1L, 10L, 9L, 16L, 4L, 5L, 11L, 19L, 6L, 20L, 2L), .Label = c(Albert, Catherine, Charles, Cyril, Damien, Daniel, Elodie, Emma, Francois, Jean, Jean-Michel, Jim, John, Julie, Michel, Pierre, Sandra, Thierry, Vincent, Yvan), class = factor), Rapporteur1 = structure(c(3L, 1L, 3L, 5L, 1L, 5L, 1L, 2L, 5L, 4L, 2L, 4L, 2L, 3L, 5L, 4L, 4L, 2L, 3L, 1L), .Label = c(Rapporteur01, Rapporteur02, Rapporteur03, Rapporteur04, Rapporteur05), class = factor), Rapporteur2 = structure(c(1L, 3L, 4L, 4L, 2L, 4L, 5L, 1L, 2L, 3L, 3L, 3L, 5L, 5L, 1L, 1L, 2L, 5L, 4L, 2L), .Label = c(Rapporteur01, Rapporteur02, Rapporteur03, Rapporteur04, Rapporteur05), class = factor)), .Names = c(Demandeur, Rapporteur1, Rapporteur2), row.names = c(NA, -20L), class = data.frame) -- Michel ARNAUD Cirad Montpellier [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] To Add a variable from Df1 to Df2 which have a same common variable
Hello I have the two dataframes Df1 and Df2 which have the common variable AgeSexeCadNCad I would like to add the new variable Df2$Pourcent which correspond at the value of Df1$AgeSexeCadNCad. Thank you for your help. Michel Df1 - structure(list(AgeSexeCadNCad = structure(1:36, .Label = c(60-Femme-Cadre, 60-Femme-Non Cadre, 60-Homme-Cadre, 60-Homme-Non Cadre, 61-Femme-Cadre, 61-Femme-Non Cadre, 61-Homme-Cadre, 61-Homme-Non Cadre, 62-Femme-Cadre, 62-Femme-Non Cadre, 62-Homme-Cadre, 62-Homme-Non Cadre, 63-Femme-Cadre, 63-Femme-Non Cadre, 63-Homme-Cadre, 63-Homme-Non Cadre, 64-Femme-Cadre, 64-Femme-Non Cadre, 64-Homme-Cadre, 64-Homme-Non Cadre, 65-Femme-Cadre, 65-Femme-Non Cadre, 65-Homme-Cadre, 65-Homme-Non Cadre, 66-Femme-Cadre, 66-Femme-Non Cadre, 66-Homme-Cadre, 66-Homme-Non Cadre, 67-Femme-Cadre, 67-Femme-Non Cadre, 67-Homme-Cadre, 67-Homme-Non Cadre, 68-Femme-Cadre, 68-Femme-Non Cadre, 68-Homme-Cadre, 68-Homme-Non Cadre ), class = factor), Pourcent = c(0.157849638357511, 0.157849638357511, 0.0562149664637629, 0.419279916358023, 0.180720729132166, 0.180720729132166, 0.092720981524322, 0.272158156192425, 0.145668562090518, 0.145668562090518, 0.101319648271574, 0.159207521192769, 0.0997898095090109, 0.0997898095090109, 0.110753346057845, 0.0193586234067497, 0.0795236495990374, 0.0795236495990374, 0.18014205547984, 0.00968491550180694, 0.0750838561972432, 0.0750838561972432, 0.237072554382218, 0.0650665901855087, 0.0587392216209752, 0.0587392216209752, 0.126427289344211, 0.00961707878904615, 0.0409034699088397, 0.0409034699088397, 0.0537806700836756, 3.11172383820597e-05, 0.0285360029533433, 0.0285360029533433, 0.0220930854712636, 2.20203747900568e-09)), .Names = c(AgeSexeCadNCad, Pourcent), row.names = c(28L, 19L, 10L, 1L, 29L, 20L, 11L, 2L, 30L, 21L, 12L, 3L, 31L, 22L, 13L, 4L, 32L, 23L, 14L, 5L, 33L, 24L, 15L, 6L, 34L, 25L, 16L, 7L, 35L, 26L, 17L, 8L, 36L, 27L, 18L, 9L), class = data.frame) Df2 - structure(list(Matricule = c(410, 453, 501, 544, 653, 765, 833, 851, 927, 1050, 1074, 1278, 1379, 1428, 359, 379, 408, 417, 424, 426, 483, 490, 528, 538, 567, 596, 603, 604, 647, 675, 677, 681, 735, 743, 787, 817, 823, 896, 917, 1071, 1144, 1157, 1823, 2497, 2868, 3556, 3614, 3632, 3646, 3656, 3660, 4162, 4503, 4711, 5531, 330, 447, 467, 546, 627, 637, 780, 892, 1487, 1492, 3324, 4873, 409, 415, 441, 579, 619, 697, 716, 719, 728, 737, 807, 832, 989, 1299, 1320, 1352, 1427, 1484, 1548, 2447, 2914, 2929, 2941, 3524, 3527, 3631, 4324, 400, 572, 1095, 1097, 1105, 2966, 392, 418, 440, 457, 466, 472, 488, 491, 506, 533, 543, 547, 552, 553, 920, 1034, 1179, 1454, 1485, 1540, 3620, 4672, 13899, 342, 1089, 1208, 1234, 2153, 3545, 253, 504, 529, 558, 578, 745, 933, 935, 2099, 16785, 356, 460, 634, 959, 1429, 1591, 1720, 3602, 3644, 322, 361, 404, 430, 525, 706, 804, 1010, 1012, 1108, 1185, 1294, 2264, 3567, 3633, 4990, 264, 298, 352, 388, 503, 508, 691, 1509, 2192, 3060, 3683, 877, 1130, 1963, 188, 327, 331, 363, 437, 445, 462, 723, 1259, 1381, 3617, 427, 1402, 3624, 141, 256, 308, 377, 414, 640, 157, 560), AgeSexeCadNCad = c(60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 62-Femme-Non Cadre,
Re: [R] To Add a variable from Df1 to Df2 which have a same common variable
Thank you to Marc Schwartz, Rui Barrada and Sarah Goslee Michel Le 19/09/2014 19:46, Marc Schwartz a écrit : On Sep 19, 2014, at 12:15 PM, Arnaud Michel michel.arn...@cirad.fr wrote: Hello I have the two dataframes Df1 and Df2 which have the common variable AgeSexeCadNCad I would like to add the new variable Df2$Pourcent which correspond at the value of Df1$AgeSexeCadNCad. Thank you for your help. Michel Df1 - structure(list(AgeSexeCadNCad = structure(1:36, .Label = c(60-Femme-Cadre, 60-Femme-Non Cadre, 60-Homme-Cadre, 60-Homme-Non Cadre, 61-Femme-Cadre, 61-Femme-Non Cadre, 61-Homme-Cadre, 61-Homme-Non Cadre, 62-Femme-Cadre, 62-Femme-Non Cadre, 62-Homme-Cadre, 62-Homme-Non Cadre, 63-Femme-Cadre, 63-Femme-Non Cadre, 63-Homme-Cadre, 63-Homme-Non Cadre, 64-Femme-Cadre, 64-Femme-Non Cadre, 64-Homme-Cadre, 64-Homme-Non Cadre, 65-Femme-Cadre, 65-Femme-Non Cadre, 65-Homme-Cadre, 65-Homme-Non Cadre, 66-Femme-Cadre, 66-Femme-Non Cadre, 66-Homme-Cadre, 66-Homme-Non Cadre, 67-Femme-Cadre, 67-Femme-Non Cadre, 67-Homme-Cadre, 67-Homme-Non Cadre, 68-Femme-Cadre, 68-Femme-Non Cadre, 68-Homme-Cadre, 68-Homme-Non Cadre ), class = factor), Pourcent = c(0.157849638357511, 0.157849638357511, 0.0562149664637629, 0.419279916358023, 0.180720729132166, 0.180720729132166, 0.092720981524322, 0.272158156192425, 0.145668562090518, 0.145668562090518, 0.101319648271574, 0.159207521192769, 0.0997898095090109, 0.0997898095090109, 0.110753346057845, 0.0193586234067497, 0.0795236495990374, 0.0795236495990374, 0.18014205547984, 0.00968491550180694, 0.0750838561972432, 0.0750838561972432, 0.237072554382218, 0.0650665901855087, 0.0587392216209752, 0.0587392216209752, 0.126427289344211, 0.00961707878904615, 0.0409034699088397, 0.0409034699088397, 0.0537806700836756, 3.11172383820597e-05, 0.0285360029533433, 0.0285360029533433, 0.0220930854712636, 2.20203747900568e-09)), .Names = c(AgeSexeCadNCad, Pourcent), row.names = c(28L, 19L, 10L, 1L, 29L, 20L, 11L, 2L, 30L, 21L, 12L, 3L, 31L, 22L, 13L, 4L, 32L, 23L, 14L, 5L, 33L, 24L, 15L, 6L, 34L, 25L, 16L, 7L, 35L, 26L, 17L, 8L, 36L, 27L, 18L, 9L), class = data.frame) Df2 - structure(list(Matricule = c(410, 453, 501, 544, 653, 765, 833, 851, 927, 1050, 1074, 1278, 1379, 1428, 359, 379, 408, 417, 424, 426, 483, 490, 528, 538, 567, 596, 603, 604, 647, 675, 677, 681, 735, 743, 787, 817, 823, 896, 917, 1071, 1144, 1157, 1823, 2497, 2868, 3556, 3614, 3632, 3646, 3656, 3660, 4162, 4503, 4711, 5531, 330, 447, 467, 546, 627, 637, 780, 892, 1487, 1492, 3324, 4873, 409, 415, 441, 579, 619, 697, 716, 719, 728, 737, 807, 832, 989, 1299, 1320, 1352, 1427, 1484, 1548, 2447, 2914, 2929, 2941, 3524, 3527, 3631, 4324, 400, 572, 1095, 1097, 1105, 2966, 392, 418, 440, 457, 466, 472, 488, 491, 506, 533, 543, 547, 552, 553, 920, 1034, 1179, 1454, 1485, 1540, 3620, 4672, 13899, 342, 1089, 1208, 1234, 2153, 3545, 253, 504, 529, 558, 578, 745, 933, 935, 2099, 16785, 356, 460, 634, 959, 1429, 1591, 1720, 3602, 3644, 322, 361, 404, 430, 525, 706, 804, 1010, 1012, 1108, 1185, 1294, 2264, 3567, 3633, 4990, 264, 298, 352, 388, 503, 508, 691, 1509, 2192, 3060, 3683, 877, 1130, 1963, 188, 327, 331, 363, 437, 445, 462, 723, 1259, 1381, 3617, 427, 1402, 3624, 141, 256, 308, 377, 414, 640, 157, 560), AgeSexeCadNCad = c(60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Femme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 60-Homme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Femme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre, 61-Homme-Non Cadre
[R] Plotrix and twoord.plot with date on x-axe
Hello With package plotrix and twoord.plot function, I would like to put the labels of the ticks values of x-axe which are date ( c(2006 Jan, 2007 Jan, 2008 Jan, 2009 Jan, 2010 Jan, 2011 Jan, 2012 Jan) with if possible, the angle of labels and x-axe = 40 Any idea ? Thank you for your help My dataframe is TT - structure(list(x1 = structure(c(1L, 8L, 15L, 22L, 29L, 36L, 43L, 50L, 57L, 64L, 71L, 77L, 2L, 9L, 16L, 23L, 30L, 37L, 44L, 51L, 58L, 65L, 3L, 10L, 17L, 24L, 31L, 38L, 45L, 52L, 59L, 66L, 72L, 78L, 4L, 11L, 18L, 25L, 32L, 39L, 46L, 53L, 60L, 67L, 73L, 79L, 5L, 12L, 19L, 26L, 33L, 40L, 47L, 54L, 61L, 68L, 74L, 80L, 6L, 13L, 20L, 27L, 34L, 41L, 48L, 55L, 62L, 69L, 75L, 81L, 7L, 14L, 21L, 28L, 35L, 42L, 49L, 56L, 63L, 70L, 76L, 82L), .Label = c(01/01/2006, 01/01/2007, 01/01/2008, 01/01/2009, 01/01/2010, 01/01/2011, 01/01/2012, 01/02/2006, 01/02/2007, 01/02/2008, 01/02/2009, 01/02/2010, 01/02/2011, 01/02/2012, 01/03/2006, 01/03/2007, 01/03/2008, 01/03/2009, 01/03/2010, 01/03/2011, 01/03/2012, 01/04/2006, 01/04/2007, 01/04/2008, 01/04/2009, 01/04/2010, 01/04/2011, 01/04/2012, 01/05/2006, 01/05/2007, 01/05/2008, 01/05/2009, 01/05/2010, 01/05/2011, 01/05/2012, 01/06/2006, 01/06/2007, 01/06/2008, 01/06/2009, 01/06/2010, 01/06/2011, 01/06/2012, 01/07/2006, 01/07/2007, 01/07/2008, 01/07/2009, 01/07/2010, 01/07/2011, 01/07/2012, 01/08/2006, 01/08/2007, 01/08/2008, 01/08/2009, 01/08/2010, 01/08/2011, 01/08/2012, 01/09/2006, 01/09/2007, 01/09/2008, 01/09/2009, 01/09/2010, 01/09/2011, 01/09/2012, 01/10/2006, 01/10/2007, 01/10/2008, 01/10/2009, 01/10/2010, 01/10/2011, 01/10/2012, 01/11/2006, 01/11/2008, 01/11/2009, 01/11/2010, 01/11/2011, 01/11/2012, 01/12/2006, 01/12/2008, 01/12/2009, 01/12/2010, 01/12/2011, 01/12/2012), class = factor), y1 = c(2.592356082, 2.345800476, 0.583585821, 5.475129414, 5.475129414, 3.718656646, 3.089374967, 2.301938832, 2.937799758, 2.194943003, 2.54038668, NA, 1.644678741, 1.449029225, 0.956848412, 0.859023655, 0.987578146, 0.843738536, 1.247265662, 1.284694265, 0.980520409, 0.835670652, 0.566612694, 0.401453666, 0.439134806, 1.15846577, 0.687179049, 0.494573266, 0.96702963, 0.504182954, 1.03521455, 0.41886541, 0.401638417, 0.143972524, 0.84561927, 0.551221244, 0.916415951, 1.455635055, 2.866544524, 1.709780054, 1.846827755, 1.451262182, 0.807575275, 1.590211883, 1.542348196, 0.937123964, 1.399639685, 1.091996771, 1.404915044, 1.816230935, 1.468899879, 1.34414673, 1.480941894, 1.325698257, 1.462174427, 1.294317244, 1.348639226, 0.743242205, 1.326535615, 1.288495608, 0.825648931, 0.520199099, 0.631660841, 0.4838471, 0.514975576, 0.644832626, 0.789323515, 0.380945025, 0.386553999, 0.342014493, 0.351664055, 0.188006956, 0.238740258, 0.223667802, 0.063293682, 0.254889318, 0.324071093, 0.323446397, 0.291533728, 0.152346111, 0.429059919, 0.237783277), y2 = c(3.59979021, 3.760067114, 3.668671329, 4.08590604, 3.772684564, 3.833825503, 3.770872483, 3.801879195, 3.776442953, 3.807248322, 3.623087248, 3.85, 3.723986014, 3.67041958, 3.594405594, 3.595244755, 3.457832168, 3.582684564, 3.580839161, 3.555804196, 3.409060403, 3.39885906, 2.707342657, 2.483986014, 2.548881119, 2.797202797, 2.66034965, 2.792447552, 2.597482517, 2.243426573, 2.300629371, 2.323356643, 2.057832168, 2.142167832, 2.829300699, 2.45013986, 2.418531469, 2.77027972, 3.384335664, 3.32048951, 3.297762238, 3.008531469, 2.645034965, 2.965104895, 2.826853147, 2.747342657, 3.053986014, 3.327132867, 2.845874126, 2.961748252, 3.104335664, 3.076153846, 3.259090909, 3.08013986, 3.022727273, 3.016433566, 3.261468531, 2.942237762, 3.217972028, 2.904685315, 2.674755245, 2.436783217, 2.466853147, 2.557272727, 2.696853147, 3.308111888, 3.121468531, 3.144195804, 3.017132867, 3.069160839, 2.985384615, 2.661258741, 2.681188811, 2.746433566, 2.797202797, 2.666713287, 2.474335664, 2.769230769, 2.623356643, 2.764195804, 2.658391608, 2.665594406 ), date = structure(1:82, .Label = c(2006-01-01, 2006-02-01, 2006-03-01, 2006-04-01, 2006-05-01, 2006-06-01, 2006-07-01, 2006-08-01, 2006-09-01, 2006-10-01, 2006-11-01, 2006-12-01, 2007-01-01, 2007-02-01, 2007-03-01, 2007-04-01, 2007-05-01, 2007-06-01, 2007-07-01, 2007-08-01, 2007-09-01, 2007-10-01, 2008-01-01, 2008-02-01, 2008-03-01, 2008-04-01, 2008-05-01, 2008-06-01, 2008-07-01, 2008-08-01, 2008-09-01, 2008-10-01, 2008-11-01, 2008-12-01, 2009-01-01, 2009-02-01, 2009-03-01, 2009-04-01, 2009-05-01, 2009-06-01, 2009-07-01, 2009-08-01, 2009-09-01, 2009-10-01, 2009-11-01, 2009-12-01, 2010-01-01, 2010-02-01, 2010-03-01, 2010-04-01, 2010-05-01, 2010-06-01, 2010-07-01, 2010-08-01, 2010-09-01, 2010-10-01, 2010-11-01, 2010-12-01, 2011-01-01, 2011-02-01, 2011-03-01, 2011-04-01, 2011-05-01, 2011-06-01, 2011-07-01, 2011-08-01, 2011-09-01, 2011-10-01, 2011-11-01, 2011-12-01, 2012-01-01, 2012-02-01, 2012-03-01, 2012-04-01, 2012-05-01, 2012-06-01, 2012-07-01, 2012-08-01, 2012-09-01, 2012-10-01, 2012-11-01, 2012-12-01 ), class = factor), datepos = structure(c(2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L,
Re: [R] Plotrix and twoord.plot with date on x-axe
Perfect Jim, It's fine ! Thank you Michel Le 26/07/2014 12:16, Jim Lemon a écrit : On Sat, 26 Jul 2014 09:36:49 AM Arnaud Michel wrote: Hello With package plotrix and twoord.plot function, I would like to put the labels of the ticks values of x-axe which are date ( c(2006 Jan, 2007 Jan, 2008 Jan, 2009 Jan, 2010 Jan, 2011 Jan, 2012 Jan) with if possible, the angle of labels and x-axe = 40 Any idea ? Thank you for your help My dataframe is TT - structure(list(x1 = structure(c(1L, 8L, 15L, 22L, 29L, 36L, 43L, 50L, 57L, 64L, 71L, 77L, 2L, 9L, 16L, 23L, 30L, 37L, 44L, 51L, 58L, 65L, 3L, 10L, 17L, 24L, 31L, 38L, 45L, 52L, 59L, 66L, 72L, 78L, 4L, 11L, 18L, 25L, 32L, 39L, 46L, 53L, 60L, 67L, 73L, 79L, 5L, 12L, 19L, 26L, 33L, 40L, 47L, 54L, 61L, 68L, 74L, 80L, 6L, 13L, 20L, 27L, 34L, 41L, 48L, 55L, 62L, 69L, 75L, 81L, 7L, 14L, 21L, 28L, 35L, 42L, 49L, 56L, 63L, 70L, 76L, 82L), .Label = c(01/01/2006, 01/01/2007, 01/01/2008, 01/01/2009, 01/01/2010, 01/01/2011, 01/01/2012, 01/02/2006, 01/02/2007, 01/02/2008, 01/02/2009, 01/02/2010, 01/02/2011, 01/02/2012, 01/03/2006, 01/03/2007, 01/03/2008, 01/03/2009, 01/03/2010, 01/03/2011, 01/03/2012, 01/04/2006, 01/04/2007, 01/04/2008, 01/04/2009, 01/04/2010, 01/04/2011, 01/04/2012, 01/05/2006, 01/05/2007, 01/05/2008, 01/05/2009, 01/05/2010, 01/05/2011, 01/05/2012, 01/06/2006, 01/06/2007, 01/06/2008, 01/06/2009, 01/06/2010, 01/06/2011, 01/06/2012, 01/07/2006, 01/07/2007, 01/07/2008, 01/07/2009, 01/07/2010, 01/07/2011, 01/07/2012, 01/08/2006, 01/08/2007, 01/08/2008, 01/08/2009, 01/08/2010, 01/08/2011, 01/08/2012, 01/09/2006, 01/09/2007, 01/09/2008, 01/09/2009, 01/09/2010, 01/09/2011, 01/09/2012, 01/10/2006, 01/10/2007, 01/10/2008, 01/10/2009, 01/10/2010, 01/10/2011, 01/10/2012, 01/11/2006, 01/11/2008, 01/11/2009, 01/11/2010, 01/11/2011, 01/11/2012, 01/12/2006, 01/12/2008, 01/12/2009, 01/12/2010, 01/12/2011, 01/12/2012), class = factor), y1 = c(2.592356082, 2.345800476, 0.583585821, 5.475129414, 5.475129414, 3.718656646, 3.089374967, 2.301938832, 2.937799758, 2.194943003, 2.54038668, NA, 1.644678741, 1.449029225, 0.956848412, 0.859023655, 0.987578146, 0.843738536, 1.247265662, 1.284694265, 0.980520409, 0.835670652, 0.566612694, 0.401453666, 0.439134806, 1.15846577, 0.687179049, 0.494573266, 0.96702963, 0.504182954, 1.03521455, 0.41886541, 0.401638417, 0.143972524, 0.84561927, 0.551221244, 0.916415951, 1.455635055, 2.866544524, 1.709780054, 1.846827755, 1.451262182, 0.807575275, 1.590211883, 1.542348196, 0.937123964, 1.399639685, 1.091996771, 1.404915044, 1.816230935, 1.468899879, 1.34414673, 1.480941894, 1.325698257, 1.462174427, 1.294317244, 1.348639226, 0.743242205, 1.326535615, 1.288495608, 0.825648931, 0.520199099, 0.631660841, 0.4838471, 0.514975576, 0.644832626, 0.789323515, 0.380945025, 0.386553999, 0.342014493, 0.351664055, 0.188006956, 0.238740258, 0.223667802, 0.063293682, 0.254889318, 0.324071093, 0.323446397, 0.291533728, 0.152346111, 0.429059919, 0.237783277), y2 = c(3.59979021, 3.760067114, 3.668671329, 4.08590604, 3.772684564, 3.833825503, 3.770872483, 3.801879195, 3.776442953, 3.807248322, 3.623087248, 3.85, 3.723986014, 3.67041958, 3.594405594, 3.595244755, 3.457832168, 3.582684564, 3.580839161, 3.555804196, 3.409060403, 3.39885906, 2.707342657, 2.483986014, 2.548881119, 2.797202797, 2.66034965, 2.792447552, 2.597482517, 2.243426573, 2.300629371, 2.323356643, 2.057832168, 2.142167832, 2.829300699, 2.45013986, 2.418531469, 2.77027972, 3.384335664, 3.32048951, 3.297762238, 3.008531469, 2.645034965, 2.965104895, 2.826853147, 2.747342657, 3.053986014, 3.327132867, 2.845874126, 2.961748252, 3.104335664, 3.076153846, 3.259090909, 3.08013986, 3.022727273, 3.016433566, 3.261468531, 2.942237762, 3.217972028, 2.904685315, 2.674755245, 2.436783217, 2.466853147, 2.557272727, 2.696853147, 3.308111888, 3.121468531, 3.144195804, 3.017132867, 3.069160839, 2.985384615, 2.661258741, 2.681188811, 2.746433566, 2.797202797, 2.666713287, 2.474335664, 2.769230769, 2.623356643, 2.764195804, 2.658391608, 2.665594406 ), date = structure(1:82, .Label = c(2006-01-01, 2006-02-01, 2006-03-01, 2006-04-01, 2006-05-01, 2006-06-01, 2006-07-01, 2006-08-01, 2006-09-01, 2006-10-01, 2006-11-01, 2006-12-01, 2007-01-01, 2007-02-01, 2007-03-01, 2007-04-01, 2007-05-01, 2007-06-01, 2007-07-01, 2007-08-01, 2007-09-01, 2007-10-01, 2008-01-01, 2008-02-01, 2008-03-01, 2008-04-01, 2008-05-01, 2008-06-01, 2008-07-01, 2008-08-01, 2008-09-01, 2008-10-01, 2008-11-01, 2008-12-01, 2009-01-01, 2009-02-01, 2009-03-01, 2009-04-01, 2009-05-01, 2009-06-01, 2009-07-01, 2009-08-01, 2009-09-01, 2009-10-01, 2009-11-01, 2009-12-01, 2010-01-01, 2010-02-01, 2010-03-01, 2010-04-01, 2010-05-01, 2010-06-01, 2010-07-01, 2010-08-01, 2010-09-01, 2010-10-01, 2010-11-01, 2010-12-01, 2011-01-01, 2011-02-01, 2011-03-01, 2011-04-01, 2011-05-01, 2011-06-01, 2011-07-01, 2011-08-01, 2011-09-01, 2011-10-01, 2011-11
Re: [R] Plotrix and twoord.plot with date on x-axe
Perfect Jim, It's fine ! Michel Le 26/07/2014 12:16, Jim Lemon a écrit : On Sat, 26 Jul 2014 09:36:49 AM Arnaud Michel wrote: Hello With package plotrix and twoord.plot function, I would like to put the labels of the ticks values of x-axe which are date ( c(2006 Jan, 2007 Jan, 2008 Jan, 2009 Jan, 2010 Jan, 2011 Jan, 2012 Jan) with if possible, the angle of labels and x-axe = 40 Any idea ? Thank you for your help My dataframe is TT - structure(list(x1 = structure(c(1L, 8L, 15L, 22L, 29L, 36L, 43L, 50L, 57L, 64L, 71L, 77L, 2L, 9L, 16L, 23L, 30L, 37L, 44L, 51L, 58L, 65L, 3L, 10L, 17L, 24L, 31L, 38L, 45L, 52L, 59L, 66L, 72L, 78L, 4L, 11L, 18L, 25L, 32L, 39L, 46L, 53L, 60L, 67L, 73L, 79L, 5L, 12L, 19L, 26L, 33L, 40L, 47L, 54L, 61L, 68L, 74L, 80L, 6L, 13L, 20L, 27L, 34L, 41L, 48L, 55L, 62L, 69L, 75L, 81L, 7L, 14L, 21L, 28L, 35L, 42L, 49L, 56L, 63L, 70L, 76L, 82L), .Label = c(01/01/2006, 01/01/2007, 01/01/2008, 01/01/2009, 01/01/2010, 01/01/2011, 01/01/2012, 01/02/2006, 01/02/2007, 01/02/2008, 01/02/2009, 01/02/2010, 01/02/2011, 01/02/2012, 01/03/2006, 01/03/2007, 01/03/2008, 01/03/2009, 01/03/2010, 01/03/2011, 01/03/2012, 01/04/2006, 01/04/2007, 01/04/2008, 01/04/2009, 01/04/2010, 01/04/2011, 01/04/2012, 01/05/2006, 01/05/2007, 01/05/2008, 01/05/2009, 01/05/2010, 01/05/2011, 01/05/2012, 01/06/2006, 01/06/2007, 01/06/2008, 01/06/2009, 01/06/2010, 01/06/2011, 01/06/2012, 01/07/2006, 01/07/2007, 01/07/2008, 01/07/2009, 01/07/2010, 01/07/2011, 01/07/2012, 01/08/2006, 01/08/2007, 01/08/2008, 01/08/2009, 01/08/2010, 01/08/2011, 01/08/2012, 01/09/2006, 01/09/2007, 01/09/2008, 01/09/2009, 01/09/2010, 01/09/2011, 01/09/2012, 01/10/2006, 01/10/2007, 01/10/2008, 01/10/2009, 01/10/2010, 01/10/2011, 01/10/2012, 01/11/2006, 01/11/2008, 01/11/2009, 01/11/2010, 01/11/2011, 01/11/2012, 01/12/2006, 01/12/2008, 01/12/2009, 01/12/2010, 01/12/2011, 01/12/2012), class = factor), y1 = c(2.592356082, 2.345800476, 0.583585821, 5.475129414, 5.475129414, 3.718656646, 3.089374967, 2.301938832, 2.937799758, 2.194943003, 2.54038668, NA, 1.644678741, 1.449029225, 0.956848412, 0.859023655, 0.987578146, 0.843738536, 1.247265662, 1.284694265, 0.980520409, 0.835670652, 0.566612694, 0.401453666, 0.439134806, 1.15846577, 0.687179049, 0.494573266, 0.96702963, 0.504182954, 1.03521455, 0.41886541, 0.401638417, 0.143972524, 0.84561927, 0.551221244, 0.916415951, 1.455635055, 2.866544524, 1.709780054, 1.846827755, 1.451262182, 0.807575275, 1.590211883, 1.542348196, 0.937123964, 1.399639685, 1.091996771, 1.404915044, 1.816230935, 1.468899879, 1.34414673, 1.480941894, 1.325698257, 1.462174427, 1.294317244, 1.348639226, 0.743242205, 1.326535615, 1.288495608, 0.825648931, 0.520199099, 0.631660841, 0.4838471, 0.514975576, 0.644832626, 0.789323515, 0.380945025, 0.386553999, 0.342014493, 0.351664055, 0.188006956, 0.238740258, 0.223667802, 0.063293682, 0.254889318, 0.324071093, 0.323446397, 0.291533728, 0.152346111, 0.429059919, 0.237783277), y2 = c(3.59979021, 3.760067114, 3.668671329, 4.08590604, 3.772684564, 3.833825503, 3.770872483, 3.801879195, 3.776442953, 3.807248322, 3.623087248, 3.85, 3.723986014, 3.67041958, 3.594405594, 3.595244755, 3.457832168, 3.582684564, 3.580839161, 3.555804196, 3.409060403, 3.39885906, 2.707342657, 2.483986014, 2.548881119, 2.797202797, 2.66034965, 2.792447552, 2.597482517, 2.243426573, 2.300629371, 2.323356643, 2.057832168, 2.142167832, 2.829300699, 2.45013986, 2.418531469, 2.77027972, 3.384335664, 3.32048951, 3.297762238, 3.008531469, 2.645034965, 2.965104895, 2.826853147, 2.747342657, 3.053986014, 3.327132867, 2.845874126, 2.961748252, 3.104335664, 3.076153846, 3.259090909, 3.08013986, 3.022727273, 3.016433566, 3.261468531, 2.942237762, 3.217972028, 2.904685315, 2.674755245, 2.436783217, 2.466853147, 2.557272727, 2.696853147, 3.308111888, 3.121468531, 3.144195804, 3.017132867, 3.069160839, 2.985384615, 2.661258741, 2.681188811, 2.746433566, 2.797202797, 2.666713287, 2.474335664, 2.769230769, 2.623356643, 2.764195804, 2.658391608, 2.665594406 ), date = structure(1:82, .Label = c(2006-01-01, 2006-02-01, 2006-03-01, 2006-04-01, 2006-05-01, 2006-06-01, 2006-07-01, 2006-08-01, 2006-09-01, 2006-10-01, 2006-11-01, 2006-12-01, 2007-01-01, 2007-02-01, 2007-03-01, 2007-04-01, 2007-05-01, 2007-06-01, 2007-07-01, 2007-08-01, 2007-09-01, 2007-10-01, 2008-01-01, 2008-02-01, 2008-03-01, 2008-04-01, 2008-05-01, 2008-06-01, 2008-07-01, 2008-08-01, 2008-09-01, 2008-10-01, 2008-11-01, 2008-12-01, 2009-01-01, 2009-02-01, 2009-03-01, 2009-04-01, 2009-05-01, 2009-06-01, 2009-07-01, 2009-08-01, 2009-09-01, 2009-10-01, 2009-11-01, 2009-12-01, 2010-01-01, 2010-02-01, 2010-03-01, 2010-04-01, 2010-05-01, 2010-06-01, 2010-07-01, 2010-08-01, 2010-09-01, 2010-10-01, 2010-11-01, 2010-12-01, 2011-01-01, 2011-02-01, 2011-03-01, 2011-04-01, 2011-05-01, 2011-06-01, 2011-07-01, 2011-08-01, 2011-09-01, 2011-10-01, 2011-11-01, 2011-12
[R] To map population of European countries from Eurostat
Hello I would like to map the population of the European countries in 2011. I am using the spatial shapefiles of Europe published by EUROSTAT. I applyed the script below of Markus Kainubut but I had a problem with the map. Any ideas ? Thanks for your help # # Importing the data library(SmarterPoland) df - getEurostatRaw(kod = tps1) # rename 1ere variable names(df) - c(xx, 2002:2013) # new variable df$unit et df$geo.time df$unit - lapply(strsplit(as.character(df$xx), ,), [, 1) df$geo.time - lapply(strsplit(as.character(df$xx), ,), [, 2) # file df.l df.l - melt(data = df, id.vars = geo.time, measure.vars = c(2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013)) df.l$geo.time - unlist(df.l$geo.time) # unlist the geo.time variable # Load the shapefile GISCO, subset it to NUTS2-level # and merge the Eurostat # attribute data with it into single SpatialPolygonDataFrame Filetodownload - http://epp.eurostat.ec.europa.eu/cache/GISCO/geodatafiles/NUTS_2010_60M_SH.zip; download.file(Filetodownload, destfile = NUTS_2010_60M_SH.zip) rm(Filetodownload) # Unzip NUTS_2010_60M_SH.zip to SpatialPolygonsDataFrame unzip(NUTS_2010_60M_SH.zip) library(rgdal) # Lecture du fichier map - readOGR(dsn = ./NUTS_2010_60M_SH/data, layer = NUTS_RG_60M_2010) # OGR data source with driver: ESRI Shapefile # Source: ./NUTS_2010_60M_SH/data, layer: NUTS_RG_60M_2010 # with 1920 features and 4 fields # Feature type: wkbPolygon with 2 dimensions names(map) #[1] NUTS_IDSTAT_LEVL_ SHAPE_Leng SHAPE_Area # as the data is at NUTS2-level, we subset the # spatialpolygondataframe map_nuts2 - subset(map, STAT_LEVL_ = 2) # dim show how many regions are in the spatialpolygondataframe dim(map_nuts2) ## [1] 467 4 # dim show how many regions are in the data.frame dim(df) ## [1] 43 15 # Spatial dataframe has 467 rows and attribute data 43. We need to make # attribute data to have similar number of rows NUTS_ID - as.character(map_nuts2$NUTS_ID) VarX - rep(empty, 467) dat - data.frame(NUTS_ID, VarX) # then we shall merge this with Eurostat data.frame dat2 - merge(dat, df, by.x = NUTS_ID, by.y = geo.time, all.x = TRUE) ## merge this manipulated attribute data with the spatialpolygondataframe ## there are still duplicates in the data, remove them dat2$dup - duplicated(dat2$NUTS_ID) dat3 - subset(dat2, dup == FALSE) ## rownames row.names(dat3) - dat3$NUTS_ID row.names(map_nuts2) - as.character(map_nuts2$NUTS_ID) ## order data dat3 - dat3[order(row.names(dat3)), ] map_nuts2 - map_nuts2[order(row.names(map_nuts2)), ] ## join library(maptools) shape - spCbind(map_nuts2, dat3) ## # Munging the shapefile into data.frame and ready for ggplot-plotting ## fortify spatialpolygondataframe into data.frame library(ggplot2) library(rgeos) shape$id - rownames(shape@data) map.points - fortify(shape, region = id) map.df - merge(map.points, shape, by = id) ### # Only year 2011 # As we want to plot map faceted by years from 2003 to 2011 we have to # melt it into long format map.df - map.df[, -c(15:23,25,26)] library(reshape2) map.df.l - melt(data = map.df, id.vars = c(id, long, lat, group), measure.vars = c(X2011)) # year variable (variable) is class string and type X20xx. Lets remove # the X and convert it to numerical library(stringr) map.df.l$variable - str_replace_all(map.df.l$variable, X, ) map.df.l$variable - factor(map.df.l$variable) map.df.l$variable - as.numeric(levels(map.df.l$variable))[map.df.l$variable] # And finally the plot using ggplot2 #Map shows proportion of materially deprived households at the NUTS2 level. #Grey color indicates missing data. library(ggplot2) ggplot(map.df.l, aes(long,lat,group=group)) + geom_polygon(aes(fill = value)) + geom_polygon(data = map.df.l, aes(long,lat), fill=NA, color = white, size=0.1) + # white borders coord_map(project=orthographic, xlim=c(-22,34), ylim=c(35,70)) + # proj labs(title = Cartographie) + theme_minimal() -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] to modify a dataframe
Dear All, From the dataframe df1 df1 - structure(list(Nom = structure(1:9, .Label = c(A1, A2, A3, B1, B2, C1, C2, C3, C4), class = factor), Pays1 = c(1, 1, 0, 0, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 0, 1, 0, 1), Pays3 = c(0, 0, 0, 0, 1, 0, 0, 0, 0), Pays4 = c(1, 0, 0, 0, 0, 0, 1, 0, 1), Pays5 = c(1, 1, 0, 0, 0, 0, 0, 0, 0)), .Names = c(Nom, Pays1, Pays2, Pays3, Pays4, Pays5), row.names = c(1L, 3L, 4L, 2L, 5L, 6L, 7L, 8L, 9L), class = data.frame) I look for a way to build the new dataframe df2 df2 - structure(list(Nom = structure(1:9, .Label = c(A1, A2, A3, B1, B2, C1, C2, C3, C4), class = factor), Pays1 = c(1, 1, 1, 1, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 1, 1, 1, 1), Pays3 = c(0, 0, 0, 1, 1, 0, 0, 0, 0), Pays4 = c(1, 1, 1, 0, 0, 1, 1, 1, 1), Pays5 = c(1, 1, 1, 0, 0, 0, 0, 0, 0)), .Names = c(Nom, Pays1, Pays2, Pays3, Pays4, Pays5), row.names = c(NA, -9L), class = data.frame) The purpose is to transform df1 it df2 by giving for every group of lines A, B and C the value 1 if there is at least a value equal to 1 or a value 0 if there is no value equal to 1 Thanks for your helps -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] to replace the for loop
Hello
I would like to replace the for loop this below
T - as.matrix(T)
for(i in 1: nrow(TEMP)){
for(j in 1: nrow(TEMP)){if (i = j) T[i, j] - 0 }}
I don't find the function in the doc.
Thanks in advance for your help.
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] To transform a vector of qualitatives values into a dataframe of quantitatives values
Hi From the vector X - c(A, A, B, C, B, A, C) I would like to build the Dataframe : data.frame( A=c(1,1,0,0,0,1,0), B=c(0,0,1,0,1,0,0), C=c(0,0,0,1,0,0,1)) Any ideas ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] To transform a vector
Dear R Users I have the vector X - c( 6 , 4 ,12 , 3) I would like to build a new vector by to transform it into Y - c(rep(X[1], X[1]), rep(X[2], X[2]), rep(X[3], X[3]), rep(X[4], X[4])) Have you a more elegant answer ? PS : Sorry for this basic question -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] To transform a vector
Thank you Michel Le 09/12/2013 08:14, Berend Hasselman a écrit : On 09-12-2013, at 08:04, Arnaud Michel michel.arn...@cirad.fr wrote: Dear R Users I have the vector X - c( 6 , 4 ,12 , 3) I would like to build a new vector by to transform it into Y - c(rep(X[1], X[1]), rep(X[2], X[2]), rep(X[3], X[3]), rep(X[4], X[4])) Have you a more elegant answer ? Have a good read of ?rep. Try this: rep(X,times=X) Berend -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] To transform an adjacency matrix
Hi I have the following problem I would like to build, from a matrix filled with 0 and with 1, a matrix or a data.frame which contains, in every line, the number of the line and the number of the column of the matrix for which the value is equal to 1. Exemple : dput(m) structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L)) Result 1 5 2 3 2 4 4 1 4 3 Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] To transform an adjacency matrix
Thank you Pascal Its fine Michel Le 20/11/2013 11:55, Pascal Oettli a écrit : Hello, One approach is: m - structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L)) out - which(m==1, arr.ind=TRUE) out[order(out[,1]),] Regards, Pascal On 20 November 2013 19:28, Arnaud Michel michel.arn...@cirad.fr wrote: Hi I have the following problem I would like to build, from a matrix filled with 0 and with 1, a matrix or a data.frame which contains, in every line, the number of the line and the number of the column of the matrix for which the value is equal to 1. Exemple : dput(m) structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L)) Result 1 5 2 3 2 4 4 1 4 3 Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] To transform an adjacency matrix
Thank you also for your help Michel Le 20/11/2013 19:04, Dennis Murphy a écrit : Hi: which(m == 1L, arr.ind = TRUE) Dennis On Wed, Nov 20, 2013 at 2:28 AM, Arnaud Michel michel.arn...@cirad.fr wrote: Hi I have the following problem I would like to build, from a matrix filled with 0 and with 1, a matrix or a data.frame which contains, in every line, the number of the line and the number of the column of the matrix for which the value is equal to 1. Exemple : dput(m) structure(c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0), .Dim = c(5L, 5L)) Result 1 5 2 3 2 4 4 1 4 3 Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] add points on an existant ggplot from another dataframe
Hello I had draw the results of PCA (Principal Components Analysis) (1) Is it possible to put on this graphic the 75% ellipse confidence of each sex calculated by dataEllipse (library(car)) ? My code is 1) PCA data frame with 169 rows and 2 columns library(ggplot2) p - ggplot(PCA, aes(x=F1, y=F2 )) p + geom_point(aes(colour = Sexe, shape = Sexe), size=3) + geom_hline(yintercept = 0) + geom_vline(xintercept = 0) + labs(title = ACP, x = Facteur 1, y = Facteur 2,fill = Sexe) + theme(legend.position = c(0.95,0.95), legend.background = element_rect(colour = black)) 2) library(car) XH - dataEllipse( X1[ACP$Sexe==H], X2[ACP$Sexe==H], levels=0.75, lty=2, add=TRUE, plot.points=FALSE, center.cex=0, col=4) XF - dataEllipse( X1[ACP$Sexe==F], X2[ACP$Sexe==F], levels=0.75, lty=2, add=TRUE, plot.points=FALSE, center.cex=0, col=2) XH and XF are two matrix with 2 columns and 52 rows Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ggplot2 : to change the title of legend
Hello I don't arrive to change the title of the legend My code is : library(ggplot2) ma - max(General$AgeChangCat) ; mi - min(General$AgeChangCat) Test$Recrutement - factor(Test$CadNonCadRecrut) p - ggplot(Test, aes(x=factor(Cat1), y=AgeChangCat )) + ylim(mi,ma) + geom_point() + geom_boxplot(aes(fill = Recrutement)) + labs(title = Age, x=catégorie, y=Age) + theme(legend.position = c(0.1,0.9), legend.background = element_rect(colour = black)) p any idea ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2 : to change the title of legend
OK It is right Thank you Petr Michel Le 11/10/2013 14:58, PIKAL Petr a écrit : Hi I usually use scale something like scale_fill_discrete(name = Fancy Title) shall do the trick Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Arnaud Michel Sent: Friday, October 11, 2013 11:02 AM To: R help Subject: [R] ggplot2 : to change the title of legend Hello I don't arrive to change the title of the legend My code is : library(ggplot2) ma - max(General$AgeChangCat) ; mi - min(General$AgeChangCat) Test$Recrutement - factor(Test$CadNonCadRecrut) p - ggplot(Test, aes(x=factor(Cat1), y=AgeChangCat )) + ylim(mi,ma) + geom_point() + geom_boxplot(aes(fill = Recrutement)) + labs(title = Age, x=catégorie, y=Age) + theme(legend.position = c(0.1,0.9), legend.background = element_rect(colour = black)) p any idea ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] delete some lines of a dataframe
Merci Arun Michel Le 17/09/2013 22:41, arun a écrit : Hi Arnaud, You could also try: indx- Df1$Mat[-1]==Df1$Mat[-nrow(Df1)] indx1-c(indx,FALSE) indx2-c(FALSE,indx) Df1[indx1,] Df1[indx2,] A.K. From: arun smartpink...@yahoo.com To: Arnaud Michel michel.arn...@cirad.fr Cc: R help r-help@r-project.org Sent: Tuesday, September 17, 2013 4:21 PM Subject: Re: [R] delete some lines of a dataframe Hi Arnaud, In that case: Try: Df1[duplicated(Df1$Mat),] Df1[duplicated(Df1$Mat,fromLast=TRUE),] #or A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Tuesday, September 17, 2013 4:00 PM Subject: Re: [R] delete some lines of a dataframe Thank you Arun but the values of other columns may be different !!! Michel Le 17/09/2013 20:56, arun a écrit : Hi, Try: Df1[duplicated(Df1),] Df1[duplicated(Df1,fromLast=TRUE),] A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: R help r-help@r-project.org Cc: Sent: Tuesday, September 17, 2013 2:14 PM Subject: [R] delete some lines of a dataframe Hi I have a dataframe Df1 dput(Df1) structure(list(Mat = c(141, 141, 157, 157, 188, 188, 232, 232, 253, 253, 253, 254, 254, 254, 254, 256, 256, 264, 264), Prenom = c(Pierre, Pierre, Jean-Claude, Jean-Claude, Jean-Louis, Jean-Louis, Philippe, Philippe, Christophe, Christophe, Christophe, Dominique, Dominique, Dominique, Dominique, Pierre-Luc, Pierre-Luc, Alain, Alain), Sexe = c(Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin), DateNais = c(23/08/1946, 23/08/1946, 11/08/1945, 11/08/1945, 09/04/1948, 09/04/1948, 01/05/1946, 01/05/1946, 11/02/1951, 11/02/1951, 11/02/1951, 21/10/1949, 21/10/1949, 21/10/1949, 21/10/1949, 25/06/1946, 25/06/1946, 13/03/1949, 13/03/1949 )), .Names = c(Mat, Prenom, Sexe, DateNais), row.names = c(207, 208, 232, 233, 288, 289, 373, 374, 412, 413, 414, 415, 416, 417, 418, 420, 421, 436, 437 ), class = data.frame) I want to extract of Df1 2 other dataframes : 1) delete the first line for each values of Mat Mat Prenom Sexe DateNais 208 141 Pierre Masculin 23/08/1946 233 157 Jean-Claude Masculin 11/08/1945 289 188 Jean-Louis Masculin 09/04/1948 374 232Philippe Masculin 01/05/1946 413 253 Christophe Masculin 11/02/1951 414 253 Christophe Masculin 11/02/1951 416 254 Dominique Masculin 21/10/1949 417 254 Dominique Masculin 21/10/1949 418 254 Dominique Masculin 21/10/1949 421 256 Pierre-Luc Masculin 25/06/1946 437 264 Alain Masculin 13/03/1949 2) delete the last line for each values of Mat Mat Prenom Sexe DateNais 207 141 Pierre Masculin 23/08/1946 232 157 Jean-Claude Masculin 11/08/1945 288 188 Jean-Louis Masculin 09/04/1948 373 232Philippe Masculin 01/05/1946 412 253 Christophe Masculin 11/02/1951 413 253 Christophe Masculin 11/02/1951 415 254 Dominique Masculin 21/10/1949 416 254 Dominique Masculin 21/10/1949 417 254 Dominique Masculin 21/10/1949 420 256 Pierre-Luc Masculin 25/06/1946 436 264 Alain Masculin 13/03/1949 Any ideas ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] delete some lines of a dataframe
Hi I have a dataframe Df1 dput(Df1) structure(list(Mat = c(141, 141, 157, 157, 188, 188, 232, 232, 253, 253, 253, 254, 254, 254, 254, 256, 256, 264, 264), Prenom = c(Pierre, Pierre, Jean-Claude, Jean-Claude, Jean-Louis, Jean-Louis, Philippe, Philippe, Christophe, Christophe, Christophe, Dominique, Dominique, Dominique, Dominique, Pierre-Luc, Pierre-Luc, Alain, Alain), Sexe = c(Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin), DateNais = c(23/08/1946, 23/08/1946, 11/08/1945, 11/08/1945, 09/04/1948, 09/04/1948, 01/05/1946, 01/05/1946, 11/02/1951, 11/02/1951, 11/02/1951, 21/10/1949, 21/10/1949, 21/10/1949, 21/10/1949, 25/06/1946, 25/06/1946, 13/03/1949, 13/03/1949 )), .Names = c(Mat, Prenom, Sexe, DateNais), row.names = c(207, 208, 232, 233, 288, 289, 373, 374, 412, 413, 414, 415, 416, 417, 418, 420, 421, 436, 437 ), class = data.frame) I want to extract of Df1 2 other dataframes : 1) delete the first line for each values of Mat Mat Prenom Sexe DateNais 208 141 Pierre Masculin 23/08/1946 233 157 Jean-Claude Masculin 11/08/1945 289 188 Jean-Louis Masculin 09/04/1948 374 232Philippe Masculin 01/05/1946 413 253 Christophe Masculin 11/02/1951 414 253 Christophe Masculin 11/02/1951 416 254 Dominique Masculin 21/10/1949 417 254 Dominique Masculin 21/10/1949 418 254 Dominique Masculin 21/10/1949 421 256 Pierre-Luc Masculin 25/06/1946 437 264 Alain Masculin 13/03/1949 2) delete the last line for each values of Mat Mat Prenom Sexe DateNais 207 141 Pierre Masculin 23/08/1946 232 157 Jean-Claude Masculin 11/08/1945 288 188 Jean-Louis Masculin 09/04/1948 373 232Philippe Masculin 01/05/1946 412 253 Christophe Masculin 11/02/1951 413 253 Christophe Masculin 11/02/1951 415 254 Dominique Masculin 21/10/1949 416 254 Dominique Masculin 21/10/1949 417 254 Dominique Masculin 21/10/1949 420 256 Pierre-Luc Masculin 25/06/1946 436 264 Alain Masculin 13/03/1949 Any ideas ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] delete some lines of a dataframe
Thank you Arun but the values of other columns may be different !!! Michel Le 17/09/2013 20:56, arun a écrit : Hi, Try: Df1[duplicated(Df1),] Df1[duplicated(Df1,fromLast=TRUE),] A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: R help r-help@r-project.org Cc: Sent: Tuesday, September 17, 2013 2:14 PM Subject: [R] delete some lines of a dataframe Hi I have a dataframe Df1 dput(Df1) structure(list(Mat = c(141, 141, 157, 157, 188, 188, 232, 232, 253, 253, 253, 254, 254, 254, 254, 256, 256, 264, 264), Prenom = c(Pierre, Pierre, Jean-Claude, Jean-Claude, Jean-Louis, Jean-Louis, Philippe, Philippe, Christophe, Christophe, Christophe, Dominique, Dominique, Dominique, Dominique, Pierre-Luc, Pierre-Luc, Alain, Alain), Sexe = c(Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin), DateNais = c(23/08/1946, 23/08/1946, 11/08/1945, 11/08/1945, 09/04/1948, 09/04/1948, 01/05/1946, 01/05/1946, 11/02/1951, 11/02/1951, 11/02/1951, 21/10/1949, 21/10/1949, 21/10/1949, 21/10/1949, 25/06/1946, 25/06/1946, 13/03/1949, 13/03/1949 )), .Names = c(Mat, Prenom, Sexe, DateNais), row.names = c(207, 208, 232, 233, 288, 289, 373, 374, 412, 413, 414, 415, 416, 417, 418, 420, 421, 436, 437 ), class = data.frame) I want to extract of Df1 2 other dataframes : 1) delete the first line for each values of Mat Mat Prenom Sexe DateNais 208 141 Pierre Masculin 23/08/1946 233 157 Jean-Claude Masculin 11/08/1945 289 188 Jean-Louis Masculin 09/04/1948 374 232Philippe Masculin 01/05/1946 413 253 Christophe Masculin 11/02/1951 414 253 Christophe Masculin 11/02/1951 416 254 Dominique Masculin 21/10/1949 417 254 Dominique Masculin 21/10/1949 418 254 Dominique Masculin 21/10/1949 421 256 Pierre-Luc Masculin 25/06/1946 437 264 Alain Masculin 13/03/1949 2) delete the last line for each values of Mat Mat Prenom Sexe DateNais 207 141 Pierre Masculin 23/08/1946 232 157 Jean-Claude Masculin 11/08/1945 288 188 Jean-Louis Masculin 09/04/1948 373 232Philippe Masculin 01/05/1946 412 253 Christophe Masculin 11/02/1951 413 253 Christophe Masculin 11/02/1951 415 254 Dominique Masculin 21/10/1949 416 254 Dominique Masculin 21/10/1949 417 254 Dominique Masculin 21/10/1949 420 256 Pierre-Luc Masculin 25/06/1946 436 264 Alain Masculin 13/03/1949 Any ideas ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Change the color of the line inside of the function lines
Hi I have the following problem : I have 3 vectors xx, yy, zz : xx - c(5479, 6209, 6940, 7670, 8766, 9496, 10227, 11048, 11778, 12509, 13239, 13970, 14700, 15340, 15948) yy - c( 267, 275, 281, 287, 296, 306, 316, 325, 334, 351, 365, 377, 389, 419, 419) zz - c( 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6) I would like a line wich join the points (xx, yy) with stair steps (as type = s) plot(xx, yy, type=n) lines(xx, yy, type =s) but I want to change the color according to the value of zz (exemple : col = 1 if zz =3 ; col =2 if zz= 4 ; col =3 if zz= 5 ; col =4 if zz= 6) Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change the color of the line inside of the function lines
Hi Tsjerk Thank you but the color always remains black ! I would want that the color changes on the same graph (color = 3 on the 4 first steps, col = 4 on 5 following steps Michel Le 16/09/2013 09:01, Tsjerk Wassenaar a écrit : Hi Michel, lines(xx,yy,col=zz-2,type=s) If you use a color vector, say cols, then you can also do lines(xx,yy,col=cols[zz-2],type=s) Hope it helps, Tsjerk On Mon, Sep 16, 2013 at 8:42 AM, Arnaud Michel michel.arn...@cirad.fr mailto:michel.arn...@cirad.fr wrote: Hi I have the following problem : I have 3 vectors xx, yy, zz : xx - c(5479, 6209, 6940, 7670, 8766, 9496, 10227, 11048, 11778, 12509, 13239, 13970, 14700, 15340, 15948) yy - c( 267, 275, 281, 287, 296, 306, 316, 325, 334, 351, 365, 377, 389, 419, 419) zz - c( 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6) I would like a line wich join the points (xx, yy) with stair steps (as type = s) plot(xx, yy, type=n) lines(xx, yy, type =s) but I want to change the color according to the value of zz (exemple : col = 1 if zz =3 ; col =2 if zz= 4 ; col =3 if zz= 5 ; col =4 if zz= 6) Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Tsjerk A. Wassenaar, Ph.D. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change the color of the line inside of the function lines
Thanks Pascal and Tsjerk Michel Le 16/09/2013 09:42, Pascal Oettli a écrit : Hi, Maybe the following might help you: s - seq(length(xx)-1) plot(xx, yy, type=n) segments(xx[s], yy[s], xx[s+1], yy[s], col=zz, lwd=2) segments(xx[s+1], yy[s], xx[s+1], yy[s+1], col='grey') Regards, Pascal On 16/09/2013 15:42, Arnaud Michel wrote: Hi I have the following problem : I have 3 vectors xx, yy, zz : xx - c(5479, 6209, 6940, 7670, 8766, 9496, 10227, 11048, 11778, 12509, 13239, 13970, 14700, 15340, 15948) yy - c( 267, 275, 281, 287, 296, 306, 316, 325, 334, 351, 365, 377, 389, 419, 419) zz - c( 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6) I would like a line wich join the points (xx, yy) with stair steps (as type = s) plot(xx, yy, type=n) lines(xx, yy, type =s) but I want to change the color according to the value of zz (exemple : col = 1 if zz =3 ; col =2 if zz= 4 ; col =3 if zz= 5 ; col =4 if zz= 6) Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] to avoid a do loop
Hello I have a large dataframe (nrow=55000). This below df1 an extract of the original dataframe dput(df1) structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7), Ech = c(8, 9, 10, 11, 12, 12, 13, 13, 11, 12, 13, 14, 14, 14, 9, 10, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 10, 11, 11, 11, 4, 5, 6, 7, 8, 8, 8, 9, 9, 8, 9, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 1, 2, 3, 4, 5, 6, 6, 6, 7, 7, 7, 5, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 15, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 13, 13, 14, 15, 15, 15, 10, 1, 2, 3, 4, 5)), .Names = c(Cat, Ech ), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110), class = data.frame) I do not manage to avoid a do loop because very slow I want to obtain a new dataframe df2 with a new variable CatEch. CatEch is the paste of the 2 variables Cat and Ech dput(df2) structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7), Ech = c(8, 9, 10, 11, 12, 12, 13, 13, 11, 12, 13, 14, 14, 14, 9, 10, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 10, 11, 11, 11, 4, 5, 6, 7, 8, 8, 8, 9, 9, 8, 9, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 1, 2, 3, 4, 5, 6, 6, 6, 7, 7, 7, 5, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 15, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 13, 13, 14, 15, 15, 15, 10, 1, 2, 3, 4, 5), CatEch = c(6.08, 6.09, 6.10, 6.11, 6.12, 6.12, 6.13, 6.13, 4.11, 4.12, 4.13, 4.14, 4.14, 4.14, 8.09, 8.10, 9.05, 9.06, 9.07, 9.07, 9.07, 9.07, 9.07, 6.07, 6.08, 6.09, 6.10, 6.11, 6.11, 6.11, 7.04, 7.05, 7.06, 7.07, 7.08, 7.08, 7.08, 7.09, 7.09, 7.08, 7.09, 8.02, 8.03, 8.04, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.01, 8.02, 8.03, 8.04, 8.05, 8.06, 8.06, 8.06, 8.07, 8.07, 8.07, 6.05, 6.06, 6.07, 6.08, 6.09, 6.09, 6.10, 6.10, 6.11, 6.11, 6.11, 6.11, 6.11, 6.11, 7.15, 8.05, 8.06, 8.07, 8.07, 8.08, 8.08, 8.08, 8.08, 8.09, 8.09, 2.13, 2.13, 2.14, 2.15, 2.15, 2.15, 6.10, 7.01, 7.02, 7.03, 7.04, 7.05)), .Names = c(Cat, Ech, CatEch), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110), class = data.frame) Any idea ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] to avoid a do loop
Thanks to all three for your fast answer Michel Le 08/09/2013 18:41, Renaud Lancelot a écrit : paste(df1$Cat, formatC(df1$Ech, flag = 0, width = max(nchar(df1$Ech))), sep = .) 2013/9/8 Arnaud Michel michel.arn...@cirad.fr mailto:michel.arn...@cirad.fr Hello I have a large dataframe (nrow=55000). This below df1 an extract of the original dataframe dput(df1) structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7), Ech = c(8, 9, 10, 11, 12, 12, 13, 13, 11, 12, 13, 14, 14, 14, 9, 10, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 10, 11, 11, 11, 4, 5, 6, 7, 8, 8, 8, 9, 9, 8, 9, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 1, 2, 3, 4, 5, 6, 6, 6, 7, 7, 7, 5, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 15, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 13, 13, 14, 15, 15, 15, 10, 1, 2, 3, 4, 5)), .Names = c(Cat, Ech ), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110), class = data.frame) I do not manage to avoid a do loop because very slow I want to obtain a new dataframe df2 with a new variable CatEch. CatEch is the paste of the 2 variables Cat and Ech dput(df2) structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7), Ech = c(8, 9, 10, 11, 12, 12, 13, 13, 11, 12, 13, 14, 14, 14, 9, 10, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 10, 11, 11, 11, 4, 5, 6, 7, 8, 8, 8, 9, 9, 8, 9, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 1, 2, 3, 4, 5, 6, 6, 6, 7, 7, 7, 5, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 15, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 13, 13, 14, 15, 15, 15, 10, 1, 2, 3, 4, 5), CatEch = c(6.08, 6.09, 6.10, 6.11, 6.12, 6.12, 6.13, 6.13, 4.11, 4.12, 4.13, 4.14, 4.14, 4.14, 8.09, 8.10, 9.05, 9.06, 9.07, 9.07, 9.07, 9.07, 9.07, 6.07, 6.08, 6.09, 6.10, 6.11, 6.11, 6.11, 7.04, 7.05, 7.06, 7.07, 7.08, 7.08, 7.08, 7.09, 7.09, 7.08, 7.09, 8.02, 8.03, 8.04, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.01, 8.02, 8.03, 8.04, 8.05, 8.06, 8.06, 8.06, 8.07, 8.07, 8.07, 6.05, 6.06, 6.07, 6.08, 6.09, 6.09, 6.10, 6.10, 6.11, 6.11, 6.11, 6.11, 6.11, 6.11, 7.15, 8.05, 8.06, 8.07, 8.07, 8.08, 8.08, 8.08, 8.08, 8.09, 8.09, 2.13, 2.13, 2.14, 2.15, 2.15, 2.15, 6.10, 7.01, 7.02, 7.03, 7.04, 7.05)), .Names = c(Cat, Ech, CatEch), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110), class = data.frame) Any idea ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Renaud Lancelot Directeur adjoint / Deputy director CIRAD, UMR15, Campus International de Baillarguet TA A-DIR / B F34398 Montpellier EDENext Project, coordinator: http://www.edenext.eu/ Tel. +33 4 67 59 37 17 - Fax +33 4 67 59 37 95 Secr. +33 4 67 59 37 37 - Cell. +33 6 77 52 08 69 -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read
Re: [R] to avoid a do loop
Hi This below system.time of the 3 solutions with a large dataframe df1 (nrow=55000). # Arun system.time(df1$CatEch1 - paste0(df1$Cat,.,sprintf(%02d,df1$Ech))) # user system elapsed # 0.060.000.06 # Rui Barradas system.time(df1$CatEch2 - pastedf1$Cat, sprintf(%02d, df1$Ech), sep = .) ) # user system elapsed # 0.170.000.17 # R Lancelot system.time(df1$CatEch3 - paste(df1$Cat, formatC(df1$Ech, flag = 0, width = max(nchar(df1$Ech))), sep = .)) # user system elapsed # 0.340.000.35 Thanks Michel Le 08/09/2013 19:15, Arnaud Michel a écrit : Thanks to all three for your fast answer Michel Le 08/09/2013 18:41, Renaud Lancelot a écrit : paste(df1$Cat, formatC(df1$Ech, flag = 0, width = max(nchar(df1$Ech))), sep = .) 2013/9/8 Arnaud Michel michel.arn...@cirad.fr mailto:michel.arn...@cirad.fr Hello I have a large dataframe (nrow=55000). This below df1 an extract of the original dataframe dput(df1) structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7), Ech = c(8, 9, 10, 11, 12, 12, 13, 13, 11, 12, 13, 14, 14, 14, 9, 10, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 10, 11, 11, 11, 4, 5, 6, 7, 8, 8, 8, 9, 9, 8, 9, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 1, 2, 3, 4, 5, 6, 6, 6, 7, 7, 7, 5, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 15, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 13, 13, 14, 15, 15, 15, 10, 1, 2, 3, 4, 5)), .Names = c(Cat, Ech ), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110), class = data.frame) I do not manage to avoid a do loop because very slow I want to obtain a new dataframe df2 with a new variable CatEch. CatEch is the paste of the 2 variables Cat and Ech dput(df2) structure(list(Cat = c(6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 8, 8, 9, 9, 9, 9, 9, 9, 9, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7), Ech = c(8, 9, 10, 11, 12, 12, 13, 13, 11, 12, 13, 14, 14, 14, 9, 10, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 10, 11, 11, 11, 4, 5, 6, 7, 8, 8, 8, 9, 9, 8, 9, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 1, 2, 3, 4, 5, 6, 6, 6, 7, 7, 7, 5, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 15, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 13, 13, 14, 15, 15, 15, 10, 1, 2, 3, 4, 5), CatEch = c(6.08, 6.09, 6.10, 6.11, 6.12, 6.12, 6.13, 6.13, 4.11, 4.12, 4.13, 4.14, 4.14, 4.14, 8.09, 8.10, 9.05, 9.06, 9.07, 9.07, 9.07, 9.07, 9.07, 6.07, 6.08, 6.09, 6.10, 6.11, 6.11, 6.11, 7.04, 7.05, 7.06, 7.07, 7.08, 7.08, 7.08, 7.09, 7.09, 7.08, 7.09, 8.02, 8.03, 8.04, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.05, 8.01, 8.02, 8.03, 8.04, 8.05, 8.06, 8.06, 8.06, 8.07, 8.07, 8.07, 6.05, 6.06, 6.07, 6.08, 6.09, 6.09, 6.10, 6.10, 6.11, 6.11, 6.11, 6.11, 6.11, 6.11, 7.15, 8.05, 8.06, 8.07, 8.07, 8.08, 8.08, 8.08, 8.08, 8.09, 8.09, 2.13, 2.13, 2.14, 2.15, 2.15, 2.15, 6.10, 7.01, 7.02, 7.03, 7.04, 7.05)), .Names = c(Cat, Ech, CatEch), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110), class = data.frame) Any idea ? -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code
Re: [R] To represent on the same plot the relation (y1, x) and (y2, x)
Thank you Arun Michel Le 01/09/2013 07:36, arun a écrit : Hi Arnaud, No problem. Try, x- 1:10 set.seed(28) y1- rnorm(10) set.seed(485) y2- rnorm(10,25) library(plotrix) twoord.plot(x,y1,y2,lylim=c(-2,2),rylim=c(20,30),ylab=y1,rylab=y2,lcol=2,rcol=4,main=y1, y2 vs. x) A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Sunday, September 1, 2013 1:18 AM Subject: Re: [R] To represent on the same plot the relation (y1, x) and (y2, x) Thank you Arun But have you a solution if y1 and y2 have not the same unit (ex : the unit of y1 is meter and the unit of y2 is Kg) and if I want the axe of y1 at the left of the plot and the axe of y2 at the rigth of the plot Michel Le 31/08/2013 21:27, arun a écrit : Hi, May be this helps: x- 1:10 set.seed(28) y1- rnorm(10) set.seed(485) y2- rnorm(10) plot(x,y1,col=red,type=b,ylab=y1:y2) lines(y2,col=blue,type=b) legend(topleft, legend = c(y1, y2),text.col=c(red,blue)) #or library(ggplot2) dat1- data.frame(x,y1,y2) ggplot(dat1,aes(x))+geom_line(aes(y=y1,colour=y1))+ geom_line(aes(y=y2,colour=y2)) +ylab(y1:y2) A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: R help r-help@r-project.org Cc: Sent: Saturday, August 31, 2013 1:57 PM Subject: [R] To represent on the same plot the relation (y1, x) and (y2, x) Hello, I have 3 vectors x, y1 and y2 I would like to represent on the same plot the two graph (y1, x) and (y2, x). Is it possible with ggplot ? other package ? Thanks for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] To represent on the same plot the relation (y1, x) and (y2, x)
Thank you Arun But have you a solution if y1 and y2 have not the same unit (ex : the unit of y1 is meter and the unit of y2 is Kg) and if I want the axe of y1 at the left of the plot and the axe of y2 at the rigth of the plot Michel Le 31/08/2013 21:27, arun a écrit : Hi, May be this helps: x- 1:10 set.seed(28) y1- rnorm(10) set.seed(485) y2- rnorm(10) plot(x,y1,col=red,type=b,ylab=y1:y2) lines(y2,col=blue,type=b) legend(topleft, legend = c(y1, y2),text.col=c(red,blue)) #or library(ggplot2) dat1- data.frame(x,y1,y2) ggplot(dat1,aes(x))+geom_line(aes(y=y1,colour=y1))+ geom_line(aes(y=y2,colour=y2)) +ylab(y1:y2) A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: R help r-help@r-project.org Cc: Sent: Saturday, August 31, 2013 1:57 PM Subject: [R] To represent on the same plot the relation (y1, x) and (y2, x) Hello, I have 3 vectors x, y1 and y2 I would like to represent on the same plot the two graph (y1, x) and (y2, x). Is it possible with ggplot ? other package ? Thanks for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] To represent on the same plot the relation (y1, x) and (y2, x)
Hello, I have 3 vectors x, y1 and y2 I would like to represent on the same plot the two graph (y1, x) and (y2, x). Is it possible with ggplot ? other package ? Thanks for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change values in a dateframe-Speed TEST
Hi
For a dataframe with name PaysContrat1 and with
nrow(PaysContrat1)
[1] 52366
the test of system.time is :
system.time(droplevels(do.call(rbind,lapply(split(PaysContrat1,PaysContrat1$Matricule),
FUN=function(x) {x[,c(Nom,Prénom)] -
x[nrow(x),c(Nom,Prénom),drop=TRUE];x}
user system elapsed
14.030.00 14.04
system.time(droplevels(PaysContrat1[with(PaysContrat1,ave(seq_along(Matricule),Matricule,FUN=min))
,] ))
user system elapsed
0.2 0.0 0.2
Michel
Le 24/07/2013 15:29, arun a écrit :
Hi Michel,
You could try:
df1New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=min)),])
row.names(df1New)-1:nrow(df1New)
df2New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=max)),])
row.names(df2New)-1:nrow(df2New)
identical(df1New,df1)
#[1] TRUE
identical(df2New,df2)
#[1] TRUE
A.K.
- Original Message -
From: Arnaud Michel michel.arn...@cirad.fr
To: R help r-help@r-project.org
Cc:
Sent: Wednesday, July 24, 2013 2:39 AM
Subject: [R] Change values in a dateframe
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents values of Nom or differents values of Prenom
but the values of Matricule or Sexe or Date.de.naissance are the same.
TEST - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L,
5L, 6L, 9L, 3L, 3L, 7L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT
), class = factor), Prenom = structure(c(8L, 3L, 4L, 5L, 1L,
2L, 2L, 9L, 6L, 7L, 7L), .Label = c(Edgar, Elodie, Jeanine,
Jeannine, Michel, Michele, Michèle, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class =
factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
I would want to make homogeneous the information and would like built 2
dataframes :
df1 wich has the value of Nom and Prenom of the first lines of TEST when
there are different values. The other values (Matricule or Sexe or
Date.de.naissance) are unchanged
df1 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L,
5L, 5L, 7L, 3L, 3L, 3L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, TRU, VINCENT), class = factor), Prenom =
structure(c(6L,
3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar,
Elodie, Jeanine, Michel, Michele, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class =
factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
df2 wich has the value of Nom and Prenom of the last lines of TEST when
there are different values. The other values (Matricule or Sexe or
Date.de.naissance) are unchanged.
df2 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L,
4L, 4L, 7L, 5L, 5L, 5L), .Label = c(CHICHE, GEOF, JACQUE,
LANGUE-LOPEZ, RIVIER, TRU, VINCENT), class = factor),
Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L,
5L, 5L), .Label = c(Edgar, Elodie, Jeannine, Michel,
Michèle, Michelle, Victor), class = factor), Sexe =
structure(c(1L,
1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin,
Masculin), class = factor), Date.de.naissance = structure(c(4L,
2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940,
04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936,
30/03/1935), class = factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
Thank for your helps
Michel
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change values in a dateframe-Speed TEST
But I just noticed that the two solutions are not comparable :
the change concern only Nom and Prenom (solution Berend) and not also
Sexe or Date.de.naissance orother variables (solution Arun) that can
changed. But my question was badly put.
Michel
Le 25/07/2013 08:06, Arnaud Michel a écrit :
Hi
For a dataframe with name PaysContrat1 and with
nrow(PaysContrat1)
[1] 52366
the test of system.time is :
system.time(droplevels(do.call(rbind,lapply(split(PaysContrat1,PaysContrat1$Matricule),
FUN=function(x) {x[,c(Nom,Prénom)] -
x[nrow(x),c(Nom,Prénom),drop=TRUE];x}
user system elapsed
14.030.00 14.04
system.time(droplevels(PaysContrat1[with(PaysContrat1,ave(seq_along(Matricule),Matricule,FUN=min))
,] ))
user system elapsed
0.2 0.0 0.2
Michel
Le 24/07/2013 15:29, arun a écrit :
Hi Michel,
You could try:
df1New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=min)),])
row.names(df1New)-1:nrow(df1New)
df2New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=max)),])
row.names(df2New)-1:nrow(df2New)
identical(df1New,df1)
#[1] TRUE
identical(df2New,df2)
#[1] TRUE
A.K.
- Original Message -
From: Arnaud Michel michel.arn...@cirad.fr
To: R help r-help@r-project.org
Cc:
Sent: Wednesday, July 24, 2013 2:39 AM
Subject: [R] Change values in a dateframe
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents values of Nom or differents values of Prenom
but the values of Matricule or Sexe or Date.de.naissance are the same.
TEST - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L,
5L, 6L, 9L, 3L, 3L, 7L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT
), class = factor), Prenom = structure(c(8L, 3L, 4L, 5L, 1L,
2L, 2L, 9L, 6L, 7L, 7L), .Label = c(Edgar, Elodie, Jeanine,
Jeannine, Michel, Michele, Michèle, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946,
07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class =
factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
I would want to make homogeneous the information and would like built 2
dataframes :
df1 wich has the value of Nom and Prenom of the first lines of TEST when
there are different values. The other values (Matricule or Sexe or
Date.de.naissance) are unchanged
df1 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L,
5L, 5L, 7L, 3L, 3L, 3L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, TRU, VINCENT), class = factor), Prenom =
structure(c(6L,
3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar,
Elodie, Jeanine, Michel, Michele, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946,
07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class =
factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
df2 wich has the value of Nom and Prenom of the last lines of TEST when
there are different values. The other values (Matricule or Sexe or
Date.de.naissance) are unchanged.
df2 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L,
4L, 4L, 7L, 5L, 5L, 5L), .Label = c(CHICHE, GEOF, JACQUE,
LANGUE-LOPEZ, RIVIER, TRU, VINCENT), class = factor),
Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L,
5L, 5L), .Label = c(Edgar, Elodie, Jeannine, Michel,
Michèle, Michelle, Victor), class = factor), Sexe =
structure(c(1L,
1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin,
Masculin), class = factor), Date.de.naissance =
structure(c(4L,
2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940,
04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947,
29/12/1936,
30/03/1935), class = factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
Thank for your helps
Michel
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
Re: [R] Change values in a dateframe-Speed TEST
Le 25/07/2013 08:50, Berend Hasselman a écrit :
On 25-07-2013, at 08:35, Arnaud Michel michel.arn...@cirad.fr wrote:
But I just noticed that the two solutions are not comparable :
the change concern only Nom and Prenom (solution Berend) and not also Sexe or
Date.de.naissance orother variables (solution Arun) that can changed. But my
question was badly put.
Indeed:-)
But that can be remedied with (small correction w.r.t. initial solution:
drop=TRUE removed; not relevant here)
r1 - droplevels(do.call(rbind,lapply(split(TEST,TEST$Matricule),
FUN=function(x) {x[,1:ncol(x)] - x[1,1:ncol(x)];x})))
and
r2 - droplevels(do.call(rbind,lapply(split(TEST,TEST$Matricule),
FUN=function(x) {x[,1:ncol(x)] -
x[nrow(x),1:ncol(x)];x})))
Thank you but I keep
{x[,c(Nom,Prénom)] - x[nrow(x),c(Nom,Prénom)];x} because in the
dataframe there are other variables that I do not want to change. I want
change only Nom and Prénom
PS : ?w.r.t.
Michel
Less elegant than alternative with ave
Berend
Michel
Le 25/07/2013 08:06, Arnaud Michel a écrit :
Hi
For a dataframe with name PaysContrat1 and with
nrow(PaysContrat1)
[1] 52366
the test of system.time is :
system.time(droplevels(do.call(rbind,lapply(split(PaysContrat1,PaysContrat1$Matricule),
FUN=function(x) {x[,c(Nom,Prénom)] -
x[nrow(x),c(Nom,Prénom),drop=TRUE];x}
user system elapsed
14.030.00 14.04
system.time(droplevels(PaysContrat1[with(PaysContrat1,ave(seq_along(Matricule),Matricule,FUN=min))
,] ))
user system elapsed
0.2 0.0 0.2
Michel
Le 24/07/2013 15:29, arun a écrit :
Hi Michel,
You could try:
df1New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=min)),])
row.names(df1New)-1:nrow(df1New)
df2New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=max)),])
row.names(df2New)-1:nrow(df2New)
identical(df1New,df1)
#[1] TRUE
identical(df2New,df2)
#[1] TRUE
A.K.
- Original Message -
From: Arnaud Michel michel.arn...@cirad.fr
To: R help r-help@r-project.org
Cc:
Sent: Wednesday, July 24, 2013 2:39 AM
Subject: [R] Change values in a dateframe
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents values of Nom or differents values of Prenom
but the values of Matricule or Sexe or Date.de.naissance are the same.
TEST - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L,
5L, 6L, 9L, 3L, 3L, 7L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT
), class = factor), Prenom = structure(c(8L, 3L, 4L, 5L, 1L,
2L, 2L, 9L, 6L, 7L, 7L), .Label = c(Edgar, Elodie, Jeanine,
Jeannine, Michel, Michele, Michèle, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class =
factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
I would want to make homogeneous the information and would like built 2
dataframes :
df1 wich has the value of Nom and Prenom of the first lines of TEST when
there are different values. The other values (Matricule or Sexe or
Date.de.naissance) are unchanged
df1 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L,
5L, 5L, 7L, 3L, 3L, 3L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, TRU, VINCENT), class = factor), Prenom =
structure(c(6L,
3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar,
Elodie, Jeanine, Michel, Michele, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class =
factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class =
factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame,
row.names = c(NA,
-11L))
df2 wich has the value of Nom and Prenom of the last lines of TEST when
there are different values. The other values (Matricule or Sexe or
Date.de.naissance) are unchanged.
df2 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L,
4L, 4L, 7L, 5L, 5L, 5L), .Label = c(CHICHE, GEOF, JACQUE,
LANGUE-LOPEZ, RIVIER, TRU, VINCENT), class = factor),
Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L,
5L, 5L), .Label = c(Edgar, Elodie, Jeannine, Michel,
Michèle, Michelle, Victor), class = factor), Sexe =
structure(c(1L,
1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L
[R] Change values in a dateframe
Hello I have the following problem : The dataframe TEST has multiple lines for a same person because : there are differents values of Nom or differents values of Prenom but the values of Matricule or Sexe or Date.de.naissance are the same. TEST - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L, 5L, 6L, 9L, 3L, 3L, 7L), .Label = c(CHICHE, GEOF, GUTIER, JACQUE, LANGUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT ), class = factor), Prenom = structure(c(8L, 3L, 4L, 5L, 1L, 2L, 2L, 9L, 6L, 7L, 7L), .Label = c(Edgar, Elodie, Jeanine, Jeannine, Michel, Michele, Michèle, Michelle, Victor ), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor), Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names = c(Matricule, Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names = c(NA, -11L)) I would want to make homogeneous the information and would like built 2 dataframes : df1 wich has the value of Nom and Prenom of the first lines of TEST when there are different values. The other values (Matricule or Sexe or Date.de.naissance) are unchanged df1 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L, 5L, 5L, 7L, 3L, 3L, 3L), .Label = c(CHICHE, GEOF, GUTIER, JACQUE, LANGUE, TRU, VINCENT), class = factor), Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar, Elodie, Jeanine, Michel, Michele, Michelle, Victor ), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor), Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names = c(Matricule, Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names = c(NA, -11L)) df2 wich has the value of Nom and Prenom of the last lines of TEST when there are different values. The other values (Matricule or Sexe or Date.de.naissance) are unchanged. df2 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L, 4L, 4L, 7L, 5L, 5L, 5L), .Label = c(CHICHE, GEOF, JACQUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT), class = factor), Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar, Elodie, Jeannine, Michel, Michèle, Michelle, Victor), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor), Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names = c(Matricule, Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names = c(NA, -11L)) Thank for your helps Michel -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change values in a dateframe
Thank you Berend
It is exactly what I wanted.
Michel
Le 24/07/2013 09:48, Berend Hasselman a écrit :
On 24-07-2013, at 08:39, Arnaud Michel michel.arn...@cirad.fr wrote:
Hello
I have the following problem :
The dataframe TEST has multiple lines for a same person because :
there are differents values of Nom or differents values of Prenom
but the values of Matricule or Sexe or Date.de.naissance are the same.
TEST - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L,
5L, 6L, 9L, 3L, 3L, 7L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT
), class = factor), Prenom = structure(c(8L, 3L, 4L, 5L, 1L,
2L, 2L, 9L, 6L, 7L, 7L), .Label = c(Edgar, Elodie, Jeanine,
Jeannine, Michel, Michele, Michèle, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names =
c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names
= c(NA,
-11L))
I would want to make homogeneous the information and would like built 2
dataframes :
df1 wich has the value of Nom and Prenom of the first lines of TEST when there
are different values. The other values (Matricule or Sexe or Date.de.naissance)
are unchanged
df1 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L,
5L, 5L, 7L, 3L, 3L, 3L), .Label = c(CHICHE, GEOF, GUTIER,
JACQUE, LANGUE, TRU, VINCENT), class = factor), Prenom =
structure(c(6L,
3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar,
Elodie, Jeanine, Michel, Michele, Michelle, Victor
), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L,
1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor),
Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L,
1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947,
18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names =
c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names
= c(NA,
-11L))
df2 wich has the value of Nom and Prenom of the last lines of TEST when there
are different values. The other values (Matricule or Sexe or Date.de.naissance)
are unchanged.
df2 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L,
91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L,
4L, 4L, 7L, 5L, 5L, 5L), .Label = c(CHICHE, GEOF, JACQUE,
LANGUE-LOPEZ, RIVIER, TRU, VINCENT), class = factor),
Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L,
5L, 5L), .Label = c(Edgar, Elodie, Jeannine, Michel,
Michèle, Michelle, Victor), class = factor), Sexe = structure(c(1L,
1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin,
Masculin), class = factor), Date.de.naissance = structure(c(4L,
2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940,
04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936,
30/03/1935), class = factor)), .Names = c(Matricule,
Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names
= c(NA,
-11L))
Something like this
r1 - droplevels(do.call(rbind,lapply(split(TEST,TEST$Matricule),
FUN=function(x) {x[,c(Nom,Prenom)] -
x[1,c(Nom,Prenom),drop=TRUE];x})))
rownames(r1) - NULL
r1
r2 - droplevels(do.call(rbind,lapply(split(TEST,TEST$Matricule),
FUN=function(x) {x[,c(Nom,Prenom)] -
x[nrow(x),c(Nom,Prenom),drop=TRUE];x})))
rownames(r2) - NULL
r2
# identical(r1,df1)
#[1] TRUE
# identical(r2,df2)
#[1] TRUE
Note: I had to change the Prenom and Sexe columns because of encoding issues.
but that shouldn't have any influence on the above.
Berend
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change values in a dateframe
Hi Arun, Merci à toi Bien amicalement Michel Le 24/07/2013 15:29, arun a écrit : Hi Michel, You could try: df1New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=min)),]) row.names(df1New)-1:nrow(df1New) df2New-droplevels(TEST[with(TEST,ave(seq_along(Matricule),Matricule,FUN=max)),]) row.names(df2New)-1:nrow(df2New) identical(df1New,df1) #[1] TRUE identical(df2New,df2) #[1] TRUE A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: R help r-help@r-project.org Cc: Sent: Wednesday, July 24, 2013 2:39 AM Subject: [R] Change values in a dateframe Hello I have the following problem : The dataframe TEST has multiple lines for a same person because : there are differents values of Nom or differents values of Prenom but the values of Matricule or Sexe or Date.de.naissance are the same. TEST - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L, 5L, 6L, 9L, 3L, 3L, 7L), .Label = c(CHICHE, GEOF, GUTIER, JACQUE, LANGUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT ), class = factor), Prenom = structure(c(8L, 3L, 4L, 5L, 1L, 2L, 2L, 9L, 6L, 7L, 7L), .Label = c(Edgar, Elodie, Jeanine, Jeannine, Michel, Michele, Michèle, Michelle, Victor ), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor), Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names = c(Matricule, Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names = c(NA, -11L)) I would want to make homogeneous the information and would like built 2 dataframes : df1 wich has the value of Nom and Prenom of the first lines of TEST when there are different values. The other values (Matricule or Sexe or Date.de.naissance) are unchanged df1 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 6L, 5L, 5L, 7L, 3L, 3L, 3L), .Label = c(CHICHE, GEOF, GUTIER, JACQUE, LANGUE, TRU, VINCENT), class = factor), Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar, Elodie, Jeanine, Michel, Michele, Michelle, Victor ), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor), Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names = c(Matricule, Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names = c(NA, -11L)) df2 wich has the value of Nom and Prenom of the last lines of TEST when there are different values. The other values (Matricule or Sexe or Date.de.naissance) are unchanged. df2 - structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 3L, 6L, 4L, 4L, 7L, 5L, 5L, 5L), .Label = c(CHICHE, GEOF, JACQUE, LANGUE-LOPEZ, RIVIER, TRU, VINCENT), class = factor), Prenom = structure(c(6L, 3L, 3L, 4L, 1L, 2L, 2L, 7L, 5L, 5L, 5L), .Label = c(Edgar, Elodie, Jeannine, Michel, Michèle, Michelle, Victor), class = factor), Sexe = structure(c(1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L), .Label = c(Féminin, Masculin), class = factor), Date.de.naissance = structure(c(4L, 2L, 2L, 7L, 6L, 5L, 5L, 1L, 3L, 3L, 3L), .Label = c(03/09/1940, 04/03/1946, 07/12/1947, 18/11/1945, 27/09/1947, 29/12/1936, 30/03/1935), class = factor)), .Names = c(Matricule, Nom, Prenom, Sexe, Date.de.naissance), class = data.frame, row.names = c(NA, -11L)) Thank for your helps Michel -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Calculate interaction for a big dataframe
Hi To calculate the value of the interaction between factors of a dataframe df, does exist any function which could replace the function when the dataframe df has the numbers of rows of df is large (~55000) and also the numbers of combinaison of the three factors is large. The calcul abort. The function to calculate the interaction is : as.numeric(interaction(df [,c(1:3)],drop=TRUE)) To complete the question and to calculate interaction beetween 3 factors f1, f2, f3, does it possible to calculate first f12 = interaction (f1,f2) and after calculate interaction (f12, f3). It seems to me that yes. Thanks for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculate interaction for a big dataframe
Thank you Petr paste is better than interaction for long vectors But now a new problem/question is appeared. Now, I would like transform the vector v1 - c( 4162.France, 4162.France, 4162.France, 4162.Mali, 4162.Mali, 4162.France, 4162.France, 4162.France, 4162.France, 4162.Mali) into a vector V2 with the same length but with number which are creasing v2 - c(1, 1,1, 2, 2, 3, 3,3,3, 4)) Any idea (function) ? Regards Le 22/07/2013 14:45, PIKAL Petr a écrit : Hi you maybe could use paste f1-sample(letters[1:3], 10, replace=T) f2-sample(letters[4:7], 10, replace=T) f3-sample(letters[9:11], 10, replace=T) interaction(f1, f2, f3, drop=T) [1] c.e.j b.e.j a.e.j c.g.i a.f.j b.g.k a.e.i a.e.k a.d.j b.e.j Levels: a.e.i c.g.i a.d.j a.e.j b.e.j c.e.j a.f.j a.e.k b.g.k paste(f1, f2, f3, sep=.) [1] c.e.j b.e.j a.e.j c.g.i a.f.j b.g.k a.e.i a.e.k a.d.j [10] b.e.j The difference is that interaction gives you directly factor, paste gives you character vector, but it may be convenient too for your purpose. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Arnaud Michel Sent: Monday, July 22, 2013 10:57 AM To: R help Subject: [R] Calculate interaction for a big dataframe Hi To calculate the value of the interaction between factors of a dataframe df, does exist any function which could replace the function when the dataframe df has the numbers of rows of df is large (~55000) and also the numbers of combinaison of the three factors is large. The calcul abort. The function to calculate the interaction is : as.numeric(interaction(df [,c(1:3)],drop=TRUE)) To complete the question and to calculate interaction beetween 3 factors f1, f2, f3, does it possible to calculate first f12 = interaction (f1,f2) and after calculate interaction (f12, f3). It seems to me that yes. Thanks for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on plotting with googleVis
Hi
You can do a rotation and use gvisColumnChart instead gvisBarChart
plot(gvisColumnChart(MyData, xvar=Names1, yvar=c(Values1,
Values2),options=list(width=2500,height=1000)))
Michel
Le 17/07/2013 15:57, Christofer Bogaso a écrit :
Hello Arnaud,
Thank you for your pointer. However I need to more clarification.
I want to control the max. and min. values for the x-axis, as well as
number of vertical gridlines to be displayed. I tried the following:
MyData - data.frame(Names1 = paste(XXX, 1:150), Values1 = 1:150 +
10, Values2 = 1:150)
library(googleVis)
plot(gvisBarChart(MyData, xvar=Names1, yvar=c(Values1,
Values2),options=list(width=1200,height=1500,hAxis.gridlines =
{count: 10},hAxis.minValue = 0, hAxis.maxValue = 100)))
However this is not clearly working. Can someone point me what went wrong?
Thanks and regards,
On 7/16/13, Arnaud Michel michel.arn...@cirad.fr wrote:
You can try with list options :
plot(gvisBarChart(MyData, xvar=Names1, yvar=c(Values1, Values2),
options=list(width=1200,height=1500)))
Le 15/07/2013 20:00, Christofer Bogaso a écrit :
Hello again,
Let say I have following data-frame:
MyData - data.frame(Names1 = paste(XXX, 1:150), Values1 = 1:150 + 10,
Values2 = 1:150)
Now I want to plot this data-frame with googleVis. Therefore I run
following codes:
library(googleVis)
plot(gvisBarChart(MyData, xvar=Names1, yvar=c(Values1, Values2)))
However the problem is that, hardly this plot can be read. However if I
plot a fraction of my data-frame then the underlying plot is clearly
visible:
plot(gvisBarChart(MyData[1:15,], xvar=Names1, yvar=c(Values1,
Values2))) ## This is clearly visible.
I would really appreciate if someone gives me some pointer how I can
clearly plot my data-frame with googleVis. I understand that there are
many
other plotting methods available with R like ggplot, however here I want
to
use googleVis because of its strength in showing the values within the
plot
area itself if you hover your mouse.
Thanks and regards,
[[alternative HTML version deleted]]
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31
--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] simplify a dataframe
Hi Arun I have two questions always about the question of symplify a dataframe I would like 1) to transform the vector x1 into the vector y1 x1 - c(1,1,1,-1000, 1,-1000, 1,1,1,1,1,1,-1000) y1 - c(1,1,1,1,2,2, 3,3,3,3,3,3,3) 2) to transform the vectors Debut and Fin by taking into account INDX into the two vectors Deb and Fin Debut - c ( 24/01/1995, 01/05/1997 ,31/12/1997, 02/02/1995 ,28/02/1995 ,01/03/1995, 13/03/1995, 01/01/1996, 31/01/1996, 24/01/1995, 01/07/1995 ,01/09/1995, 01/07/1997, 01/01/1998, 01/08/1998, 01/01/2000, 17/01/2000,29/02/2000) Fin - c ( 30/04/1997, 30/12/1997 ,31/12/1997, 27/02/1995, 28/02/1995, 12/03/1995, 30/06/1995, 30/01/1996, 31/01/1996, 30/06/1995, 31/08/1995, 30/06/1997, 31/12/1997, 31/07/1998, 31/12/1999, 16/01/2000, 28/02/2000, 29/02/2000) INDX - c(6,6,6,11,11,11, 4,5,5) Deb - c(*24/01/1995*, *02/02/1995*, *13/03/1995*, *01/01/1996*) Fi n - c(*31/12/1997*, *12/03/1995*, *30/06/1995*, *31/01/1996*) DebutFin INDX *24/01/1995* 30/04/19976 01/05/1997 30/12/19976 31/12/1997 *31/12/1997*6 *02/02/1995* 27/02/1995 11 28/02/1995 28/02/1995 11 01/03/1995 *12/03/1995* 11 *13/03/1995* *30/06/1995*4 *01/01/1996* 30/01/19965 31/01/1996 *31/01/1996*5 Thanks for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simplify a dataframe
Thank you for the question (1) Sorry for the imprecision for the question (2) : Suppose the date frame df df1 - data.frame( Debut =c ( 24/01/1995, 01/05/1997 ,31/12/1997, 02/02/1995 ,28/02/1995 ,01/03/1995, 13/03/1995, 01/01/1996, 31/01/1996) , Fin = c ( 30/04/1997, 30/12/1997 ,31/12/1997, 27/02/1995, 28/02/1995, 12/03/1995, 30/06/1995, 30/01/1996, 31/01/1996) , INDX = c(6,6,6, 11,11,11, 4, 5,5) ) I would like replace df1 by df2 df2 - data.frame( Deb = c(24/01/1995, 02/02/1995, 13/03/1995, 01/01/1996) , Fin = c(31/12/1997, 12/03/1995, 30/06/1995, 31/01/1996) ) Explication : The lines 1, 2 3 of df1 (who have same value of index =6) are replaced by only one line with value of Debut of df2 = Debut of line 1 of df1 value of Fin of df2 = Fin of line 3 of df1 The lines 4,5,6 of df1 (who have same value of index =11) are replaced by only one line with value of Debut of df2 = Debut of line 4 of df1 and value of fin of df2 = Fin of line 6 of df1 The line 7 of df1 (who have same value of index =4) are replaced by only one line with value of Debut of df2 = Debut of line 7of df1 and value of fin of df2 = Fin of line 7of df1 == No change The lines 8,9 of df1 (who have same value of index =5) are replaced by only one line with value of Debut of df2 = Debut of line 8of df1 and value of fin of df2 = Fin of line 9 of df1 df1 DebutFin INDX 1 24/01/1995 30/04/19976 2 01/05/1997 30/12/19976 3 31/12/1997 31/12/19976 4 02/02/1995 27/02/1995 11 5 28/02/1995 28/02/1995 11 6 01/03/1995 12/03/1995 11 7 13/03/1995 30/06/19954 8 01/01/1996 30/01/19965 9 31/01/1996 31/01/19965 DebFin 1 24/01/1995 31/12/1997 2 02/02/1995 12/03/1995 3 13/03/1995 30/06/1995 4 01/01/1996 31/01/1996 Thank you for your helps Michel Le 17/07/2013 19:57, Rui Barradas a écrit : Hello, As for question (1), try the following. y2 - cumsum(c(TRUE, diff(x1) 0)) identical(as.integer(y1), y2) # y1 is of class numeric As for question (2) I'm not understanding it. Hope this helps, Rui Barradas Em 17-07-2013 18:21, Arnaud Michel escreveu: Hi Arun I have two questions always about the question of symplify a dataframe I would like 1) to transform the vector x1 into the vector y1 x1 - c(1,1,1,-1000, 1,-1000, 1,1,1,1,1,1,-1000) y1 - c(1,1,1,1,2,2, 3,3,3,3,3,3,3) 2) to transform the vectors Debut and Fin by taking into account INDX into the two vectors Deb and Fin Debut - c ( 24/01/1995, 01/05/1997 ,31/12/1997, 02/02/1995 ,28/02/1995 ,01/03/1995, 13/03/1995, 01/01/1996, 31/01/1996, 24/01/1995, 01/07/1995 ,01/09/1995, 01/07/1997, 01/01/1998, 01/08/1998, 01/01/2000, 17/01/2000,29/02/2000) Fin - c ( 30/04/1997, 30/12/1997 ,31/12/1997, 27/02/1995, 28/02/1995, 12/03/1995, 30/06/1995, 30/01/1996, 31/01/1996, 30/06/1995, 31/08/1995, 30/06/1997, 31/12/1997, 31/07/1998, 31/12/1999, 16/01/2000, 28/02/2000, 29/02/2000) INDX - c(6,6,6,11,11,11, 4,5,5) Deb - c(*24/01/1995*, *02/02/1995*, *13/03/1995*, *01/01/1996*) Fi n - c(*31/12/1997*, *12/03/1995*, *30/06/1995*, *31/01/1996*) DebutFin INDX *24/01/1995* 30/04/19976 01/05/1997 30/12/19976 31/12/1997 *31/12/1997*6 *02/02/1995* 27/02/1995 11 28/02/1995 28/02/1995 11 01/03/1995 *12/03/1995* 11 *13/03/1995* *30/06/1995*4 *01/01/1996* 30/01/19965 31/01/1996 *31/01/1996*5 Thanks for your help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simplify a dataframe
Thanks Arun and Rui for your helps Michel Le 17/07/2013 22:20, arun a écrit : #or library(plyr) res-ddply(df1,.(INDX),summarize,Debut=head(Debut,1),Fin=tail(Fin,1)) res$INDX-factor(res$INDX,levels=unique(df1$INDX)) res[order(res$INDX),-1] # DebutFin #3 24/01/1995 31/12/1997 #4 02/02/1995 12/03/1995 #1 13/03/1995 30/06/1995 #2 01/01/1996 31/01/1996 A.K. - Original Message - From: arun smartpink...@yahoo.com To: Arnaud Michel michel.arn...@cirad.fr Cc: R help r-help@r-project.org; Rui Barradas ruipbarra...@sapo.pt Sent: Wednesday, July 17, 2013 4:14 PM Subject: Re: [R] simplify a dataframe Hi, You could try: df1[,1:2]-lapply(df1[,1:2],as.character) df2New- data.frame(Deb=unique(with(df1,ave(Debut,INDX,FUN=function(x) head(x,1,Fin=unique(with(df1,ave(Fin,INDX,FUN=function(x) tail(x,1) identical(df2New,df2) #[1] TRUE A.K. - Original Message - From: Arnaud Michel michel.arn...@cirad.fr To: Rui Barradas ruipbarra...@sapo.pt; R help r-help@r-project.org; arun smartpink...@yahoo.com Cc: Sent: Wednesday, July 17, 2013 4:03 PM Subject: Re: [R] simplify a dataframe Thank you for the question (1) Sorry for the imprecision for the question (2) : Suppose the date frame df df1 - data.frame( Debut =c ( 24/01/1995, 01/05/1997 ,31/12/1997, 02/02/1995 ,28/02/1995 ,01/03/1995, 13/03/1995, 01/01/1996, 31/01/1996) , Fin = c ( 30/04/1997, 30/12/1997 ,31/12/1997, 27/02/1995, 28/02/1995, 12/03/1995, 30/06/1995, 30/01/1996, 31/01/1996) , INDX = c(6,6,6, 11,11,11, 4, 5,5) ) I would like replace df1 by df2 df2 - data.frame( Deb = c(24/01/1995, 02/02/1995, 13/03/1995, 01/01/1996) , Fin = c(31/12/1997, 12/03/1995, 30/06/1995, 31/01/1996) ) Explication : The lines 1, 2 3 of df1 (who have same value of index =6) are replaced by only one line with value of Debut of df2 = Debut of line 1 of df1 value of Fin of df2 = Fin of line 3 of df1 The lines 4,5,6 of df1 (who have same value of index =11) are replaced by only one line with value of Debut of df2 = Debut of line 4 of df1 and value of fin of df2 = Fin of line 6 of df1 The line 7 of df1 (who have same value of index =4) are replaced by only one line with value of Debut of df2 = Debut of line 7of df1 and value of fin of df2 = Fin of line 7of df1 == No change The lines 8,9 of df1 (who have same value of index =5) are replaced by only one line with value of Debut of df2 = Debut of line 8of df1 and value of fin of df2 = Fin of line 9 of df1 df1 DebutFin INDX 1 24/01/1995 30/04/19976 2 01/05/1997 30/12/19976 3 31/12/1997 31/12/19976 4 02/02/1995 27/02/1995 11 5 28/02/1995 28/02/1995 11 6 01/03/1995 12/03/1995 11 7 13/03/1995 30/06/19954 8 01/01/1996 30/01/19965 9 31/01/1996 31/01/19965 DebFin 1 24/01/1995 31/12/1997 2 02/02/1995 12/03/1995 3 13/03/1995 30/06/1995 4 01/01/1996 31/01/1996 Thank you for your helps Michel Le 17/07/2013 19:57, Rui Barradas a écrit : Hello, As for question (1), try the following. y2 - cumsum(c(TRUE, diff(x1) 0)) identical(as.integer(y1), y2) # y1 is of class numeric As for question (2) I'm not understanding it. Hope this helps, Rui Barradas Em 17-07-2013 18:21, Arnaud Michel escreveu: Hi Arun I have two questions always about the question of symplify a dataframe I would like 1) to transform the vector x1 into the vector y1 x1 - c(1,1,1,-1000, 1,-1000, 1,1,1,1,1,1,-1000) y1 - c(1,1,1,1,2,2, 3,3,3,3,3,3,3) 2) to transform the vectors Debut and Fin by taking into account INDX into the two vectors Deb and Fin Debut - c ( 24/01/1995, 01/05/1997 ,31/12/1997, 02/02/1995 ,28/02/1995 ,01/03/1995, 13/03/1995, 01/01/1996, 31/01/1996, 24/01/1995, 01/07/1995 ,01/09/1995, 01/07/1997, 01/01/1998, 01/08/1998, 01/01/2000, 17/01/2000,29/02/2000) Fin - c ( 30/04/1997, 30/12/1997 ,31/12/1997, 27/02/1995, 28/02/1995, 12/03/1995, 30/06/1995, 30/01/1996, 31/01/1996, 30/06/1995, 31/08/1995, 30/06/1997, 31/12/1997, 31/07/1998, 31/12/1999, 16/01/2000, 28/02/2000, 29/02/2000) INDX - c(6,6,6,11,11,11, 4,5,5) Deb - c(*24/01/1995*, *02/02/1995*, *13/03/1995*, *01/01/1996*) Fi n - c(*31/12/1997*, *12/03/1995*, *30/06/1995*, *31/01/1996*) DebutFin INDX *24/01/1995* 30/04/19976 01/05/1997 30/12/19976 31/12/1997 *31/12/1997*6 *02/02/1995* 27/02/1995 11 28/02/1995 28/02/1995 11 01/03/1995 *12/03/1995* 11 *13/03/1995* *30/06/1995*4 *01/01/1996* 30/01/19965 31/01/1996 *31/01/1996*5 Thanks for your help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38
Re: [R] Question on plotting with googleVis
You can try with list options : plot(gvisBarChart(MyData, xvar=Names1, yvar=c(Values1, Values2), options=list(width=1200,height=1500))) Le 15/07/2013 20:00, Christofer Bogaso a écrit : Hello again, Let say I have following data-frame: MyData - data.frame(Names1 = paste(XXX, 1:150), Values1 = 1:150 + 10, Values2 = 1:150) Now I want to plot this data-frame with googleVis. Therefore I run following codes: library(googleVis) plot(gvisBarChart(MyData, xvar=Names1, yvar=c(Values1, Values2))) However the problem is that, hardly this plot can be read. However if I plot a fraction of my data-frame then the underlying plot is clearly visible: plot(gvisBarChart(MyData[1:15,], xvar=Names1, yvar=c(Values1, Values2))) ## This is clearly visible. I would really appreciate if someone gives me some pointer how I can clearly plot my data-frame with googleVis. I understand that there are many other plotting methods available with R like ggplot, however here I want to use googleVis because of its strength in showing the values within the plot area itself if you hover your mouse. Thanks and regards, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simplify a dataframe
Hi,
Excuse me for the indistinctness
Le 13/07/2013 17:18, arun a écrit :
Hi,
when the value of Debut of lines i = value Fin of lines i-1
That part is not clear esp. when it is looked upon with the expected output
(df2).
I want to group the lines which have the same caracteristics (Matricule,
Nom, Sexe, DateNaissance, Contrat, Pays) and with period of time
(Debut/start and Fin/end) without interruption of time.
For exemple :
The following three lines : Debut/Start Fin/End
1 VERON Féminin02/09/1935 CDI commun France
*24/01/1995* 30/04/1997
1 VERON Féminin02/09/1935 CDI commun France
01/05/1997 30/12/1997
1 VERON Féminin02/09/1935 CDI commun France
31/12/1997 *31/12/1997*
are transformed into 1 line
1 VERON Féminin02/09/1935 CDI commun France
*24/01/1995* *31/12/1997*
because same caracteristicsand period of time without interruption of
time (from *24/01/1995* to *31/12/1997)*
The following six lines :
6 BENARD Masculin01/04/1935 CDI commun Philippines
*02/02/1995* 27/02/1995
6 BENARD Masculin01/04/1935 CDI commun Philippines
28/02/1995 28/02/1995
6 BENARD Masculin01/04/1935 CDI commun Philippines
01/03/1995 *12/03/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*13/03/1995* *30/06/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*01/01/1996* 30/01/1996
6 BENARD Masculin01/04/1935 CDI commun France
31/01/1996 *31/01/1996*
are transformed into
6 BENARD Masculin01/04/1935 CDI commun Philippines
*02/02/1995* *12/03/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*13/03/1995* *30/06/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*01/01/1996* *31/01/1996*
because
lines 1-3 identical for caracteristics and without interruption in time
lines 4 and lines 5-6 are not grouped because there is an interruption
in time beetween *30/06/1995 *and *01/01/1996*
Thank you for your help
Michel
Also, in your example dataset:
df1$contrat[grep(^CDD,df1$contrat)]
#[1] CDD détaché ext. Cirad CDD détaché ext. Cirad CDD détaché ext.
Cirad
#[4] CDD détaché ext. Cirad CDD détaché ext.Cirad CDD détaché ext.
Cirad
#[7] CDD détaché ext. Cirad CDD détaché ext.Cirad CDD détaché ext.
Cirad
##Looks like there are extra spaces in some of them. I guess these are the
same
df1$contrat[grep(^CDD,df1$contrat)]- CDD détaché ext. Cirad
I tried this:
indx-as.numeric(interaction(df1[,1:6],drop=FALSE))
df1New- df1
res2-unique(within(df1New,{Debut-ave(seq_along(indx),indx,FUN=function(x)
Debut[head(x,1)]);Fin- ave(seq_along(indx),indx,FUN=function(x)
Fin[tail(x,1)])}))
row.names(res2)- 1:nrow(res2)
res2[,c(1,2,7:8)]
MatriculeNom DebutFin
1 1 VERON 24/01/1995 31/12/1997
2 6 BENARD 02/02/1995 12/03/1995
3 6 BENARD 13/03/1995 31/01/1996 ###here not correct
4 8 DALNIC 24/01/1995 31/08/1995
5 8 DALNIC 01/09/1995 29/02/2000
6934 FORNI 26/01/1995 31/08/2001
7934 FORNI 01/09/2001 31/08/2004
8934 FORNI 01/09/2004 31/08/2007
9934 FORNI 01/09/2007 04/09/2012
10 934 FORNI 05/09/2012 31/12/4712
df2[,c(1,2,7:8)]
MatNom DebutFin
11 VERON 24/01/1995 31/12/1997
26 BENARD 02/02/1995 12/03/1995
36 BENARD 13/03/1995 30/06/1995
46 BENARD 01/01/1996 31/01/1996 #missing this row
58 DALNIC 24/01/1995 31/08/1995
68 DALNIC 01/09/1995 29/02/2000
7 934 FORNI 26/01/1995 31/08/2001
8 934 FORNI 01/09/2001 31/08/2004
9 934 FORNI 01/09/2004 31/08/2007
10 934 FORNI 01/09/2007 04/09/2012
11 934 FORNI 05/09/2012 31/12/4712
Here, the dates look similar to the ones on df2 except for one row in df2.
A.K.
- Original Message -
From: Arnaud Michel michel.arn...@cirad.fr
To: R help r-help@r-project.org
Cc:
Sent: Friday, July 12, 2013 3:45 PM
Subject: [R] simplify a dataframe
Hello
I have the following problem : group the lines of a dataframe when no
information change (Matricule, Nom, Sexe, DateNaissance, Contrat, Pays)
and when the value of Debut of lines i = value Fin of lines i-1
I can obtain it with a do loop. Is it possible to avoid the loop ?
The dataframe initial is df1
dput(df1)
structure(list(Matricule = c(1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L,
6L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 934L, 934L, 934L, 934L,
934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L,
934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L,
934L, 934L, 934L, 934L), Nom = c(VERON, VERON, VERON, BENARD,
BENARD, BENARD, BENARD, BENARD, BENARD, DALNIC, DALNIC,
DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC,
FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI,
FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI,
FORNI, FORNI, FORNI, FORNI
Re: [R] simplify a dataframe
An other remark :
If I calculate
df1$D- as.numeric(as.Date(paste0(substr(df1$Debut,7,10),-,
substr(df1$Debut,4,5),-,substr(df1$Debut,1,2
and
df1$F - as.numeric(as.Date(paste0(substr(df1$Fin,7,10),-,
substr(df1$Fin,4,5),-,substr(df1$Fin,1,2
and if there is no interruption of time for the lines i and i+1
then df1$F[i] + 1 == df1$D[i+1]
Michel
Le 14/07/2013 18:17, Arnaud Michel a écrit :
Hi,
Excuse me for the indistinctness
Le 13/07/2013 17:18, arun a écrit :
Hi,
when the value of Debut of lines i = value Fin of lines i-1
That part is not clear esp. when it is looked upon with the expected output
(df2).
I want to group the lines which have the same caracteristics (Matricule,
Nom, Sexe, DateNaissance, Contrat, Pays) and with period of time
(Debut/start and Fin/end) without interruption of time.
For exemple :
The following three lines : Debut/Start Fin/End
1 VERON Féminin02/09/1935 CDI commun France
*24/01/1995* 30/04/1997
1 VERON Féminin02/09/1935 CDI commun France
01/05/1997 30/12/1997
1 VERON Féminin02/09/1935 CDI commun France
31/12/1997 *31/12/1997*
are transformed into 1 line
1 VERON Féminin02/09/1935 CDI commun France
*24/01/1995* *31/12/1997*
because same caracteristicsand period of time without interruption of
time (from *24/01/1995* to *31/12/1997)*
The following six lines :
6 BENARD Masculin01/04/1935 CDI commun Philippines
*02/02/1995* 27/02/1995
6 BENARD Masculin01/04/1935 CDI commun Philippines
28/02/1995 28/02/1995
6 BENARD Masculin01/04/1935 CDI commun Philippines
01/03/1995 *12/03/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*13/03/1995* *30/06/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*01/01/1996* 30/01/1996
6 BENARD Masculin01/04/1935 CDI commun France
31/01/1996 *31/01/1996*
are transformed into
6 BENARD Masculin01/04/1935 CDI commun Philippines
*02/02/1995* *12/03/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*13/03/1995* *30/06/1995*
6 BENARD Masculin01/04/1935 CDI commun France
*01/01/1996* *31/01/1996*
because
lines 1-3 identical for caracteristics and without interruption in time
lines 4 and lines 5-6 are not grouped because there is an interruption
in time beetween *30/06/1995 *and *01/01/1996*
Thank you for your help
Michel
Also, in your example dataset:
df1$contrat[grep(^CDD,df1$contrat)]
#[1] CDD détaché ext. Cirad CDD détaché ext. Cirad CDD détaché ext.
Cirad
#[4] CDD détaché ext. Cirad CDD détaché ext.Cirad CDD détaché ext.
Cirad
#[7] CDD détaché ext. Cirad CDD détaché ext.Cirad CDD détaché ext.
Cirad
##Looks like there are extra spaces in some of them. I guess these are the
same
df1$contrat[grep(^CDD,df1$contrat)]- CDD détaché ext. Cirad
I tried this:
indx-as.numeric(interaction(df1[,1:6],drop=FALSE))
df1New- df1
res2-unique(within(df1New,{Debut-ave(seq_along(indx),indx,FUN=function(x)
Debut[head(x,1)]);Fin- ave(seq_along(indx),indx,FUN=function(x)
Fin[tail(x,1)])}))
row.names(res2)- 1:nrow(res2)
res2[,c(1,2,7:8)]
MatriculeNom DebutFin
1 1 VERON 24/01/1995 31/12/1997
2 6 BENARD 02/02/1995 12/03/1995
3 6 BENARD 13/03/1995 31/01/1996 ###here not correct
4 8 DALNIC 24/01/1995 31/08/1995
5 8 DALNIC 01/09/1995 29/02/2000
6934 FORNI 26/01/1995 31/08/2001
7934 FORNI 01/09/2001 31/08/2004
8934 FORNI 01/09/2004 31/08/2007
9934 FORNI 01/09/2007 04/09/2012
10 934 FORNI 05/09/2012 31/12/4712
df2[,c(1,2,7:8)]
MatNom DebutFin
11 VERON 24/01/1995 31/12/1997
26 BENARD 02/02/1995 12/03/1995
36 BENARD 13/03/1995 30/06/1995
46 BENARD 01/01/1996 31/01/1996 #missing this row
58 DALNIC 24/01/1995 31/08/1995
68 DALNIC 01/09/1995 29/02/2000
7 934 FORNI 26/01/1995 31/08/2001
8 934 FORNI 01/09/2001 31/08/2004
9 934 FORNI 01/09/2004 31/08/2007
10 934 FORNI 01/09/2007 04/09/2012
11 934 FORNI 05/09/2012 31/12/4712
Here, the dates look similar to the ones on df2 except for one row in df2.
A.K.
- Original Message -
From: Arnaud Michel michel.arn...@cirad.fr
To: R help r-help@r-project.org
Cc:
Sent: Friday, July 12, 2013 3:45 PM
Subject: [R] simplify a dataframe
Hello
I have the following problem : group the lines of a dataframe when no
information change (Matricule, Nom, Sexe, DateNaissance, Contrat, Pays)
and when the value of Debut of lines i = value Fin of lines i-1
I can obtain it with a do loop. Is it possible to avoid the loop ?
The dataframe initial is df1
dput(df1)
structure(list(Matricule = c(1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L,
6L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 934L, 934L, 934L, 934L,
934L
Re: [R] simplify a dataframe
Super !!!
Thank you very much Arun
Michel
Le 15/07/2013 03:47, arun a écrit :
HI Michel,
This gives the same order as that of df2.
df1$contrat[grep(^CDD,df1$contrat)]- CDD détaché ext. Cirad
df1[48,8]- 31/12/2013
indx-as.numeric(interaction(df1[,1:6],drop=TRUE))
lst1-split(df1,indx)
lst2-lst1[match(unique(indx),names(lst1))]
res-do.call(rbind,lapply(lst2,function(x){x1- as.Date(x$Debut,format=%d/%m/%Y);x2-
as.Date(x$Fin,format=%d/%m/%Y);do.call(rbind,lapply(split(x,cumsum(c(FALSE,(x1[-1]-x2[-nrow(x)])!=1))),function(x)
data.frame(x[1,1:6],Debut=head(x$Debut,1),Fin=tail(x$Fin,1),stringsAsFactors=FALSE)))}))
row.names(res)- 1:nrow(res)
df2[11,8]- 31/12/2013
names(res)[1]- Mat
identical(res,df2)
#[1] TRUE
A.K.
- Original Message -
From: arun smartpink...@yahoo.com
To: Arnaud Michel michel.arn...@cirad.fr
Cc: R help r-help@r-project.org
Sent: Sunday, July 14, 2013 2:39 PM
Subject: Re: [R] simplify a dataframe
Hi,
May be this helps you.
df1$contrat[grep(^CDD,df1$contrat)]- CDD détaché ext. Cirad
df1[48,8]
[1] 31/12/4712 #strange value
df1[48,8]- 31/12/2013 #changed
indx-as.numeric(interaction(df1[,1:6],drop=TRUE))
res-do.call(rbind,lapply(split(df1,indx),function(x) {x1-
as.Date(x$Debut,format=%d/%m/%Y);x2-
as.Date(x$Fin,format=%d/%m/%Y);do.call(rbind,lapply(split(x,cumsum(c(FALSE,(x1[-1]-x2[-nrow(x)])!=1))),function(x)
data.frame(x[1,1:6],Debut=head(x$Debut,1),Fin=tail(x$Fin,1),stringsAsFactors=FALSE)))}))
res[order(res$Matricule),] #the order of rows is a bit different than df2.
MatriculeNom Sexe DateNaissancecontratPays
5 1 VERON Féminin02/09/1935 CDI commun France
4.0 6 BENARD Masculin01/04/1935 CDI commun France
4.1 6 BENARD Masculin01/04/1935 CDI commun France
10 6 BENARD Masculin01/04/1935 CDI commun Philippines
6 8 DALNIC Féminin19/02/1940 CDI commun France
9 8 DALNIC Féminin19/02/1940 CDI commun Martinique
1 934 FORNI Masculin10/07/1961 CDD détaché ext. CiradCameroun
2 934 FORNI Masculin10/07/1961 CDI commun Congo
3 934 FORNI Masculin10/07/1961CDI Détachés Autres Congo
7 934 FORNI Masculin10/07/1961CDI Détachés Autres France
8 934 FORNI Masculin10/07/1961 CDI commun Gabon
DebutFin
5 24/01/1995 31/12/1997
4.0 13/03/1995 30/06/1995
4.1 01/01/1996 31/01/1996
10 02/02/1995 12/03/1995
6 24/01/1995 31/08/1995
9 01/09/1995 29/02/2000
1 26/01/1995 31/08/2001
2 05/09/2012 31/12/2013
3 01/09/2004 31/08/2007
7 01/09/2001 31/08/2004
8 01/09/2007 04/09/2012
A.K.
From: Arnaud Michel michel.arn...@cirad.fr
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; jholt...@gmail.com; Rui Barradas
ruipbarra...@sapo.pt
Sent: Sunday, July 14, 2013 12:17 PM
Subject: Re: [R] simplify a dataframe
Hi,
Excuse me for the indistinctness
Le 13/07/2013 17:18, arun a écrit :
Hi,
when the value of Debut of lines i = value Fin of lines i-1
That part is not clear esp. when it is looked upon with the expected output
(df2).
I want to group the lines which have the same caracteristics (Matricule, Nom,
Sexe, DateNaissance, Contrat, Pays) and with period of time (Debut/start and
Fin/end) without interruption of time.
For exemple :
The following three lines
:
Debut/Start Fin/End
1 VERON Féminin02/09/1935 CDI commun France 24/01/1995
30/04/1997
1 VERON Féminin02/09/1935 CDI commun France
01/05/1997 30/12/1997
1 VERON Féminin02/09/1935 CDI commun France
31/12/1997 31/12/1997
are transformed into 1 line
1 VERON Féminin02/09/1935 CDI commun France 24/01/1995
31/12/1997
because same caracteristicsand period of time without interruption
of time (from 24/01/1995 to 31/12/1997)
The following six lines :
6 BENARD Masculin01/04/1935 CDI commun Philippines 02/02/1995
27/02/1995
6 BENARD Masculin01/04/1935 CDI commun Philippines
28/02/1995 28/02/1995
6 BENARD Masculin01/04/1935 CDI commun Philippines
01/03/1995 12/03/1995
6 BENARD Masculin01/04/1935 CDI commun France 13/03/1995
30/06/1995
6 BENARD Masculin01/04/1935 CDI commun France 01/01/1996
30/01/1996
6 BENARD Masculin01/04/1935 CDI commun France
31/01/1996 31/01/1996
are transformed into
6 BENARD Masculin01/04/1935 CDI commun Philippines 02/02/1995
12/03/1995
6 BENARD Masculin01/04/1935 CDI commun France 13/03/1995
30/06/1995
6 BENARD Masculin01/04/1935 CDI commun France 01/01/1996
31/01/1996
because
lines 1-3 identical
[R] simplify a dataframe
Hello I have the following problem : group the lines of a dataframe when no information change (Matricule, Nom, Sexe, DateNaissance, Contrat, Pays) and when the value of Debut of lines i = value Fin of lines i-1 I can obtain it with a do loop. Is it possible to avoid the loop ? The dataframe initial is df1 dput(df1) structure(list(Matricule = c(1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L, 6L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L, 934L), Nom = c(VERON, VERON, VERON, BENARD, BENARD, BENARD, BENARD, BENARD, BENARD, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, DALNIC, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI, FORNI), Sexe = c(Féminin, Féminin, Féminin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Féminin, Féminin, Féminin, Féminin, Féminin, Féminin, Féminin, Féminin, Féminin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin, Masculin), DateNaissance = c(02/09/1935, 02/09/1935, 02/09/1935, 01/04/1935, 01/04/1935, 01/04/1935, 01/04/1935, 01/04/1935, 01/04/1935, 19/02/1940, 19/02/1940, 19/02/1940, 19/02/1940, 19/02/1940, 19/02/1940, 19/02/1940, 19/02/1940, 19/02/1940, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961, 10/07/1961), contrat = c(CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDD détaché ext. Cirad, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI Détachés Autres, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun, CDI commun), Pays = c(France, France, France, Philippines, Philippines, Philippines, France, France, France, France, France, Martinique, Martinique, Martinique, Martinique, Martinique, Martinique, Martinique, Cameroun, Cameroun, Cameroun, Cameroun, Cameroun, Cameroun, Cameroun, Cameroun, Cameroun, France, France, France, France, France, France, France, Congo, Congo, Congo, Congo, Congo, Congo, Gabon, Gabon, Gabon, Gabon, Gabon, Gabon, Congo, Congo), Debut = c(24/01/1995, 01/05/1997, 31/12/1997, 02/02/1995, 28/02/1995, 01/03/1995, 13/03/1995, 01/01/1996, 31/01/1996, 24/01/1995, 01/07/1995, 01/09/1995, 01/07/1997, 01/01/1998, 01/08/1998, 01/01/2000, 17/01/2000, 29/02/2000, 26/01/1995, 01/07/1996, 16/09/1997, 01/01/1998, 01/07/1998, 04/11/1999, 01/01/2001, 01/04/2001, 31/08/2001, 01/09/2001, 02/09/2001, 01/12/2001, 01/02/2003, 01/04/2003, 01/01/2004, 01/03/2004, 01/09/2004, 01/01/2005, 01/04/2005, 28/10/2006, 01/01/2007, 01/04/2007, 01/09/2007, 01/01/2009, 01/04/2009, 01/01/2010, 01/01/2011, 01/04/2011, 05/09/2012, 01/01/2013 ), Fin = c(30/04/1997, 30/12/1997, 31/12/1997, 27/02/1995, 28/02/1995, 12/03/1995, 30/06/1995, 30/01/1996, 31/01/1996, 30/06/1995, 31/08/1995, 30/06/1997, 31/12/1997, 31/07/1998, 31/12/1999, 16/01/2000, 28/02/2000, 29/02/2000, 30/06/1996, 15/09/1997, 31/12/1997, 30/06/1998, 03/11/1999, 31/12/2000, 31/03/2001, 30/08/2001, 31/08/2001, 01/09/2001, 30/11/2001, 31/01/2003, 31/03/2003, 31/12/2003, 29/02/2004, 31/08/2004, 31/12/2004, 31/03/2005, 27/10/2006, 31/12/2006, 31/03/2007, 31/08/2007, 31/12/2008, 31/03/2009, 31/12/2009, 31/12/2010, 31/03/2011, 04/09/2012, 31/12/2012, 31/12/4712)), .Names = c(Matricule, Nom, Sexe, DateNaissance, contrat, Pays, Debut, Fin ), class = data.frame, row.names = c(NA, -48L)) The dataframe to be obtained is df2 dput(df2) structure(list(Mat = c(1L, 6L, 6L, 6L, 8L, 8L, 934L, 934L, 934L, 934L, 934L), Nom = c(VERON, BENARD, BENARD, BENARD, DALNIC, DALNIC, FORNI, FORNI, FORNI, FORNI, FORNI), Sexe = c(Féminin, Masculin, Masculin, Masculin, Féminin, Féminin, Masculin, Masculin, Masculin, Masculin, Masculin), DateNaissance = c(02/09/1935, 01/04/1935, 01/04/1935, 01/04/1935,
[R] To transform a dataframe
Hello I would like to transform the dataframe df1 into df2 (ie copy the data from several lines for a men/women to only one line by individu men/women) dput(df1) structure(list(Mat = c(934L, 934L, 934L, 935L, 935L, 936L, 936L, 936L, 936L, 937L, 937L, 937L, 937L), Nom = structure(c(2L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L), .Label = c(, , , ), class = factor), Sexe = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = Masculin, class = factor), Cat = c(6L, 7L, 8L, 7L, 8L, 6L, 7L, 8L, 9L, 3L, 4L, 6L, 7L ), Début = structure(c(5L, 7L, 2L, 12L, 8L, 3L, 4L, 10L, 11L, 13L, 1L, 6L, 9L), .Label = c(01/01/1990, 01/01/2011, 01/02/1986, 01/02/1990, 01/07/1986, 01/07/1993, 01/07/1994, 01/07/1996, 01/10/2003, 01/11/2002, 01/11/2011, 13/01/1986, 23/01/1986), class = factor), Fin = structure(c(2L, 9L, 10L, 3L, 10L, 5L, 6L, 7L, 10L, 8L, 1L, 4L, 10L), .Label = c(30/06/1993, 30/06/1994, 30/06/1996, 30/09/2003, 31/01/1990, 31/10/2002, 31/10/2011, 31/12/1989, 31/12/2010, 31/12/4712), class = factor)), .Names = c(Mat, Nom, Sexe, Cat, Début, Fin), class = data.frame, row.names = c(NA, -13L)) dput(df2) structure(list(Mat = 934:937, Nom = structure(c(2L, 3L, 4L, 1L ), .Label = c(, , , ), class = factor), Sexe = structure(c(1L, 1L, 1L, 1L), .Label = Masculin, class = factor), Cat4 = c(NA, NA, NA, 4L), Début4 = structure(c(1L, 1L, 1L, 2L), .Label = c(, 01/01/1990), class = factor), Fin4 = structure(c(1L, 1L, 1L, 2L), .Label = c(, 30/06/1993), class = factor), Cat5 = c(NA, NA, NA, NA), Début5 = c(NA, NA, NA, NA), Fin5 = c(NA, NA, NA, NA), Cat6 = c(6L, NA, 6L, 6L), Début6 = structure(c(3L, 1L, 2L, 4L), .Label = c(, 01/02/1986, 01/07/1986, 01/07/1993 ), class = factor), Fin6 = structure(c(2L, 1L, 4L, 3L), .Label = c(, 30/06/1994, 30/09/2003, 31/01/1990), class = factor), Cat7 = c(7L, 7L, 7L, 7L), Début7 = structure(c(2L, 4L, 1L, 3L), .Label = c(01/02/1990, 01/07/1994, 01/10/2003, 13/01/1986), class = factor), Fin7 = structure(c(3L, 1L, 2L, 4L), .Label = c(30/06/1996, 31/10/2002, 31/12/2010, 31/12/4712), class = factor), Cat8 = c(8L, 8L, 8L, NA ), Début8 = structure(c(2L, 3L, 4L, 1L), .Label = c(, 01/01/2011, 01/07/1996, 01/11/2002), class = factor), Fin8 = structure(c(3L, 3L, 2L, 1L), .Label = c(, 31/10/2011, 31/12/4712), class = factor), Cat9 = c(NA, NA, 9L, NA), Début9 = structure(c(1L, 1L, 2L, 1L), .Label = c(, 01/11/2011), class = factor), Fin9 = structure(c(1L, 1L, 2L, 1L), .Label = c(, 31/12/4712), class = factor)), .Names = c(Mat, Nom, Sexe, Cat4, Début4, Fin4, Cat5, Début5, Fin5, Cat6, Début6, Fin6, Cat7, Début7, Fin7, Cat8, Début8, Fin8, Cat9, Début9, Fin9), class = data.frame, row.names = c(NA, -4L)) Any idea ? Thank you -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] a question to transform a dataframe empty 0/1
Hello I have the following dataframe : df - data.frame( Country=c(Zimbabwe,Burkina Faso,South Africa,Madagascar,Tanzania, Mali,Mozambique,Madagascar,Ghana,Nigeria,Kenya,Burkina Faso, South Africa,Tanzania,Kenya,Ethiopia ) , Iso=c(ZW,BF,ZA,MG,TZ,ML,MZ,MG,GH,NG,KE,BF, ZA,TZ,KE,ET) , Abaco=c(1,0,1,1,0,1,0,0,0,0,1,1,0,0,0,0) , Adaptclone= c(0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0) ) There is a lot of column like Abaco, Adaptclone,... I would like to built a dataframe wich transforms the initial dataframe of 4 columns into a dataframe of 3 columns as the following dataframe Country IsoProject Zimbabwe ZW Abaco South Africa ZA Abaco Madagascar MG Abaco Tanzania TZ Adaptclone Mali ML Abaco Mali ML Adaptclone .. Any idea ? Michel -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a question to transform a dataframe empty 0/1
Thank you arun ! I don't know the library reshape2 Michel Le 20/06/2013 19:55, arun a écrit : Hi, Not sure if you wanted the entries with 0. library(reshape2) dfMelt-melt(df,id.var=c(Country,Iso)) #subset those with 1 dfNew- subset(dfMelt,value==1,select=-4) row.names(dfNew)- 1:nrow(dfNew) dfNew #Country Iso variable #1 Zimbabwe ZW Abaco #2 South Africa ZA Abaco #3Madagascar MG Abaco #4 Mali ML Abaco #5 Kenya KE Abaco #6 Burkina Faso BF Abaco #7 Tanzania TZ Adaptclone #8 Mali ML Adaptclone #9Mozambique MZ Adaptclone #10 Madagascar MG Adaptclone #11Ghana GH Adaptclone #12 Nigeria NG Adaptclone A.K. Hello I have the following dataframe : df - data.frame( Country=c(Zimbabwe,Burkina Faso,South Africa,Madagascar,Tanzania, Mali,Mozambique,Madagascar,Ghana,Nigeria,Kenya,Burkina Faso, South Africa,Tanzania,Kenya,Ethiopia ) , Iso=c(ZW,BF,ZA,MG,TZ,ML,MZ,MG,GH,NG,KE,BF, ZA,TZ,KE,ET) , Abaco=c(1,0,1,1,0,1,0,0,0,0,1,1,0,0,0,0) , Adaptclone= c(0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0) ) There is a lot of column like Abaco, Adaptclone,... I would like to built a dataframe wich transforms the initial dataframe of 4 columns into a dataframe of 3 columns as the following dataframe Country IsoProject Zimbabwe ZW Abaco South Africa ZA Abaco Madagascar MG Abaco Tanzania TZ Adaptclone Mali ML Abaco Mali ML Adaptclone .. Any idea ? Michel -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] to build a data.frame
Hello I have the following dataframe df - data.frame( Project=c(Abaco,Abaco,Abac,Abaco,Abaco,Abaco, Abaco,Adaptclone,Adaptclone,Adaptclone,Adaptclone,Adaptclone, Adaptclone,Adopt,Adopt,Adopt), Country=c(Zimbabwe,Burkina Faso,South Africa,Madagascar,Tanzania, Mali,Mozambique,Madagascar,Ghana,Nigeria,Kenya,Burkina Faso, South Africa,Tanzania,Kenya,Ethiopia ), Iso=c(ZW,BF,ZA,MG,TZ,ML,MZ,MG,GH,NG,KE,BF, ZA,TZ,KE,ET)) I would like to build a other dataframe with name Country where column 1 is country, the column 2 is number of project (in the country) the column 3 is the code Iso (wich correspond with the country : Ex ZW is ISO of Zimbabwe) PaysPaysIso NbrProj Zimbabwe ZW 1 Burkina FasoBF2 I know associate Country and Number of projets but how associate Iso Any idea ? Thank you for your help -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] transform 3 numeric vectors empty of 0/1
Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to Vec2 : transform 1 to B and 0 to Vec3 : transform 1 to C and 0 to to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] a title on the map (function gvisGeoChart package googleVis )
Hi I am using the package googleVis and the function gvisGeoChart Is it possible to put a title on the map ? Here is the call of the function : library(googleVis) G1 - gvisGeoChart(PaysProjets, locationvar='Pays', colorvar='NbProj', options=list( region= world, displayMode=regions, height=347*1.5, width= 556*1.5 )) plot(G1) Thank you -- Michel ARNAUD Chargé de mission auprès du DRH DGDRD-Drh - TA 174/04 Av Agropolis 34398 Montpellier cedex 5 tel : 04.67.61.75.38 fax : 04.67.61.57.87 port: 06.47.43.55.31 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
