[R] Correlation

2014-07-27 Thread Nico Met 
Dear all,

I have written the following code for correlation calculations. I want to
create a new matrix  from corData) with correlation more than 0.5 only and
the rest of the columns should be removed. How can I do it?


set.seed(1234)
data<-matrix(rnorm(100),nrow=10)
data[,1]<-100*(data[,2]+data[,5]+data[,9])
corData<-cor(data)

Thanks a lot

Nico

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[R] average and median values for each of the class

2014-04-26 Thread Nico Met 
Dear all,



I have a matrix (dimension, 16 x 12) where  2nd column represents class
 (1,1,1,1,1,2,2,2, etc) information. I want to estimate average  and median
values for each of the class and add this information as a row at end of
the each classes.


for example:

dput(dat)

structure(list(class = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,

3L, 3L, 3L, 4L, 4L, 4L, 5L), name1 = c(2.554923977, 2.371586762,

2.497293431, 2.464827875, 2.981934845, 2.228995664, 2.099640729,

1.900314302, 2.630005966, 2.632590262, 2.581887814, 2.408797563,

2.098761103, 3.070460716, 1.436980716, 1.645121806), name2 = c(1.297412278,

1.104804244, 1.30621114, 1.126009533, 1.466740841, 1.012041118,

0.923466541, 0.840575023, 1.285530176, 1.041909333, 1.194917856,

1.085015826, 1.047492703, 1.587558217, 0.593340012, 0.723630088

), name3 = c(0.587160798, 0.596127884, 0.623760721, 0.549016135,

0.686642084, 0.487523394, 0.458620467, 0.397974913, 0.615928976,

0.546005649, 0.657383069, 0.546613129, 0.476503461, 0.749062102,

0.304160587, 0.29037358), name4 = c(2.833441759, 2.713374426,

2.532626548, 2.409093102, 3.014912721, 2.113507947, 2.017291324,

1.667744912, 2.602560666, 2.31649643, 2.761204809, 2.433963493,

2.229911767, 3.191646399, 1.269919241, 1.387479858), name5 = c(2.172365295,

1.955695471, 2.141072829, 1.975743278, 2.377018372, 1.791300389,

1.669079382, 1.500209628, 2.164401874, 1.830038378, 2.106750025,

1.92888294, 1.707217549, 2.585082653, 1.114841754, 1.315712452

), name6 = c(0.715129844, 0.688186262, 0.70133748, 0.709362008,

0.712145174, 0.563593885, 0.532109761, 0.472197304, 0.690165016,

0.65635473, 0.615835066, 0.64310098, 0.562974891, 0.900622255,

0.408546784, 0.416284408), name7 = c(1.995505133, 1.860095899,

1.843151597, 1.709861774, 2.155993511, 1.506409746, 1.315405587,

1.234544153, 1.96629927, 1.74879757, 1.93994009, 1.660173854,

1.556735295, 2.355723318, 0.866634243, 1.013367677), name8 = c(0.275484997,

0.233856392, 0.294021245, 0.315504347, 0.251906585, 0.250263636,

0.348599173, 0.273806933, 0.32067937, 0.278581115, 0.293726291,

0.308350808, 0.201297444, 0.351927886, 0.204230625, 0.185681471

), name9 = c(2.461066627, 2.210756164, 2.289047888, 2.253988252,

2.668184733, 1.911697836, 1.793443775, 1.560027186, 2.36941155,

1.96191, 2.391501376, 2.002215107, 1.932144233, 2.73705052,

1.15580754, 1.807697999), name10 = c(0.723025351, 0.613147422,

0.805399925, 0.65651577, 0.779389048, 0.54260459, 0.492283542,

0.507969501, 0.749700016, 0.644231327, 0.810319215, 0.620331891,

0.600240557, 0.884775748, 0.40006142, 0.391661912), name11 = c(0.308565619,

0.453808281, 0.363716904, 0.376332596, 0.324998876, 0.361013073,

0.430744786, 0.468818055, 0.166072668, 0.369262627, 0.297666411,

0.256091173, 0.123021464, 0.308188684, 0.646436241, 0.722972632

)), .Names = c("class", "name1", "name2", "name3", "name4", "name5",

"name6", "name7", "name8", "name9", "name10", "name11"), class = "data.frame",
row.names = c("ara1",

"ara2", "ara3", "ara4", "ara5", "ara6", "ara7", "ara8", "ara9",

"ara10", "ara11", "ara12", "ara13", "ara14", "ara15", "ara16"

))


I wrote this:



 avg<-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat["class"],
function(x) mean(x,na.rm=T)) )


med<-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat["class"], function(x)
median(x,na.rm=T)) )


# avg

#  classname1 name2 name3name4name5 name6name7
name#8name9name10name11

#1 1 2.574113 1.2602356 0.6085415 2.700690 2.124379 0.7052322 1.912922
#0.2741547 2.376609 0.7154955 0.3654845

#2 2 2.214739 1.0154032 0.4900119 2.100276 1.781248 0.5645165 1.505665
#0.2983373 1.908645 0.5731394 0.3566621

#3 3 2.541092 1.1072810 0.589 2.503888 1.955224 0.6384303 1.782971
#0.2935527 2.118543 0.6916275 0.3076734

#4 4 2.202068 1.0761303 0.5099087 2.230492 1.802381 0.6240480 1.593031
#0.2524853 1.941667 0.6283592 0.3592155

#5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368
#0.1856815 1.807698 0.3916619 0.7229726

#> med

#  classname1 name2 name3name4name5 name6name7
name#8name9name10name11

#1 1 2.497293 1.2974123 0.5961279 2.713374 2.141073 0.7093620 1.860096
#0.2754850 2.289048 0.7230254 0.3637169

#2 2 2.164318 0.9677538 0.4730719 2.065400 1.730190 0.5478518 1.410908
#0.2972432 1.852571 0.5252870 0.3958789

#3 3 2.581888 1.0850158 0.5466131 2.433963 1.928883 0.6431010 1.748798
#0.2937263 2.002215 0.6442313 0.2976664

#4 4 2.098761 1.0474927 0.4765035 2.229912 1.707218 0.5629749 1.556735
#0.2042306 1.932144 0.6002406 0.3081887

#5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368
#0.1856815 1.807698 0.3916619 0.7229726




But I do not know how can I add this information in the original data?


For example, for class 1, the output will look like this:

dput(res1)

structure(list(class = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), name1 =
c(2.554923977,

2.371586762, 2.497293431, 2.464827875, 2.98193

Re: [R] Formatting with strings

2014-04-23 Thread Nico Met 
Many thanks Jim. It works for my large data set!

Best Regards

Nico


On Wed, Apr 23, 2014 at 1:45 PM, jim holtman  wrote:

> try this:
>
> > x <- readLines(textConnection("---CLUSTER 1 ---
> + 3
> + 4
> + 5
> + 6
> + ---CLUSTER 2 ---
> + 9
> + 10
> + 8
> + 11"))
> >
> > # create a list of where the 'clusters' are
> > clust <- c(grep("CLUSTER", x), length(x) + 1L)
> >
> > # get size of each cluster
> > clustSize <- diff(clust) - 1L
> >
> > # get cluster number
> > clustNum <- gsub("[^0-9]+", "", x[grep("CLUSTER", x)])
> >
> > result <- do.call(rbind, lapply(seq(length(clustNum)), function(.cl){
> + cbind(Object = x[seq(clust[.cl] + 1L, length = clustSize[.cl])]
> + , Cluster = .cl
> + )
> + }))
> >
> > result
>  Object Cluster
> [1,] "3""1"
> [2,] "4""1"
> [3,] "5""1"
> [4,] "6""1"
> [5,] "9""2"
> [6,] "10"   "2"
> [7,] "8""2"
> [8,] "11"   "2"
> >
>
>
>
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.
>
>
> On Wed, Apr 23, 2014 at 5:58 AM, Nico Met  wrote:
>
>> Dear all, I have object where I stored clusters in the following manner:
>>
>> ---CLUSTER 1 ---
>> 3
>> 4
>> 5
>> 6
>> ---CLUSTER 2 ---
>> 9
>> 10
>> 8
>> 11
>>
>> Now want to format the data in the following way:
>>
>> Object  Cluster
>> 31
>> 41
>> 51
>> 61
>> 9 2
>> 10   2
>> 8 2
>> 11   2
>>
>>
>> How can I do this in R?
>>
>> Thanks
>>
>> Nico
>>
>> [[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.
>>
>
>

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and provide commented, minimal, self-contained, reproducible code.

[R] Formatting with strings

2014-04-23 Thread Nico Met 
Dear all, I have object where I stored clusters in the following manner:

---CLUSTER 1 ---
3
4
5
6
---CLUSTER 2 ---
9
10
8
11

Now want to format the data in the following way:

Object  Cluster
31
41
51
61
9 2
10   2
8 2
11   2


How can I do this in R?

Thanks

Nico

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and provide commented, minimal, self-contained, reproducible code.

[R] Column value comparison and colouring

2014-04-06 Thread Nico Met 
Dear all,

I have a big matrix,  where I want to compare, one element of a vector with
many others and clear them.

For example:

In column sil compare elements with other columns (0.734)>-1.0770 in 3rd
column, so, color them bold else color them bold red
For example:

> dput(test)
structure(list(Class = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), sil = c(0.734,
0.734, 0.734, 0.734, 0.734, -0.03, -0.03, -0.03), M201.1637T.1 =
c(-1.077019059,
-1.025923945, -2.025907674, -1.223001413, -1.547348898, -0.376842603,
-1.33495095, -0.486811653), M203.1794T.1 = c(-0.884899777, -0.878305283,
-1.910526051, -1.446826543, -1.950515819, -0.192865129, -1.669691644,
-0.636095942), M205.1950T.1 = c(-1.307163965, -1.022407192, -2.050283307,
-1.508604131, -1.710170605, -0.197410826, -1.547129558, -0.780189201
), M207.2107T.1 = c(-1.378614081, -0.950293267, -1.714837198,
-1.827195011, -1.629677288, -0.064343778, -1.598304259, -0.502002575
)), .Names = c("Class", "sil", "M201.1637T.1", "M203.1794T.1",
"M205.1950T.1", "M207.2107T.1"), class = "data.frame", row.names =
c("52T_D06_19_42",
"500030X_B03_19_15", "500059R_G09_19_81", "500061V_H02_13_86",
"500078W_D10_6_46", "500082M_D05_18_41", "500105S_B09_15_21",
"500120W_B04_19_16")

Thanks in advance

Nico

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] matrix of mean values

2013-10-08 Thread Nico Met 
Dear Arun, thanks for the solution. however, for my data I am getting this
warning message

Warning messages:
1: In mean.default(.value[0], ...) :
  argument is not numeric or logical: returning NA

what can be the cause of it?

Regards

Nico


On Tue, Oct 8, 2013 at 4:56 PM, arun  wrote:

> To make it matrix:
>
>
> row.names(res)<- res[,1]
>  res1<- as.matrix(res[,-1])
> A.K.
>
>
>
> - Original Message -
> From: arun 
> To: Nico Met 
> Cc: R help 
> Sent: Tuesday, October 8, 2013 10:54 AM
> Subject: Re: [R] matrix of mean values
>
> Try:
>
> d1$value<-as.numeric(gsub(",","",as.character(d1$value)))
> library(reshape2)
> res <- dcast(d1,loc~variables,value.var="value",mean)
> A.K.
>
>
>
> - Original Message -
> From: Nico Met 
> To: R help 
> Cc:
> Sent: Tuesday, October 8, 2013 10:47 AM
> Subject: [R] matrix of mean values
>
> Dear all, I have a data set where I want to make a matrix of the mean
> values using 3rd column (value). Where rows are unique “loc” and columns
> would be “variables”. If there is one observation or no observation just
> use “NA”.
>
>
>
> > dput(d1)
>
> structure(list(loc = structure(c(6L, 9L, 9L, 9L, 9L, 4L, 4L,
>
> 4L, 4L, 4L, 4L, 13L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 7L,
>
> 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 10L,
>
> 10L, 10L, 10L, 10L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 12L, 1L,
>
> 3L, 3L, 3L, 2L), .Label = c("AX09_126", "AX09_628", "AX09_924",
>
> "IALA08_3299", "JA08_246", "JA08_3299", "JA08_408", "JE08_246",
>
> "JE08_3299", "JE08_408", "NO08_246", "NO08_408", "PP08_3299"), class =
> "factor"),
>
> variables = structure(c(5L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L,
>
> 7L, 7L, 7L, 7L, 4L, 4L, 4L, 4L, 4L, 2L, 2L, 2L, 2L, 3L, 3L,
>
> 3L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 1L, 5L, 5L, 5L, 6L, 6L, 7L,
>
> 7L, 7L, 7L, 7L, 7L, 7L, 3L, 7L, 4L, 4L, 4L, 7L, 7L, 7L, 7L
>
> ), .Label = c("i", "r", "s", "t", "x", "y", "z"), class = "factor"),
>
> value = structure(c(1L, 21L, 2L, 1L, 30L, 32L, 4L, 32L, 5L,
>
> 20L, 5L, 1L, 24L, 12L, 26L, 24L, 12L, 15L, 26L, 14L, 15L,
>
> 23L, 13L, 11L, 23L, 25L, 23L, 6L, 7L, 3L, 8L, 19L, 3L, 19L,
>
> 28L, 16L, 18L, 18L, 17L, 31L, 29L, 10L, 27L, 22L, 9L, 22L,
>
> 22L, 9L, 9L, 35L, 33L, 33L, 33L, 34L), .Label = c("0", "0,",
>
> "142,1", "1506,6", "1520,4", "170,36", "18,6185",
>
> "19,2933", "20,2125", "20,7423", "225,94", "23,2245", "24,3526",
>
> "24,3685", "24,4593", "242,84", "25,7422", "26,34", "26,5989",
>
> "30,6284", "36,6608", "39,913", "41,4933", "43,9104",
>
> "47,1241", "49,3467", "49,5715", "50,0037", "50,7974", "53,6667",
>
> "56,2778", "76", "76,5", "77,6667", "82"), class = "factor")), .Names =
> c("loc",
>
> "variables", "value"), class = "data.frame", row.names = c(NA,
>
> -54L))
>
>
>
> Thanks
>
>
>
> Nico
>
> [[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.
>
>

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and provide commented, minimal, self-contained, reproducible code.

[R] matrix of mean values

2013-10-08 Thread Nico Met 
Dear all, I have a data set where I want to make a matrix of the mean
values using 3rd column (value). Where rows are unique “loc” and columns
would be “variables”. If there is one observation or no observation just
use “NA”.



> dput(d1)

structure(list(loc = structure(c(6L, 9L, 9L, 9L, 9L, 4L, 4L,

4L, 4L, 4L, 4L, 13L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 7L,

7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 10L,

10L, 10L, 10L, 10L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 12L, 1L,

3L, 3L, 3L, 2L), .Label = c("AX09_126", "AX09_628", "AX09_924",

"IALA08_3299", "JA08_246", "JA08_3299", "JA08_408", "JE08_246",

"JE08_3299", "JE08_408", "NO08_246", "NO08_408", "PP08_3299"), class =
"factor"),

variables = structure(c(5L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L,

7L, 7L, 7L, 7L, 4L, 4L, 4L, 4L, 4L, 2L, 2L, 2L, 2L, 3L, 3L,

3L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 1L, 5L, 5L, 5L, 6L, 6L, 7L,

7L, 7L, 7L, 7L, 7L, 7L, 3L, 7L, 4L, 4L, 4L, 7L, 7L, 7L, 7L

), .Label = c("i", "r", "s", "t", "x", "y", "z"), class = "factor"),

value = structure(c(1L, 21L, 2L, 1L, 30L, 32L, 4L, 32L, 5L,

20L, 5L, 1L, 24L, 12L, 26L, 24L, 12L, 15L, 26L, 14L, 15L,

23L, 13L, 11L, 23L, 25L, 23L, 6L, 7L, 3L, 8L, 19L, 3L, 19L,

28L, 16L, 18L, 18L, 17L, 31L, 29L, 10L, 27L, 22L, 9L, 22L,

22L, 9L, 9L, 35L, 33L, 33L, 33L, 34L), .Label = c("0", "0,",

"142,1", "1506,6", "1520,4", "170,36", "18,6185",

"19,2933", "20,2125", "20,7423", "225,94", "23,2245", "24,3526",

"24,3685", "24,4593", "242,84", "25,7422", "26,34", "26,5989",

"30,6284", "36,6608", "39,913", "41,4933", "43,9104",

"47,1241", "49,3467", "49,5715", "50,0037", "50,7974", "53,6667",

"56,2778", "76", "76,5", "77,6667", "82"), class = "factor")), .Names =
c("loc",

"variables", "value"), class = "data.frame", row.names = c(NA,

-54L))



Thanks



Nico

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[R] t-test across columns

2013-07-15 Thread Nico Met 
Dear all,

I would like to do t-test across two columns "name" with different "cat"
with overall mean ("val").

(Removing if there is a single observation)

And finally, make a matrix with t-value and p-value associated with a name
(in rows) and cat (in columns)

dput(x)
structure(list(name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("12.2", "15.9"
), class = "factor"), cat = structure(c(2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 1L, 1L, 1L, 3L, 1L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 8L, 8L,
8L, 8L, 8L, 8L, 8L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = c("p178266580",
"p178269196", "p178316310", "p191287337", "p195158904", "p196921846",
"p197427158", "p238921966"), class = "factor"), val = c(148.90772,
184.253375, 183.97486667, 191.868125, 173.30515, 187.876975,
177.453775, 184.799525, 212.39065, 205.504525, 186.152025, 194.337075,
193.2703, 204.71665, 211.4452, 202.609175, 203.72918, 193.7261,
196.1186, 202.79556, 203.48818, 191.13744, 205.23315, 198.66842,
196.81032, 200.90512, 206.13564, 205.372225, 196.22835, 211.04686,
219.9771, 224.7602, 231.6596, 211.10581667, 215.44474,
210.83514, 228.173125, 224.09034, 212.96026, 239.0085, 213.5407,
227.12115, 209.24888, 232.8964, 232.22146, 228.1643, 236.43082,
232.20792, 238.49192, 224.64014, 233.75898, 207.06138, 215.3649,
211.14802, 201.86854, 200.52278, 199.05752, 194.90904, 214.44334,
249.35726667, 239.98525, 234.50848333, 243.86508333,
233.59581667, 248.1219, 225.28941667, 248.22088333,
193.69566, 198.43578, 205.06055, 208.525975, 198.28692, 206.88496,
201.60162, 205.7943, 210.5117, 196.69886, 193.58288, 198.86094,
201.81676, 225.8266, 205.879725, 218.370475, 214.006125, 198.74038,
206.00314, 198.37446, 225.5357, 216.721025, 226.543925, 158.1011,
158.15674, 166.07518, 179.942225, 158.16046, 165.0685, 159.56146
)), .Names = c("name", "cat", "val"), class = "data.frame", row.names = c(
NA,
97L))

Thanks

Nico

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[R] Transpose of the rows

2013-06-26 Thread Nico Met 
Dear R users,

I am using a big data matrix and need to transpose rows (formatting of
input matrix). I would like write a general code for this example.

for example: my input file is "dta"

 dput(dta)
structure(list(id = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L), .Label = c("TN00016-003A", "TN00016-014A",
"TN00031-014A"), class = "factor"), name = structure(c(1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L), .Label = c("AreaCycl",
"AreaEmer", "AreaMax", "HeightMax", "nrfilledseed"), class = "factor"),
value = structure(c(3L, 7L, 8L, 14L, 12L, 1L, 2L, 9L, 5L,
13L, 11L, 10L, 4L, 6L), .Label = c("-0,584379415", "-0,842892819",
"-1,399894415", "-1200,824457", "-13,59414667", "-25,86897333",
"-3,277150907", "-3281,076515", "-7939,444248", "0,174160882",
"0,800196212", "1,825", "12,2", "8,43449"), class = "factor")),
.Names = c("id",
"name", "value"), class = "data.frame", row.names = c(NA, -14L
))

And output should look like this:

dput(dta1)
structure(list(id = structure(1:3, .Label = c("TN00016-003A",
"TN00016-014A", "TN00031-014A"), class = "factor"), AreaCycl = structure(c(
2L,
1L, 3L), .Label = c("-0,584379415", "-1,399894415", "0,800196212"
), class = "factor"), AreaEmer = structure(c(2L, 1L, 3L), .Label = c(
"-0,842892819",
"-3,277150907", "0,174160882"), class = "factor"), AreaMax = structure(c(
2L,
3L, 1L), .Label = c("-1200,824457", "-3281,076515", "-7939,444248"
), class = "factor"), HeightMax = structure(c(3L, 1L, 2L), .Label = c(
"-13,59414667",
"-25,86897333", "8,43449"), class = "factor"), nrfilledseed = structure(
c(1L,
2L, NA), .Label = c("1,825", "12,2"), class = "factor")), .Names = c("id",
"AreaCycl", "AreaEmer", "AreaMax", "HeightMax", "nrfilledseed"
), class = "data.frame", row.names = c(NA, -3L))

Thanks for your help.

Nico

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Re: [R] Fetch and merge from a data set

2013-06-25 Thread Nico Met 
Many thanks Rui,

However If I want to extract only first column (data1) from data file, then
how Can I do it?

Thanks again

Nico


On Tue, Jun 25, 2013 at 11:43 AM, Rui Barradas  wrote:

> Hello,
>
> I'm not sure I understand, but it seems as simple as
>
> merge(data1, data)
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 25-06-2013 10:34, Nico Met escreveu:
>
>> Dear all,
>>
>> I would like to fetch a list (data1) of entities from a big data file
>> (data) and merged together. for example: data is the file from where I
>> want
>> to extract
>>
>> dput(data)
>> structure(list(NAME = structure(c(6L, 6L, 7L, 6L, 6L, 7L, 6L,
>> 7L, 3L, 5L, 3L, 3L, 3L, 3L, 4L, 3L, 3L, 4L, 3L, 3L, 3L, 3L, 1L,
>> 1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L), .Label = c("CDS0008_3",
>> "CDS0008_9", "CDS0248_2", "CDS0248_3", "CDS0248_9", "CDS0644",
>> "CDS0645"), class = "factor"), QUAL = structure(c(1L, 1L, 1L,
>> 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L,
>> 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("CE",
>> "IE", "IL", "NL"), class = "factor"), DIFF = c(-1.3998944145561,
>> -3.2771509067793, -3281.0765147, 8.434493332, 1.825,
>> -0.584379415213178, -0.842892819141902, -7939.444248,
>> 0.305343667019102,
>> 11.4859716384148, 1242.7216649, 33.25400, -14.8,
>> 62.8783807080717, 0.3234355, 3325.4720195, 33.8787927988058,
>> -26.772370001, 58.6, 58.6, 0.0566947723248923,
>> 1.294147, -13.319016665, -18.2, -76.008156572, -13.319016665,
>> -18.2, -13.319016665, -18.2, -2632.77274125, -42.6985227297727,
>> -19.2272727272727, -640.535327886362), PVALUE = c(0.393708851005828,
>> 0.213217899579307, 0.59042649939326, 0.874157227673879, 0.953482720079014,
>> 0.737143165148719, 0.76195136190783, 0.27992628190224, 0.737143165148719,
>> 0.76195136190783, 0.27992628190224, 0.672772547835936, 0.86918514824479,
>> 0.86918514824479, 0.794677241773376, 0.67568899695407, 0.0792182671307693,
>> 0.53713122011077, 0.298506678908869, 0.343822055403655, 0.962553162399683,
>> 0.335590623453015, 0.962553162399683, 0.335590623453015,
>> 0.966426100547593,
>> 0.671619043425778, 0.225088848812347, 0.962553162399683,
>> 0.335590623453015,
>> 0.17524845367103, 0.205476355179601, 0.229212899348569, 0.739457737437997
>> ), MEAD = c(61.997380489015, 38.9012158419308, 42541.899878,
>> 644.34279333, 58.8, 65.4391126224113, 45.1617170900398,
>> 37470.374736,
>> 61.6954795437718, 36.4397768557611, 26367.622967, 485.3921,
>> 76, 0, 0.8717201, 28005.1589441667, 46.2203164422305,
>> 713.02971667, 64, 64, 23.0947770774977, 1.456775, 79.2102758175251,
>> 39.5498022382743, 31387.15404, 883.47568, 180, 76.8751996288758,
>> 41.0701371024372, 23960.47775025, 746.3317500025, 104.5, 17488.00655825
>> )), .Names = c("NAME", "QUAL", "DIFF", "PVALUE", "MEAD"), class =
>> "data.frame",
>> row.names = c(NA,
>> 33L))
>>
>>
>> And data1 : subset of entities
>>
>> dput(data1)
>> structure(list(NAME = structure(c(5L, 4L, 2L, 3L, 1L), .Label =
>> c("CDG981",
>> "CDS0248_2", "CDS0248_9", "CDS0644", "CDS0645"), class = "factor"),
>>  VAL = structure(c(1L, 2L, 5L, 3L, 4L), .Label = c("YU1",
>>  "YU2", "YU4", "YU5", "YU7"), class = "factor")), .Names = c("NAME",
>> "VAL"), class = "data.frame", row.names = c(NA, 5L))
>>
>>
>> Please help me how can I do it.
>>
>> Many thanks
>>
>> Nico
>>
>> [[alternative HTML version deleted]]
>>
>> __**
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help>
>> PLEASE do read the posting guide http://www.R-project.org/**
>> posting-guide.html <http://www.R-project.org/posting-guide.html>
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>

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[R] Fetch and merge from a data set

2013-06-25 Thread Nico Met 
Dear all,

I would like to fetch a list (data1) of entities from a big data file
(data) and merged together. for example: data is the file from where I want
to extract

dput(data)
structure(list(NAME = structure(c(6L, 6L, 7L, 6L, 6L, 7L, 6L,
7L, 3L, 5L, 3L, 3L, 3L, 3L, 4L, 3L, 3L, 4L, 3L, 3L, 3L, 3L, 1L,
1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L), .Label = c("CDS0008_3",
"CDS0008_9", "CDS0248_2", "CDS0248_3", "CDS0248_9", "CDS0644",
"CDS0645"), class = "factor"), QUAL = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("CE",
"IE", "IL", "NL"), class = "factor"), DIFF = c(-1.3998944145561,
-3.2771509067793, -3281.0765147, 8.434493332, 1.825,
-0.584379415213178, -0.842892819141902, -7939.444248,
0.305343667019102,
11.4859716384148, 1242.7216649, 33.25400, -14.8,
62.8783807080717, 0.3234355, 3325.4720195, 33.8787927988058,
-26.772370001, 58.6, 58.6, 0.0566947723248923,
1.294147, -13.319016665, -18.2, -76.008156572, -13.319016665,
-18.2, -13.319016665, -18.2, -2632.77274125, -42.6985227297727,
-19.2272727272727, -640.535327886362), PVALUE = c(0.393708851005828,
0.213217899579307, 0.59042649939326, 0.874157227673879, 0.953482720079014,
0.737143165148719, 0.76195136190783, 0.27992628190224, 0.737143165148719,
0.76195136190783, 0.27992628190224, 0.672772547835936, 0.86918514824479,
0.86918514824479, 0.794677241773376, 0.67568899695407, 0.0792182671307693,
0.53713122011077, 0.298506678908869, 0.343822055403655, 0.962553162399683,
0.335590623453015, 0.962553162399683, 0.335590623453015, 0.966426100547593,
0.671619043425778, 0.225088848812347, 0.962553162399683, 0.335590623453015,
0.17524845367103, 0.205476355179601, 0.229212899348569, 0.739457737437997
), MEAD = c(61.997380489015, 38.9012158419308, 42541.899878,
644.34279333, 58.8, 65.4391126224113, 45.1617170900398,
37470.374736,
61.6954795437718, 36.4397768557611, 26367.622967, 485.3921,
76, 0, 0.8717201, 28005.1589441667, 46.2203164422305,
713.02971667, 64, 64, 23.0947770774977, 1.456775, 79.2102758175251,
39.5498022382743, 31387.15404, 883.47568, 180, 76.8751996288758,
41.0701371024372, 23960.47775025, 746.3317500025, 104.5, 17488.00655825
)), .Names = c("NAME", "QUAL", "DIFF", "PVALUE", "MEAD"), class = "data.frame",
row.names = c(NA,
33L))


And data1 : subset of entities

dput(data1)
structure(list(NAME = structure(c(5L, 4L, 2L, 3L, 1L), .Label = c("CDG981",
"CDS0248_2", "CDS0248_9", "CDS0644", "CDS0645"), class = "factor"),
VAL = structure(c(1L, 2L, 5L, 3L, 4L), .Label = c("YU1",
"YU2", "YU4", "YU5", "YU7"), class = "factor")), .Names = c("NAME",
"VAL"), class = "data.frame", row.names = c(NA, 5L))


Please help me how can I do it.

Many thanks

Nico

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Re: [R] split and common variables

2013-06-05 Thread Nico Met 
Dear Dr. David,

Many thanks for your answer.

Now, if I want to see if there are common "name"in those "places" , how can
I do it? So, by common I mean, it might be compared with 2, 3, 4  all
of them. All possible combinations

Many thanks

regards

Nico


On Mon, Jun 3, 2013 at 5:20 PM, David Carlson  wrote:

> It may be easier if you convert the list you provided to a
> data.frame:
>
> > dta.df <- data.frame(place=dta$place, name=dta$name,
> value=dta$value)
> > dta.df
>
> Note that the last line is blank so you probably want to remove that
> and remove the blank factor levels:
>
> > dta.df <- dta.df[-nrow(dta.df),]
> > dta.df$place <- factor(dta.df$place)
> > dta.df$name <- factor(dta.df$name)
> > dta.df
>
> Now you can make a list containing separate data.frames by place
>
> > dta.df.sp <- split(dta.df, dta.df$place)
> > dta.df.sp
>
> -
> David L Carlson
> Associate Professor of Anthropology
> Texas A&M University
> College Station, TX 77840-4352
>
>
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Nico Met
> Sent: Monday, June 3, 2013 8:42 AM
> To: R help
> Subject: [R] split and common variables
>
> Dear all,
>
> I would like to split the data based on the "place" and then would
> like to
> see how many "names" were common in place with corresponding "value"
> .
>
> Please find a demo file.
>
> Thanks for your expert comment
>
> best
>
> Nico
>
> > dput(dta)
> structure(list(place = structure(c(3L, 2L, 5L, 6L, 4L, 3L, 2L,
> 5L, 6L, 4L, 2L, 5L, 6L, 4L, 2L, 5L, 6L, 4L, 2L, 3L, 2L, 5L, 6L,
> 4L, 3L, 2L, 5L, 6L, 4L, 5L, 6L, 4L, 5L, 6L, 1L), .Label = c("",
> "GCKT", "IKLI", "KLOI", "PLRT", "POIV"), class = "factor"), name =
> structure
> (c(2L,
> 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 2L, 3L, 4L, 5L, 6L,
> 7L, 8L, 9L, 10L, 11L, 12L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
> 11L, 12L, 12L, 1L), .Label = c("", "P_CK23", "P_CK24", "P_CK25",
> "P_CK26", "P_CK27", "P_CK28", "P_CK29", "P_CK30", "P_CK31",
> "P_CK32",
> "P_CK33"), class = "factor"), value = c(5.9464, 6.0786, -4.5155,
> 15.0241, -38.1847, -0.0861, 1.2757, -23.9914, 9.5951, -11.128,
> 6.2826, 23.5218, 20.862, 3.1626, 6.242, -20.5348, -14.0126, -13.796,
> 15.3869, -15.7409, -8.1963, -14.4522, 3.2117, -1.2738, 14.3556,
> -12.5337, 20.4308, -3.3227, -34.802, -11.1103, -9.7146, -35.9044,
> -9.4303, -28.1949, NA)), .Names = c("place", "name", "value"), class
> = "data
> .frame", row.names = c(NA,
> 35L))
> >
>
> [[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.
>
>

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[R] split and common variables

2013-06-03 Thread Nico Met 
Dear all,

I would like to split the data based on the "place" and then would like to
see how many "names" were common in place with corresponding "value" .

Please find a demo file.

Thanks for your expert comment

best

Nico

> dput(dta)
structure(list(place = structure(c(3L, 2L, 5L, 6L, 4L, 3L, 2L,
5L, 6L, 4L, 2L, 5L, 6L, 4L, 2L, 5L, 6L, 4L, 2L, 3L, 2L, 5L, 6L,
4L, 3L, 2L, 5L, 6L, 4L, 5L, 6L, 4L, 5L, 6L, 1L), .Label = c("",
"GCKT", "IKLI", "KLOI", "PLRT", "POIV"), class = "factor"), name = structure
(c(2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 2L, 3L, 4L, 5L, 6L,
7L, 8L, 9L, 10L, 11L, 12L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12L, 12L, 1L), .Label = c("", "P_CK23", "P_CK24", "P_CK25",
"P_CK26", "P_CK27", "P_CK28", "P_CK29", "P_CK30", "P_CK31", "P_CK32",
"P_CK33"), class = "factor"), value = c(5.9464, 6.0786, -4.5155,
15.0241, -38.1847, -0.0861, 1.2757, -23.9914, 9.5951, -11.128,
6.2826, 23.5218, 20.862, 3.1626, 6.242, -20.5348, -14.0126, -13.796,
15.3869, -15.7409, -8.1963, -14.4522, 3.2117, -1.2738, 14.3556,
-12.5337, 20.4308, -3.3227, -34.802, -11.1103, -9.7146, -35.9044,
-9.4303, -28.1949, NA)), .Names = c("place", "name", "value"), class = "data
.frame", row.names = c(NA,
35L))
>

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[R] Coversion from yearly to weekly data

2013-03-08 Thread Nico Met 
Dear all, I have a big data matrix and I want to convert those data into
weekly basis which means 7 days needs to be avaraged and aggregate a single
value

> dput(test)
structure(list(locid = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "EC",
class = "factor"),
Day = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
31, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
22, 23, 24, 25, 26, 27, 28, 29, 30, 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
23, 24, 25, 26, 27, 28, 29, 30, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29
), Data = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,

[R] Correlation with p value

2013-02-13 Thread Nico Met 
Dear all,

I have a data (bellow) and I want to make a correlation test with p-value

structure(list(Name = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("CTJ",
"PKR", "TTK"), class = "factor"), score = c(86.4371428571428,
89.7028571428572, 87.728, 89.99, 89.42, 85.6914285714286, 82.256,
83.9942857142857, 86.4371428571429, 84.596, 84.74, 87.8771428571428,
83.12, 81.824, 88.5457142857143, 85.7171428571429, 84.902,
89.8828571428571,
91.5542857142857, 89.42, 86.81)), .Names = c("Name", "score"), class =
"data.frame", row.names = c(NA,
21L))

And finally , I would like to display as matrix where I can fill up with
p-values


 *TTK*
 *CTJ*
 *PKR*
 *TTK*
 **
 **
 **
 *CTJ*
 **
 **
 **
 *PKR*





Many thanks for your help in advance

Nico

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[R] Average calculations

2013-01-23 Thread Nico Met 
Dear all,

I have a matrix with two columns: "Names" and "Values"

In names: there are 4 groups they are, CK113234, CK116296, CK116292 and
CK114042

I want to *sort values* (decreasing order) based on each group and
take average of the *top two numbers* in each of the groups.


> dput(x)
structure(list(Names = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 5L, 5L, 5L, 5L, 2L, 2L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("CK113234", "CK113298",
"CK114042", "CK116292", "CK116296"), class = "factor"), Values =
c(208.3360,
223.29665, 221.63255, 211.29735, 217.75112, 210.97916, 222.67365,
216.0822, 189.83685, 194.1595, 210.66298, 223.63718333, 187.74864,
192.5964, 237.593625, 216.8277, 225.8966, 228.00374, 214.99454,
211.38114, 218.7659, 209.9958, 214.8041, 215.184489473684,
224.2446, 217.559878571429, 232.02229167, 214.02384, 236.393875,
228.73522)), .Names = c("Names", "Values"), class = "data.frame", row.names
= c(NA,
30L))


The out put will look like following:

 Names Values  CK113234 222,98  CK116296 217,14  CK116292 232,79  CK114042
234,2
Your help is appreciated.

Regards

Nico

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Re: [R] xlsx file read in R

2012-12-03 Thread Nico Met 
 Dear all,

this is the code I used to read the .xlsx file but get an error message

library(rJava)
library(XLConnectJars)
 library(XLConnect)
 file<-readWorksheetFromFile("all_in.xlsx", sheet = 1)

*Error: NoClassDefFoundError (Java): Could not initialize class
org.apache.poi.POIXMLDocument*

No idea where it went wrong

Regards

Nico

On Mon, Dec 3, 2012 at 2:07 PM, jim holtman  wrote:

> use the XLConnect package.
>
> On Mon, Dec 3, 2012 at 7:59 AM, Nico Met  wrote:
> > Dear all,
> >
> > How can I read .xlsx files in R
> >
> > Regards Nico
> >
> > [[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.
>
>
>
> --
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.
>

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[R] xlsx file read in R

2012-12-03 Thread Nico Met 
Dear all,

How can I read .xlsx files in R

Regards Nico

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[R] mean value calculation

2012-10-18 Thread Nico Met 
Dear all,

I want to calculate mean values for multiple rows:

structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L), .Label = c("AKT", "CKT"), class = "factor"), val1 = c(2,
3, 2, 2, 2, 5, 3, 8, 2), val2. = c(4, 5, 4, 8, 4, 8, 4, 7, 4),
val3 = c(5, 6, 5, 9, 5, 9, 5, 9, 5)), .Names = c("Name",
"val1", "val2.", "val3"), class = "data.frame", row.names = c(NA,
9L))

For example: as an output I must get

 Name val1 val2  val3  AKT   CKT

I wrote this :


#as.matrix(tapply(dat[,2:4], dat[,1], FUN = mean))

Error in as.matrix(tapply(met[, 2:4], met[, 1], FUN = mean)) :
  error in evaluating the argument 'x' in selecting a method for function
'as.matrix': Error in tapply(met[, 2:4], met[, 1], FUN = mean) :
  arguments must have same length

Thanks for you kind advise


Nico

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[R] How to count combinations

2012-10-08 Thread Nico Met 
Dear all,

Need a help. I would like to count combination of two columns:

structure(list(V1 = structure(c(4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("23",
"55", "67", "Env"), class = "factor"), V2 = structure(c(9L, 7L,
7L, 7L, 2L, 1L, 11L, 11L, 11L, 11L, 5L, 6L, 6L, 3L, 8L, 2L, 1L,
4L, 4L, 10L, 10L, 10L), .Label = c("Ck", "CK", "Dk", "IN", "jk",
"NO", "NW", "PK", "place", "PR", "PT"), class = "factor")), .Names =
c("V1",
"V2"), class = "data.frame", row.names = c(NA, 22L))


For example: how many times 23 appear with NW and then I want to put the
out put as a matrix

  NW Ck PT jK NO DK PK IN PR  23 3  55  67

Many thanks in advance

Nico

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Re: [R] t-test

2012-10-05 Thread Nico Met 
Many many thanks for the input in this context.

Nico

On Thu, Oct 4, 2012 at 4:56 PM, Jose Iparraguirre <
jose.iparragui...@ageuk.org.uk> wrote:

> Hi,
>
> You can also have a look at this paper (no subscription needed):
>
> Herberich, E.; Sikorski, J.; and Hothorn, T. (2010). "A Robust Procedure
> for Comparing Multiple Means under Heteroscedasticity in Unbalanced
> Designs". PLoSONE, Vol. 5, Issue 3, e9788. doi:10.1371/journal.pone.0009788
>
> In this paper, Herberich et al have included a piece of R code to run
> their procedure. Their procedure is a modification of Tukey, and as it says
> in the title of the paper, it can be used regardless of whether the samples
> had different sizes or distributions.
>
> José
>
>
> José Iparraguirre
> Chief Economist
> Age UK
>
> T 020 303 31482
> E jose.iparragui...@ageuk.org.uk
> Twitter @jose.iparraguirre@ageuk
>
>
> Tavis House, 1- 6 Tavistock Square
> London, WC1H 9NB
> www.ageuk.org.uk | ageukblog.org.uk | @ageukcampaigns
>
>
> For a copy of our new Economic Monitor and the full Chief Economist's
> report, visit the Age UK Knowledge Hub
> http://www.ageuk.org.uk/professional-resources-home/knowledge-hub-evidence-statistics/
>
>
> For evidence and statistics on the older population, visit the Age UK
> Knowledge Hub
> http://www.ageuk.org.uk/professional-resources-home/knowledge-hub-evidence-statistics/
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of arun
> Sent: 04 October 2012 13:59
> To: John Kane
> Cc: R help
> Subject: Re: [R] t-test
>
> Hi John,
> You are right.  With more than two groups, the type 1 error rate should be
> a problem.
> A.K.
>
>
>
> - Original Message -
> From: John Kane 
> To: arun 
> Cc:
> Sent: Thursday, October 4, 2012 8:40 AM
> Subject: Re: [R] t-test
>
> My stats are lousy but isnt Nico doing some multiple t-tests when an anova
> with some post hoc comparisons complete with a Tukey or Bonferroni
> correction looks more suitable?
>
> Of course I have no idea of the topic area and maybe he has already done
> thi.
>
> John Kane
> Kingston ON Canada
>
>
> > -Original Message-
> > From: smartpink...@yahoo.com
> > Sent: Thu, 4 Oct 2012 05:31:55 -0700 (PDT)
> > To: nicome...@gmail.com
> > Subject: Re: [R] t-test
> >
> > HI,
> > Try this:
> > sapply(split(dat,dat$Name),function(x) t.test(x[,2],dat[,2])$p.value)
> > #CTK100 CTK103 CTK121
> > #0.86330310 0.32706859 0.02023357
> > A.K.
> >
> >
> >
> > - Original Message -
> > From: Nico Met 
> > To: Rui Barradas 
> > Cc: r-help@r-project.org; r-help 
> > Sent: Thursday, October 4, 2012 6:37 AM
> > Subject: Re: [R] t-test
> >
> > Dear Rui,
> >
> > Many thanks for help.
> >
> > "mean for CTK and all"  = comparison between mean of all groups ( which
> > means second col) vs. each groups like CTK100, CTK121 etc.
> >
> > Regards
> >
> > Nico
> >
> > On Thu, Oct 4, 2012 at 12:28 PM, Rui Barradas 
> > wrote:
> >
> >> Hello,
> >>
> >> I'm not quite sure I understand, but something like this?
> >>
> >> tapply(dat$Score, dat$Name, FUN = mean)
> >> sapply(unique(dat$Name), function(un){
> > > with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})
> >>
> >> My doubt is in what you mean by "mean for CTK and all". The ?t.test
> >> gives
> >> a confidence interval for the difference in the means, so maybe you'll
> >> have
> >> to look there for what you want.
> >>
> >> Hope this helps,
> >>
> >> Rui Barradas
> >> Em 04-10-2012 10:34, Nico Met escreveu:
> >>
> >>> Dear Group,
> >>>
> >>> I want to do a t-test calculation on a large data set.
> >>>
> >>> I am pasting some part of it
> >>>
> >>> structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
> >>> 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> >>> 2L), .Label = c("CTK100", "CTK103", "CTK121"), class = "factor"),
> > >>  Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
> > >>  206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
> > >>  249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
> > >>  196.0453, 232.6055, 225.0783, 196.045

Re: [R] t-test

2012-10-04 Thread Nico Met 
Dear Rui,

Many thanks for help.

"mean for CTK and all"  = comparison between mean of all groups ( which
means second col) vs. each groups like CTK100, CTK121 etc.

Regards

Nico

On Thu, Oct 4, 2012 at 12:28 PM, Rui Barradas  wrote:

> Hello,
>
> I'm not quite sure I understand, but something like this?
>
> tapply(dat$Score, dat$Name, FUN = mean)
> sapply(unique(dat$Name), function(un){
> with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})
>
> My doubt is in what you mean by "mean for CTK and all". The ?t.test gives
> a confidence interval for the difference in the means, so maybe you'll have
> to look there for what you want.
>
> Hope this helps,
>
> Rui Barradas
> Em 04-10-2012 10:34, Nico Met escreveu:
>
>> Dear Group,
>>
>> I want to do a t-test calculation on a large data set.
>>
>> I am pasting some part of it
>>
>> structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
>> 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
>> 2L), .Label = c("CTK100", "CTK103", "CTK121"), class = "factor"),
>>  Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
>>  206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
>>  249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
>>  196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
>>  )), .Names = c("Name", "Score"), class = "data.frame", row.names =
>> c(NA,
>> 24L))
>>
>>
>> I want to compare groups with CTK100 and with all the groups and want to
>> save the p-values and mean for each of that particular group (for example:
>> mean for CTK and all)
>> Similarly, for other groups like that CTK121 etc...
>>
>> Is there any way to automate this process?
>>
>> Thanks for your advice !
>>
>> Nico
>>
>> [[alternative HTML version deleted]]
>>
>> __**
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help>
>> PLEASE do read the posting guide http://www.R-project.org/**
>> posting-guide.html <http://www.R-project.org/posting-guide.html>
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>

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[R] t-test

2012-10-04 Thread Nico Met 
Dear Group,

I want to do a t-test calculation on a large data set.

I am pasting some part of it

structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), .Label = c("CTK100", "CTK103", "CTK121"), class = "factor"),
Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
)), .Names = c("Name", "Score"), class = "data.frame", row.names =
c(NA,
24L))


I want to compare groups with CTK100 and with all the groups and want to
save the p-values and mean for each of that particular group (for example:
mean for CTK and all)
Similarly, for other groups like that CTK121 etc...

Is there any way to automate this process?

Thanks for your advice !

Nico

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Re: [R] Sorting of columns of a matrix

2012-08-29 Thread Nico Met 
Yes,  each column shall have different sequence of row names



N

On Wed, Aug 29, 2012 at 5:21 PM, PIKAL Petr  wrote:

>  Hi
>
> ** **
>
> But in your example there are no row names? How would you structure the
> result? Each column has different order and therefore each column shall
> have different sequence of row names?
>
> ** **
>
> Send also your responses to r-help as somebody could have better answer
> for you.
>
> ** **
>
> Regards
>
> Petr****
>
> ** **
>
> ** **
>
> ** **
>
> *From:* Nico Met [mailto:nicome...@gmail.com]
> *Sent:* Wednesday, August 29, 2012 4:56 PM
> *To:* PIKAL Petr
>
> *Subject:* Re: [R] Sorting of columns of a matrix
>
>  ** **
>
> Thanks a lot. 
>
> ** **
>
> But I also want to attach row names (if available)  associated with each
> column. 
>
> ** **
>
> In that case does your second code works in the same way?
>
> ** **
>
> thanks 
>
> ** **
>
> Animesh 
>
> ** **
>
> ** **
>
> On Wed, Aug 29, 2012 at 4:46 PM, PIKAL Petr 
> wrote:
>
> Hi
>
>
>
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.org] On Behalf Of Nico Met
> > Sent: Wednesday, August 29, 2012 4:25 PM
> > To: Berend Hasselman
> > Cc: r-help@r-project.org
> > Subject: Re: [R] Sorting of columns of a matrix
> >
> > Please find the require info:
> >
> >
> >  set.seed(12345)
> >  X<-matrix(rnorm(5*10),nrow=5)
> >
>
> 
>
>
> >
> >  X[2,5]<-0
> >  X[3,1]<-0
> >  X[5,8]<-0
>
> 
>
> >
> > dput(X)
> >
> > structure(c(0.585528817843856, 0.709466017509524, 0, -
> > 0.453497173462763, 0.605887455840393, -1.81795596770373,
>
> > 0.630098551068391, -0.276184105225216, -0.284159743943371, -
> > 0.919322002474128, -0.116247806352002, 1.81731204370422,
> > 0.370627864257954, 0.520216457554957, -0.750531994502331,
> > 0.816899839520583, -0.886357521243213, -0.331577589942552,
> > 1.12071265166956, 0.298723699267293, 0.779621924555324, 0, -
> > 0.644328429231302, -1.55313740522969, -1.59770951669631,
> > 1.80509751881082, -0.481647363694637, 0.620379801298422,
> > 0.612123492650849, -0.162310976918126, 0.81187317855386,
> > 2.19683354634753, 2.04919033740619, 1.63244563948047,
> > 0.254271192814055, 0.491188279272559, -0.324086578737118, -
> > 1.66205024385863, 1.76773385087297, 0, 1.12851083359018, -
> > 2.38035806139704, -1.06026555215253, 0.937140540182908,
> > 0.854451720330554, 1.46072940310409, -1.4130987778892,
> > 0.567403253424482, 0.583187653435685, -1.30679883346442), .Dim = c(5L,
> > 10L))
> >
>
> I am not sure what do you want to do with it? Remove rows which has 0
>
> X[X==0]<-NA
> X[complete.cases(X),]
>
> remove zeroes and sort values independently in each column?
>
> X[X==0]<-NA
> apply(X, 2, sort, decreasing=TRUE)
>
> results in list as you have various number of values in columns.
>
> Regards
> Petr
>
>
> >
> > On Wed, Aug 29, 2012 at 4:16 PM, Berend Hasselman 
> > wrote:
> >
> > >
> > > On 29-08-2012, at 16:08, Nico Met wrote:
> > >
> > > > Hello john,
> > > >
> > > > thanks for the suggestion. Please find an example:
> > > >
> > > >> X<-matrix(rnorm(5*10),nrow=5)
> > > >
> > > >> dim(X)
> > > > [1]  5 10
> > > >> X
> > > >   [,1]   [,2][,3]   [,4]   [,5]
> > [,6]
> > > >   [,7]   [,8]   [,9]  [,10]
> > > > [1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799
> > > > -2.2994684
> > > > -0.28823228  0.1929301  0.1187168 -1.2078372 [2,]  0.5969187
> > > > -1.2571336  0.30292368 -0.2098139 -0.9689961  1.7105109
> > > > 0.77240398  0.1482712 -1.2486152  0.2359723 [3,] -0.9669289
> > > > -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
> > > > -0.60517647  0.5317435 -0.2556417 -0.4838956 [4,] -1.9345400
> > > > 0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
> > > > 0.49423641  0.4015327  2.8596603  0.6618432 [5,] -1.1969226
> > > > -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
> > > > -0.09497448 -0.3986624  0.1327156 -0.8935923
> > > >
> > > >> X[2,5]<-0
> > > >> X[3,1]<-0
> > > >> X[6,8]<-0
> &

Re: [R] Sorting of columns of a matrix

2012-08-29 Thread Nico Met 
Please find the require info:


 set.seed(12345)
 X<-matrix(rnorm(5*10),nrow=5)

 dim(X)

 X
   [,1]   [,2][,3]   [,4]   [,5]   [,6]
   [,7]   [,8]   [,9]  [,10]
[1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799 -2.2994684
-0.28823228  0.1929301  0.1187168 -1.2078372
[2,]  0.5969187 -1.2571336  0.30292368 -0.2098139 -0.9689961  1.7105109
 0.77240398  0.1482712 -1.2486152  0.2359723
[3,] -0.9669289 -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
-0.60517647  0.5317435 -0.2556417 -0.4838956
[4,] -1.9345400  0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
 0.49423641  0.4015327  2.8596603  0.6618432
[5,] -1.1969226 -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
-0.09497448 -0.3986624  0.1327156 -0.8935923

 X[2,5]<-0
 X[3,1]<-0
 X[5,8]<-0
 X
   [,1]   [,2][,3]   [,4]   [,5]   [,6]
   [,7]  [,8]   [,9]  [,10]
[1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799 -2.2994684
-0.28823228 0.1929301  0.1187168 -1.2078372
[2,]  0.5969187 -1.2571336  0.30292368 -0.2098139  0.000  1.7105109
 0.77240398 0.1482712 -1.2486152  0.2359723
[3,]  0.000 -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
-0.60517647 0.5317435 -0.2556417 -0.4838956
[4,] -1.9345400  0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
 0.49423641 0.4015327  2.8596603  0.6618432
[5,] -1.1969226 -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
-0.09497448 0.000  0.1327156 -0.8935923

dput(X)

structure(c(0.585528817843856, 0.709466017509524, 0, -0.453497173462763,
0.605887455840393, -1.81795596770373, 0.630098551068391,
-0.276184105225216,
-0.284159743943371, -0.919322002474128, -0.116247806352002,
1.81731204370422,
0.370627864257954, 0.520216457554957, -0.750531994502331,
0.816899839520583,
-0.886357521243213, -0.331577589942552, 1.12071265166956,
0.298723699267293,
0.779621924555324, 0, -0.644328429231302, -1.55313740522969,
-1.59770951669631, 1.80509751881082, -0.481647363694637, 0.620379801298422,
0.612123492650849, -0.162310976918126, 0.81187317855386, 2.19683354634753,
2.04919033740619, 1.63244563948047, 0.254271192814055, 0.491188279272559,
-0.324086578737118, -1.66205024385863, 1.76773385087297, 0,
1.12851083359018,
-2.38035806139704, -1.06026555215253, 0.937140540182908, 0.854451720330554,
1.46072940310409, -1.4130987778892, 0.567403253424482, 0.583187653435685,
-1.30679883346442), .Dim = c(5L, 10L))


On Wed, Aug 29, 2012 at 4:16 PM, Berend Hasselman  wrote:

>
> On 29-08-2012, at 16:08, Nico Met wrote:
>
> > Hello john,
> >
> > thanks for the suggestion. Please find an example:
> >
> >> X<-matrix(rnorm(5*10),nrow=5)
> >
> >> dim(X)
> > [1]  5 10
> >> X
> >   [,1]   [,2][,3]   [,4]   [,5]   [,6]
> >   [,7]   [,8]   [,9]  [,10]
> > [1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799 -2.2994684
> > -0.28823228  0.1929301  0.1187168 -1.2078372
> > [2,]  0.5969187 -1.2571336  0.30292368 -0.2098139 -0.9689961  1.7105109
> > 0.77240398  0.1482712 -1.2486152  0.2359723
> > [3,] -0.9669289 -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
> > -0.60517647  0.5317435 -0.2556417 -0.4838956
> > [4,] -1.9345400  0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
> > 0.49423641  0.4015327  2.8596603  0.6618432
> > [5,] -1.1969226 -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
> > -0.09497448 -0.3986624  0.1327156 -0.8935923
> >
> >> X[2,5]<-0
> >> X[3,1]<-0
> >> X[6,8]<-0
> >> X[5,8]<-0
> >> X
> >   [,1]   [,2][,3]   [,4]   [,5]   [,6]
> >   [,7]  [,8]   [,9]  [,10]
> > [1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799 -2.2994684
> > -0.28823228 0.1929301  0.1187168 -1.2078372
> > [2,]  0.5969187 -1.2571336  0.30292368 -0.2098139  0.000  1.7105109
> > 0.77240398 0.1482712 -1.2486152  0.2359723
> > [3,]  0.000 -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
> > -0.60517647 0.5317435 -0.2556417 -0.4838956
> > [4,] -1.9345400  0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
> > 0.49423641 0.4015327  2.8596603  0.6618432
> > [5,] -1.1969226 -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
> > -0.09497448 0.000  0.1327156 -0.8935923
>
> This is just as unreadable as your first try.
> Please do as suggested
>
> dput(X)
>
> and also include a set.seed() before generating the matrix to make it
> reproducible.
>
> Berend
>
>

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Re: [R] Sorting of columns of a matrix

2012-08-29 Thread Nico Met 
Hello john,

thanks for the suggestion. Please find an example:

> X<-matrix(rnorm(5*10),nrow=5)

> dim(X)
[1]  5 10
> X
   [,1]   [,2][,3]   [,4]   [,5]   [,6]
   [,7]   [,8]   [,9]  [,10]
[1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799 -2.2994684
-0.28823228  0.1929301  0.1187168 -1.2078372
[2,]  0.5969187 -1.2571336  0.30292368 -0.2098139 -0.9689961  1.7105109
 0.77240398  0.1482712 -1.2486152  0.2359723
[3,] -0.9669289 -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
-0.60517647  0.5317435 -0.2556417 -0.4838956
[4,] -1.9345400  0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
 0.49423641  0.4015327  2.8596603  0.6618432
[5,] -1.1969226 -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
-0.09497448 -0.3986624  0.1327156 -0.8935923

> X[2,5]<-0
> X[3,1]<-0
> X[6,8]<-0
> X[5,8]<-0
> X
   [,1]   [,2][,3]   [,4]   [,5]   [,6]
   [,7]  [,8]   [,9]  [,10]
[1,]  1.2774431 -1.2427735  0.81933548 -2.1098586 -1.6726799 -2.2994684
-0.28823228 0.1929301  0.1187168 -1.2078372
[2,]  0.5969187 -1.2571336  0.30292368 -0.2098139  0.000  1.7105109
 0.77240398 0.1482712 -1.2486152  0.2359723
[3,]  0.000 -0.2315229  1.04611417  0.7902772  0.5791674  0.6296305
-0.60517647 0.5317435 -0.2556417 -0.4838956
[4,] -1.9345400  0.2898647  0.08983352  0.1482130  0.8229054  0.7778114
 0.49423641 0.4015327  2.8596603  0.6618432
[5,] -1.1969226 -1.3991657 -0.76507258  0.5024922  0.2633307  0.0236774
-0.09497448 0.000  0.1327156 -0.8935923



Now I want to sort and remove "zero" based on rows.

for example for column 1:

[1,]1.2774431

[2,]0.5969187

[5,]-1.1969226

[4,]-1.9345400


Regards

Nico

On Wed, Aug 29, 2012 at 3:29 PM, John Kane  wrote:

> Perhaps you could supply the matrix using dput() ?  It is a handy way to
> supply sample data.  Just do dput(mydata), copy the results and paste into
> email.
>
> At the moment your matrix is almost unreadable.
>
> John Kane
> Kingston ON Canada
>
>
> > -Original Message-
> > From: nicome...@gmail.com
> > Sent: Wed, 29 Aug 2012 15:24:17 +0200
> > To: r-help@r-project.org
> > Subject: [R] Sorting of columns of a matrix
> >
> > Dear all,
> >
> > Please suggest me how can I do it.
> >
> >
> > I have a matrix which look like following:
> >
> >   x1 x2 x3  t1 .01 0.3 0  t2 0 0.1 0.01  t3 0 .01 .01  t4 0 0  t5 5 0 0
> > t6
> > 0 0 0  t7 0 0 0  t8 0 0 0  t9 0.6 0 0  t10 0 0 0.66  t11 0 0.6 0.11  t12
> > 0
> > 0.4 0
> >
> > I want to sort decreasing order in each column based on rows. and then to
> > display only those rows which has a value. The expected out put matrix
> > will
> > look like
> >
> >   x1 x2 x3  t9 0.6 t11 0.6 t10 0.66  t1 .01 t12 0.4 t11 0.11   t1 0.3 t2
> > .01
> > t2 0.1t3 .01
> > many thanks
> >
> > Nico
> >
> >   [[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.
>
> 
> FREE ONLINE PHOTOSHARING - Share your photos online with your friends and
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[R] Sorting of columns of a matrix

2012-08-29 Thread Nico Met 
Dear all,

Please suggest me how can I do it.


I have a matrix which look like following:

  x1 x2 x3  t1 .01 0.3 0  t2 0 0.1 0.01  t3 0 .01 .01  t4 0 0  t5 5 0 0  t6
0 0 0  t7 0 0 0  t8 0 0 0  t9 0.6 0 0  t10 0 0 0.66  t11 0 0.6 0.11  t12 0
0.4 0

I want to sort decreasing order in each column based on rows. and then to
display only those rows which has a value. The expected out put matrix will
look like

  x1 x2 x3  t9 0.6 t11 0.6 t10 0.66  t1 .01 t12 0.4 t11 0.11   t1 0.3 t2 .01
t2 0.1t3 .01
many thanks

Nico

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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.

[R] elasticnet iteration problem

2012-08-21 Thread Nico Met 
Dear all,

I am trying to iterate an elastic net regression method on a matrix

1) Each column of the matrix ( nxp) will act as a response (y) and rest of
the variables (columns) which means p-1 will act as predictor set

2) i want to store selected variables as matrix

I wrote a code. Please help me to automate this procedure

# install packages for analysis

 rm(list = ls())

 library(caret)

 library(glmnet)

 X<-matrix(rnorm(100*500),nrow=100)


 y<-X[,1]  # I want automatically it will take next column as a response
means untill 500 columns

 X1<-X[,-1] # If I use first column as a response it should delete first
column from the matrix

 dim(X1)


 ### Applicarion of the Elastic net for selecting the genes


con<-trainControl(method="cv",number=10)


fit_data<-train(X1,
y,method="glmnet",metric="RMSE",trControl=con,tuneLength = 10)

glmnetcalc<-glmnet(X1,y,alpha=fit_data$finalModel$tuneValue$.alpha)

fitcoef<-predict(glmnetcalc,s=fit_data$finalModel$tuneValue$.alpha,type="coefficients")

CoefEN_1<-as.matrix(fitcoef)

write.table(CoefEN_1, "varSelect_1.txt")  # I want to store the non zero
values in a matrix



thanks

Nico

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