Re: [R] Query about calculating the monthly average of daily data columns

[Jim Lemon]

Hi Subhamitra,
This is not the only way to do this, but if you only want the monthly
averages, it is simple:

# I had to change the "soft" tabs in your email to commas
# in order to read the data in
spdf<-read.table(text="PERMNO,DATE,Spread
111,19940103,0.025464308
111,19940104,0.064424296
111,19940105,0.018579337
111,19940106,0.018872211
111,19940107,0.065279782
111,19940110,0.063485905
111,19940111,0.018355453
111,19940112,0.064135683
111,19940113,0.063519987
111,19940114,0.018277351
111,19940117,0.018628417
111,19940118,0.065630229
111,19940119,0.018713152
111,19940120,0.019119037
111,19940121,0.068342043
111,19940124,0.020843244
111,19940125,0.019954211
111,19940126,0.018980321
111,19940127,0.066827165
111,19940128,0.067459235
111,19940131,0.068682559
111,19940201,0.02081465
111,19940202,0.068236091
111,19940203,0.068821406
111,19940204,0.020075648
111,19940207,0.066070584
111,19940208,0.066068837
111,19940209,0.019077072
111,19940210,0.065894875
111,19940211,0.018847478
111,19940214,0.065040844
111,19940215,0.01880332
111,19940216,0.018836199
111,19940217,0.06665
111,19940218,0.067116793
111,19940221,0.068809742
111,19940222,0.068230213
111,19940223,0.069502855
111,19940224,0.070383523
111,19940225,0.020430811
111,19940228,0.067087257
111,19940301,0.066776479
111,19940302,0.019959031
111,19940303,0.066596469
111,19940304,0.019131334
111,19940307,0.019312528
111,19940308,0.067349909
111,19940309,0.068916431
111,19940310,0.068620043
111,19940311,0.070494844
111,19940314,0.071056842
111,19940315,0.071042517
111,19940316,0.072401771
111,19940317,0.071940001
111,19940318,0.07352884
111,19940321,0.072671688
111,19940322,0.072652595
111,19940323,0.021352138
111,19940324,0.069933727
111,19940325,0.068717467
111,19940328,0.020470748
111,19940329,0.020003748
111,19940330,0.065833717
111,19940331,0.065268388
111,19940401,0.018762356
111,19940404,0.064914179
111,19940405,0.064706743
111,19940406,0.018764175
111,19940407,0.06524806
111,19940408,0.018593449
111,19940411,0.064913949
111,19940412,0.01872089
111,19940413,0.018729328
111,19940414,0.018978773
111,19940415,0.065477137
111,19940418,0.064614365
111,19940419,0.064184148
111,19940420,0.018553192
111,19940421,0.066872771
111,19940422,0.06680782
111,19940425,0.067467961
111,19940426,0.02014297
111,19940427,0.062464016
111,19940428,0.062357052
112,19940429,0.000233993
112,19940103,0.000815264
112,19940104,0.000238165
112,19940105,0.000813632
112,19940106,0.000236915
112,19940107,0.000809102
112,19940110,0.000801642
112,19940111,0.000797932
112,19940112,0.000795251
112,19940113,0.000795186
112,19940114,0.000231359
112,19940117,0.000232134
112,19940118,0.000233718
112,19940119,0.000233993
112,19940120,0.000234694
112,19940121,0.000235753
112,19940124,0.000808653
112,19940125,0.000235604
112,19940126,0.000805068
112,19940127,0.000802337
112,19940128,0.000801768
112,19940131,0.000233517
112,19940201,0.000797431
112,19940202,0.00028
112,19940203,0.000233826
112,19940204,0.000799519
112,19940207,0.000798105
112,19940208,0.000792245
112,19940209,0.000231113
112,19940210,0.000233413
112,19940211,0.000798168
112,19940214,0.000233282
112,19940215,0.000797848
112,19940216,0.000785165
112,19940217,0.000228426
112,19940218,0.000786783
112,19940221,0.00078343
112,19940222,0.000781459
112,19940223,0.000776264
112,19940224,0.000226399
112,19940225,0.000779066
112,19940228,0.000773603
112,19940301,0.000226487
112,19940302,0.000775233
112,19940303,0.000227017
112,19940304,0.000227854
112,19940307,0.000782814
112,19940308,0.000229164
112,19940309,0.000787033
112,19940310,0.000784049
112,19940311,0.000228984
112,19940314,0.00078697
112,19940315,0.000782567
112,19940316,0.000228516
112,19940317,0.000786347
112,19940318,0.000229236
112,19940321,0.000230107
112,19940322,0.000792689
112,19940323,0.000787284
112,19940324,0.000787221
112,19940325,0.000227978",
header=TRUE,sep=",",stringsAsFactors=FALSE)
# split the year and month out of the date string
# as you have more than one year in your complete
# data set
spdf$yrmon<-substr(spdf$DATE,1,6)
# get the mean for each PERMNO and year/month
by(spdf$Spread,spdf[,c("PERMNO","yrmon")],mean)

Jim

On Sun, Oct 20, 2019 at 11:09 PM Subhamitra Patra <
subhamitra.pa...@gmail.com> wrote:

>
> Here, I am asking one more query (just for learning purpose) that if my
> country name and its respective variable is in the panel format, and I want
> to take the monthly average for each country, how the code will be
> arranged. For your convenience, I am providing a small data sample below.
>
>
>

[[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] Query about calculating the monthly average of daily data columns

[Rui Barradas]

42863, 10.3509617168731,
9.09646558899397, 11.270647314, 11.3984335011704, 11.4808985388742,
10.5608771133999, 10.3684356806175, 10.4815588822618, 10.5818867877558,
12.2561035284691, 8.6464271477849, 10.3412351841865, 10.7577574534162,
11.1124067479261, 9.91627943243343, 10.6356898895291, 10.2107566441478,
10.0672734202575, 10.2385787014999, 11.7112606160069, 10.0453801263575,
8.84654136100724, 10.2173421609193, 9.27919801705716, 10.4755578829547,
7.69340209082122, 9.24705253848083, 10.8415406794597, 8.69603117680965,
11.2589214416702, 10.5425642239737, 10.1389355042458, 9.17267675180435,
12.3052338002213, 10.0181674985326, 12.2715476751051, 9.64516268052739,
10.6305299379912, 10.1829347684655, 9.97983942366781, 10.2559847744715,
10.309221814, 9.75215330673072, 10.250464278709, 9.31118800061454,
10.3310666767171, 9.09703848990093, 10.241195105962, 8.57290406448477,
8.98090855172704, 8.64653101832229, 12.6791587435376, 9.56000538681993,
10.4062255533723, 11.067091476284, 10.5255014737268, 10.2240941949978,
9.13081571869084, 9.5942352120783, 9.2753466212409, 10.2789293993548,
8.10255065585342, 9.48751297655077, 8.51198576785003, 9.46310532206947,
9.86727270762806, 11.5149248124739, 9.31557156735022, 9.34351230206303,
10.022139448869, 11.4111350893792, 8.57891783464065, 10.3761090924661,
9.38300408584683, 9.33694577526158, 9.2581686085, 9.29856853889735,
8.4250073823245, 8.83022950824832, 9.1510846172981, 10.2553042376765,
10.0739540955956, 9.04955917463259, 10.8927827168631, 9.44611041690694,
10.7883395708593, 10.6010088332078, 7.72560864006592, 10.1760839916637,
11.5576569894392, 11.384809257294, 8.73504353987083, 9.00585942714512,
9.62327893504013, 10.3527072699866, 10.5220100705827, 8.74921668696853,
8.56415116683662, 12.1348451793815, 10.9496674323819, 9.64443817181322,
9.52977454697087, 10.4281877186725, 8.52701721410292, 11.6911584965782,
10.2300108250139, 8.65368821276485, 11.7733431942379, 10.2060233777681,
9.57291673029552, 9.82687667895106, 10.5939736188493, 11.2510605726337,
10.3383384488323, 9.92301237292945, 10.0164623230529, 10.4939857044034,
10.5631769648289, 10.935731043532, 11.0659359187168, 8.51697010486427,
9.79512310587405, 9.35132038807071, 11.3286703149903, 10.4621597293933,
10.4099459919071, 8.86246315190942, 9.30054044639769, 9.40346575227191,
9.59278722974697)), row.names = c(NA, -260L), class = "data.frame")





From: Subhamitra Patra <mailto:subhamitra.pa...@gmail.com>
Sent: Friday, September 13, 2019 3:59 PM
To: PIKAL Petr <mailto:petr.pi...@precheza.cz>; r-help mailing list
<mailto:r-help@r-project.org>
Subject: Re: [R] Query about calculating the monthly average of daily

data

columns

Dear PIKAL,

Thank you very much for your suggestion.

I tried your previous suggested code and getting the average value for
each month for both country A, and B. But in your recent email, you are
suggesting not to change the date column to real date. If I am going
through your recently suggested code, i.e.

  "aggregate(value column, list(format(date column, "%m.%Y"), country
column), mean)"

I am getting an Error that "aggregate(value, list(format(date, "%m.%Y"),
country), mean) : object 'value' not found".

Here, my query "may I need to define the date column, country column, and
value column separately?"

Further, I need something the average value result like below in the data
frame

Month   Country A   Country B
Jan 199426.66 35.78
Feb 199426.13 29.14

so that it will be easy for me to export to excel, and to use for the
further calculations.

Please suggest me in this regard.

Thank you.








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Sender notified by



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09/13/19, 07:22:53 PM



On Fri, Sep 13, 2019 at 7:03 PM PIKAL Petr 
petr.pi...@precheza.cz>

wrote:
Hi

I am almost 100% sure that you would spare yourself much trouble if you
changed your date column to real date

?as.Date

reshape your wide format to long one
library(reshape2)
?melt

to get 3 column data.frame with one date column, one country column and
one value column

use ?aggregate and ?format to get summary value

something like
aggregate(value column, list(format(date column, "%m.%Y"), country
column), mean)

But if you insist to scratch your left ear with right hand accross your
head, you could continue your way.

Cheers
Petr


-Original Message-
From: R-help <mailto:r-help-boun...@r-project.org> On Behalf Of

Subhamitra

Patra
Sent: Friday, September 13, 2019 3:20 PM
To: Jim Lemon <mailto:drjimle...@gmail.com>; r-help mailing list


http://project.org>
Subject: Re: [R] Query about calculating the monthly average of daily

data

columns

Dear Sir,

Yes, I understood the logic. But, still, I have a few queries that I

mentioned

below your answers.

"# if

Re: [R] Query about calculating the monthly average of daily data columns

[jim holtman]

8,
> > 1.3130254639318,
> > 0.191890764940071, -0.0493619237876962, -0.55183232511689,
> > 0.470263932874487,
> > -0.217088645692971, 0.231550037620628, -0.530406537266415,
> > -0.616522469083808,
> > 0.329347084038265, 1.49420692610475, 1.91750823142859, 0.753536143872474,
> > 0.766584887163714, -0.259803384094296, -0.402463714097741,
> > -0.0229799209735185,
> > -0.259677990559218, -1.41529707261105, 0.191362852138627,
> > 1.54483266684747,
> > -1.17947655378489, -0.426265411073274, 0.723010460481118,
> > 1.37405142869537,
> > -0.374771207936141, 0.0513905365832423, -0.369432731236118,
> > -0.945441984794364,
> > 0.17950664824, 0.31971255725438, -1.25117960937319, 2.46228549580083
> > ), countryB = c(9.4028512714591, 10.7551115504652, 11.2436629228434,
> > 8.25642360904389, 10.7054313972395, 10.1714609666091, 10.3726975056206,
> > 10.6594299429162, 8.56250595443296, 10.5612153841067, 8.07612112826519,
> > 9.94704207511951, 11.392407273156, 10.4257690445268, 10.6339442533038,
> > 10.5314883825356, 10.3506665399033, 10.2426403362978, 10.8437715647564,
> > 10.8247200587034, 11.2449815064171, 9.2898697883112, 9.05418978124619,
> > 10.6080277672463, 9.19882175737148, 11.3589722806948, 10.4139334238189,
> > 9.44305216810892, 9.58426470056472, 8.84208362003176, 10.8125431356391,
> > 7.71357872650814, 8.73526671289828, 10.714693958224, 9.49976972594189,
> > 9.41896864941478, 7.33073349261249, 10.5268398982262, 9.92255671125184,
> > 10.5665378092202, 10.5035704895405, 7.93682068228084, 10.882421050424,
> > 9.3237610577468, 8.42128120513304, 9.46103753451174, 10.3472215515392,
> > 11.0483414005193, 10.3421689244599, 7.85120280889754, 11.6327644046254,
> > 9.57620745972098, 10.6553844719749, 10.8490250129346, 10.2742492933876,
> > 9.55428072119304, 9.0976351049804, 10.0731951766966, 10.6956488509516,
> > 11.1530744146062, 10.3496303024767, 9.52734980693306, 9.64478424894734,
> > 9.28301632295047, 10.9568790570688, 11.6052870914912, 9.58530202776537,
> > 11.1338134902295, 8.66685735149472, 11.0230863576875, 10.8000609212302,
> > 10.6510296259782, 11.831292326569, 9.53836122448974, 9.55805411377422,
> > 9.90336204189518, 9.36377040999133, 11.7041009433341, 9.95628297574831,
> > 10.718111342931, 10.4562688422472, 8.85976383099186, 8.94085496683824,
> > 8.19538394018434, 10.1058448260449, 9.70821289789561, 9.08757962570738,
> > 10.657541876, 10.0521137258219, 9.9924295699559, 11.8730269098299,
> > 11.2634470795907, 11.3854762443416, 9.56742053529845, 10.4101561978503,
> > 9.53376547865009, 9.75410966995361, 9.92804558924886, 8.36231430067066,
> > 10.7486459346681, 12.0143881312685, 11.0083060332839, 9.32820954213586,
> > 10.8420346742049, 9.73064414798223, 10.7593902723319, 10.976622155215,
> > 10.1039774975157, 8.36317871802524, 9.21809894958653, 10.1015362220683,
> > 11.4655736295123, 9.65528297274543, 9.67844310028008, 10.1516820910267,
> > 8.38764450852642, 10.163558398201, 11.1432463477237, 12.0509818193223,
> > 10.9896913965091, 11.1772406550953, 9.14396687337779, 9.93338627749979,
> > 10.9548864433126, 8.64911301751956, 11.706463972364, 11.1012846649741,
> > 8.7805267197408, 11.5802098773954, 10.2268513542863, 10.3509617168731,
> > 9.09646558899397, 11.270647314, 11.3984335011704, 11.4808985388742,
> > 10.5608771133999, 10.3684356806175, 10.4815588822618, 10.5818867877558,
> > 12.2561035284691, 8.6464271477849, 10.3412351841865, 10.7577574534162,
> > 11.1124067479261, 9.91627943243343, 10.6356898895291, 10.2107566441478,
> > 10.0672734202575, 10.2385787014999, 11.7112606160069, 10.0453801263575,
> > 8.84654136100724, 10.2173421609193, 9.27919801705716, 10.4755578829547,
> > 7.69340209082122, 9.24705253848083, 10.8415406794597, 8.69603117680965,
> > 11.2589214416702, 10.5425642239737, 10.1389355042458, 9.17267675180435,
> > 12.3052338002213, 10.0181674985326, 12.2715476751051, 9.64516268052739,
> > 10.6305299379912, 10.1829347684655, 9.97983942366781, 10.2559847744715,
> > 10.309221814, 9.75215330673072, 10.250464278709, 9.31118800061454,
> > 10.3310666767171, 9.09703848990093, 10.241195105962, 8.57290406448477,
> > 8.98090855172704, 8.64653101832229, 12.6791587435376, 9.56000538681993,
> > 10.4062255533723, 11.067091476284, 10.5255014737268, 10.2240941949978,
> > 9.13081571869084, 9.5942352120783, 9.2753466212409, 10.2789293993548,
> > 8.10255065585342, 9.48751297655077, 8.51198576785003, 9.46310532206947,
> > 9.86727270762806, 11.5149248124739, 9.31557156735022, 9.34351230206303,
> > 10.022139448869, 11.4111350893792, 8.57891783464065, 10.3761090924661,
> > 9.38300408584683, 9.33694577526158, 

Re: [R] Query about calculating the monthly average of daily data columns

[Subhamitra Patra]

9261, 9.91627943243343, 10.6356898895291, 10.2107566441478,
> 10.0672734202575, 10.2385787014999, 11.7112606160069, 10.0453801263575,
> 8.84654136100724, 10.2173421609193, 9.27919801705716, 10.4755578829547,
> 7.69340209082122, 9.24705253848083, 10.8415406794597, 8.69603117680965,
> 11.2589214416702, 10.5425642239737, 10.1389355042458, 9.17267675180435,
> 12.3052338002213, 10.0181674985326, 12.2715476751051, 9.64516268052739,
> 10.6305299379912, 10.1829347684655, 9.97983942366781, 10.2559847744715,
> 10.309221814, 9.75215330673072, 10.250464278709, 9.31118800061454,
> 10.3310666767171, 9.09703848990093, 10.241195105962, 8.57290406448477,
> 8.98090855172704, 8.64653101832229, 12.6791587435376, 9.56000538681993,
> 10.4062255533723, 11.067091476284, 10.5255014737268, 10.2240941949978,
> 9.13081571869084, 9.5942352120783, 9.2753466212409, 10.2789293993548,
> 8.10255065585342, 9.48751297655077, 8.51198576785003, 9.46310532206947,
> 9.86727270762806, 11.5149248124739, 9.31557156735022, 9.34351230206303,
> 10.022139448869, 11.4111350893792, 8.57891783464065, 10.3761090924661,
> 9.38300408584683, 9.33694577526158, 9.2581686085, 9.29856853889735,
> 8.4250073823245, 8.83022950824832, 9.1510846172981, 10.2553042376765,
> 10.0739540955956, 9.04955917463259, 10.8927827168631, 9.44611041690694,
> 10.7883395708593, 10.6010088332078, 7.72560864006592, 10.1760839916637,
> 11.5576569894392, 11.384809257294, 8.73504353987083, 9.00585942714512,
> 9.62327893504013, 10.3527072699866, 10.5220100705827, 8.74921668696853,
> 8.56415116683662, 12.1348451793815, 10.9496674323819, 9.64443817181322,
> 9.52977454697087, 10.4281877186725, 8.52701721410292, 11.6911584965782,
> 10.2300108250139, 8.65368821276485, 11.7733431942379, 10.2060233777681,
> 9.57291673029552, 9.82687667895106, 10.5939736188493, 11.2510605726337,
> 10.3383384488323, 9.92301237292945, 10.0164623230529, 10.4939857044034,
> 10.5631769648289, 10.935731043532, 11.0659359187168, 8.51697010486427,
> 9.79512310587405, 9.35132038807071, 11.3286703149903, 10.4621597293933,
> 10.4099459919071, 8.86246315190942, 9.30054044639769, 9.40346575227191,
> 9.59278722974697)), row.names = c(NA, -260L), class = "data.frame")
>
>
>
>
>
> From: Subhamitra Patra <mailto:subhamitra.pa...@gmail.com>
> Sent: Friday, September 13, 2019 3:59 PM
> To: PIKAL Petr <mailto:petr.pi...@precheza.cz>; r-help mailing list
> <mailto:r-help@r-project.org>
> Subject: Re: [R] Query about calculating the monthly average of daily data
> columns
>
> Dear PIKAL,
>
> Thank you very much for your suggestion.
>
> I tried your previous suggested code and getting the average value for
> each month for both country A, and B. But in your recent email, you are
> suggesting not to change the date column to real date. If I am going
> through your recently suggested code, i.e.
>
>  "aggregate(value column, list(format(date column, "%m.%Y"), country
> column), mean)"
>
> I am getting an Error that "aggregate(value, list(format(date, "%m.%Y"),
> country), mean) : object 'value' not found".
>
> Here, my query "may I need to define the date column, country column, and
> value column separately?"
>
> Further, I need something the average value result like below in the data
> frame
>
> Month   Country A   Country B
> Jan 199426.66 35.78
> Feb 199426.13 29.14
>
> so that it will be easy for me to export to excel, and to use for the
> further calculations.
>
> Please suggest me in this regard.
>
> Thank you.
>
>
>
>
>
>
> https://mailtrack.io?utm_source=gmail_medium=signature_campaign=signaturevirality5;
> Sender notified by
>
> https://mailtrack.io?utm_source=gmail_medium=signature_campaign=signaturevirality5;
> 09/13/19, 07:22:53 PM
>
>
>
> On Fri, Sep 13, 2019 at 7:03 PM PIKAL Petr <mailto:petr.pi...@precheza.cz>
> wrote:
> Hi
>
> I am almost 100% sure that you would spare yourself much trouble if you
> changed your date column to real date
>
> ?as.Date
>
> reshape your wide format to long one
> library(reshape2)
> ?melt
>
> to get 3 column data.frame with one date column, one country column and
> one value column
>
> use ?aggregate and ?format to get summary value
>
> something like
> aggregate(value column, list(format(date column, "%m.%Y"), country
> column), mean)
>
> But if you insist to scratch your left ear with right hand accross your
> head, you could continue your way.
>
> Cheers
> Petr
>
> > -Original Message-
> > From: R-help <mailto:r-help-boun...@r-project.org> On Behalf Of
> Subhamitra
> > Patra
> > Sent: Fri

Re: [R] Query about calculating the monthly average of daily data columns

[PIKAL Petr]

7040999133, 11.7041009433341, 9.95628297574831, 
10.718111342931, 10.4562688422472, 8.85976383099186, 8.94085496683824, 
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9.59278722974697)), row.names = c(NA, -260L), class = "data.frame")





From: Subhamitra Patra <mailto:subhamitra.pa...@gmail.com> 
Sent: Friday, September 13, 2019 3:59 PM
To: PIKAL Petr <mailto:petr.pi...@precheza.cz>; r-help mailing list 
<mailto:r-help@r-project.org>
Subject: Re: [R] Query about calculating the monthly average of daily data 
columns

Dear PIKAL,

Thank you very much for your suggestion.

I tried your previous suggested code and getting the average value for each 
month for both country A, and B. But in your recent email, you are suggesting 
not to change the date column to real date. If I am going through your recently 
suggested code, i.e.

 "aggregate(value column, list(format(date column, "%m.%Y"), country column), 
mean)"

I am getting an Error that "aggregate(value, list(format(date, "%m.%Y"), 
country), mean) : object 'value' not found". 

Here, my query "may I need to define the date column, country column, and value 
column separately?"

Further, I need something the average value result like below in the data frame

Month       Country A   Country B
Jan 1994    26.66         35.78
Feb 1994    26.13         29.14

so that it will be easy for me to export to excel, and to use for the further 
calculations.

Please suggest me in this regard.

Thank you.





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Sender notified by 
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 09/13/19, 07:22:53 PM 



On Fri, Sep 13, 2019 at 7:03 PM PIKAL Petr <mailto:petr.pi...@precheza.cz> 
wrote:
Hi

Re: [R] Query about calculating the monthly average of daily data columns

[Jim Lemon]

Sorry, forgot to include the list.

On Sat, Sep 14, 2019 at 10:27 AM Jim Lemon  wrote:
>
> See inline
>
> On Fri, Sep 13, 2019 at 11:20 PM Subhamitra Patra 
>  wrote:
>>
>> Dear Sir,
>>
>> Yes, I understood the logic. But, still, I have a few queries that I 
>> mentioned below your answers.
>>
>>> "# if you only have to get the monthly averages, it can be done this way
>>> spdat$month<-sapply(strsplit(spdat$dates,"-"),"[",2)
>>> spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",3)"
>>>
>>> B. Here, I need to define the no. of months, and years separately, right? 
>>> or else what 2, and 3 (in bold) indicates?
>>
>>
>> To get the grouping variable of sequential months that you want, you only 
>> need the month and year values of the dates in the first column. First I 
>> used the "strsplit" function to split the date field at the hyphens, then 
>> used "sapply" to extract ("[") the second (month) and third (year) parts as 
>> two new columns. Because you have more than one year of data, you need the 
>> year values or you will group all Januarys, all Februarys and so on. Notice 
>> how I pass both of the new columns as a list (a data frame is a type of 
>> list) in the call to get the mean of each month.
>>
>> 1. Here, as per my understanding, the "3" indicates the 3rd year, right? 
>> But, you showed an average for 2 months of the same year. Then, what "3" in 
>> the  spdat$year object indicate?
>
>
> No, as I explained in the initial email and below, the "strsplit" function 
> takes one or more strings (your dates) and breaks them at the specified 
> character ("-"), So
>
> strsplit("1-1-1994","-")
> [[1]]
> [1] "1""1""1994"
>
> That is passed to the "sapply" function that applies the extraction ("[") 
> operator to the result of "strsplit". The "3" indicates that you want to 
> extract the third element, in this case, the year.
>
> > sapply(strsplit("1-1-1994","-"),"[",3)
> [1] "1994"
>
> So by splitting the dates and extracting the second (month) and third (year) 
> element from each date, we have all the information needed to create a 
> grouping variable for monthly averages.
>
>>
>>
>>> C. From this part, I got the exact average values of both January and 
>>> February of 1994 for country A, and B. But, in code, I have a query that I 
>>> need to define  spdat$returnA, and  spdat$returnB separately before writing 
>>> this code, right? Like this, I need to define for each 84 countries 
>>> separately with their respective number of months, and years before writing 
>>> this code, right?
>>
>>
>> I don't think so. Because I don't know what your data looks like, I am 
>> guessing that for each row, it has columns for each of the 84 countries. I 
>> don't know what these columns are named, either. Maybe:
>>
>> date Australia   Belarus   ...Zambia
>> 01/01/1994   20 21 22
>> ...
>>
>> Here, due to my misunderstanding about the code, I was wrong. But, what data 
>> structure you guessed, it is absolutely right that for each row, I have 
>> columns for each of the 84 countries. So, I think, I need to define the date 
>> column with no. of months, and years once for all the countries. Therefore, 
>> I got my answer to the first and third question in the previous email (what 
>> you suggested) that I no need to define the column of each country, as the 
>> date, and no. of observations are same for all countries. But, the no. of 
>> days are different for each month, and similarly, for each year. So, I think 
>> I need to define date for each year separately.  Hence, I have given an 
>> example of 12 months, for 2 years (i.e. 1994, and 1995), and have written 
>> the following code. Please correct me in case I am wrong.
>>
>>  spdat<-data.frame(
>>   
>> dates=paste(c(1:21,1:20,1:23,1:21,1:22,1:22,1:21,1:23,1:22,1:21,1:22,1:22),c(rep(1,21),rep(2,20),rep(3,23),
>>  rep(4,21), 
>> rep(5,22),rep(6,22),rep(7,21),rep(8,23),rep(9,22),rep(10,21),rep(11,22),rep(12,22)),rep(1994,260)
>>  
>> dates1=paste(c(1:22,1:20,1:23,1:20,1:23,1:22,1:21,1:23,1:21,1:22,1:22,1:21),c(rep(1,22),rep(2,20),rep(3,23),
>>  rep(4,20), 
>> rep(5,23),rep(6,22),rep(7,21),rep(8,23),rep(9,21),rep(10,21),rep(11,22),rep(12,21)),rep(1995,259)
>>  ,sep="-")
>>
> First, you don't have to recreate the data that you already have. I did 
> because I don't have it and have to guess what it looks like. Remember 
> neither I nor any of the others who have offered help have your data or even 
> a representative sample. If you tried the code above, you surely must know 
> that it doesn't work. I could create code that would produce the dates from 
> 1-1-1994 to 31/12/1995 or any other stretch you would like, but it would only 
> confuse you more.  _You already have the dates in your data file._ What I 
> have shown you is how to use those dates to create the grouping variable that 
> you want.
>
>> Concerning the exporting of structure of the dataset to excel, I will have 
>> 12*84 matrix. But, please 

Re: [R] Query about calculating the monthly average of daily data columns

[Subhamitra Patra]

Dear PIKAL,

Thank you very much for your suggestion.

I tried your previous suggested code and getting the average value for each
month for both country A, and B. But in your recent email, you are
suggesting not to change the date column to real date. If I am going
through your recently suggested code, i.e.

 "aggregate(value column, list(format(date column, "%m.%Y"), country
column), mean)"

I am getting an Error that "*aggregate(value, list(format(date, "%m.%Y"),
country), mean) : **object 'value' not found"*.

Here, my query "*may I need to define the date column, country column, and
value column separately?"*

Further, I need something the average value result like below in the data
frame

Month   Country A   Country B
Jan 199426.66 35.78
Feb 199426.13 29.14

so that it will be easy for me to export to excel, and to use for the
further calculations.

Please suggest me in this regard.

Thank you.







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09/13/19,
07:22:53 PM

On Fri, Sep 13, 2019 at 7:03 PM PIKAL Petr  wrote:

> Hi
>
> I am almost 100% sure that you would spare yourself much trouble if you
> changed your date column to real date
>
> ?as.Date
>
> reshape your wide format to long one
> library(reshape2)
> ?melt
>
> to get 3 column data.frame with one date column, one country column and
> one value column
>
> use ?aggregate and ?format to get summary value
>
> something like
> aggregate(value column, list(format(date column, "%m.%Y"), country
> column), mean)
>
> But if you insist to scratch your left ear with right hand accross your
> head, you could continue your way.
>
> Cheers
> Petr
>
> > -Original Message-
> > From: R-help  On Behalf Of Subhamitra
> > Patra
> > Sent: Friday, September 13, 2019 3:20 PM
> > To: Jim Lemon ; r-help mailing list  > project.org>
> > Subject: Re: [R] Query about calculating the monthly average of daily
> data
> > columns
> >
> > Dear Sir,
> >
> > Yes, I understood the logic. But, still, I have a few queries that I
> mentioned
> > below your answers.
> >
> > "# if you only have to get the monthly averages, it can be done this way
> > > spdat$month<-sapply(strsplit(spdat$dates,"-"),"["*,2*)
> > > spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",*3*)"
> > >
> > > B. Here, I need to define the no. of months, and years separately,
> right?
> > > or else what 2, and 3 (in bold) indicates?
> > >
> >
> > To get the grouping variable of sequential months that you want, you only
> > need the month and year values of the dates in the first column. First I
> used
> > the "strsplit" function to split the date field at the hyphens, then used
> > "sapply" to extract ("[") the second (month) and *third (year)* parts as
> two
> > new columns. Because you have more than one year of data, you need the
> > year values or you will group all Januarys, all Februarys and so on.
> > Notice how I pass both of the new columns as a list (a data frame is a
> type of
> > list) in the call to get the mean of each month.
> >
> > 1. Here, as per my understanding, the "3" indicates the 3rd year, right?
> > But, you showed an average for 2 months of the same year. Then, what "3"
> > in the  spdat$year object indicate?
> >
> >
> > C. From this part, I got the exact average values of both January and
> > > February of 1994 for country A, and B. But, in code, I have a query
> > > that I need to define  spdat$returnA, and  spdat$returnB separately
> > > before writing this code, right? Like this, I need to define for each
> > > 84 countries separately with their respective number of months, and
> > > years before writing this code, right?
> > >
> >
> > I don't think so. Because I don't know what your data looks like, I am
> > guessing that for each row, it has columns for each of the 84 countries.
> I
> > don't know what these columns are named, either. Maybe:
> >
> > date Australia   Belarus   ...Zambia
> > 01/01/1994   20 21 22
> > ...
> >
> > Here, due to my misunderstanding about the code, I was wrong. But, what
> > data structure you guessed, it is absolutely right that for each row, I
> have
> > columns for ea

Re: [R] Query about calculating the monthly average of daily data columns

[PIKAL Petr]

Hi

I am almost 100% sure that you would spare yourself much trouble if you changed 
your date column to real date

?as.Date

reshape your wide format to long one
library(reshape2)
?melt

to get 3 column data.frame with one date column, one country column and one 
value column

use ?aggregate and ?format to get summary value

something like
aggregate(value column, list(format(date column, "%m.%Y"), country column), 
mean)

But if you insist to scratch your left ear with right hand accross your head, 
you could continue your way.

Cheers
Petr

> -Original Message-
> From: R-help  On Behalf Of Subhamitra
> Patra
> Sent: Friday, September 13, 2019 3:20 PM
> To: Jim Lemon ; r-help mailing list  project.org>
> Subject: Re: [R] Query about calculating the monthly average of daily data
> columns
>
> Dear Sir,
>
> Yes, I understood the logic. But, still, I have a few queries that I mentioned
> below your answers.
>
> "# if you only have to get the monthly averages, it can be done this way
> > spdat$month<-sapply(strsplit(spdat$dates,"-"),"["*,2*)
> > spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",*3*)"
> >
> > B. Here, I need to define the no. of months, and years separately, right?
> > or else what 2, and 3 (in bold) indicates?
> >
>
> To get the grouping variable of sequential months that you want, you only
> need the month and year values of the dates in the first column. First I used
> the "strsplit" function to split the date field at the hyphens, then used
> "sapply" to extract ("[") the second (month) and *third (year)* parts as two
> new columns. Because you have more than one year of data, you need the
> year values or you will group all Januarys, all Februarys and so on.
> Notice how I pass both of the new columns as a list (a data frame is a type of
> list) in the call to get the mean of each month.
>
> 1. Here, as per my understanding, the "3" indicates the 3rd year, right?
> But, you showed an average for 2 months of the same year. Then, what "3"
> in the  spdat$year object indicate?
>
>
> C. From this part, I got the exact average values of both January and
> > February of 1994 for country A, and B. But, in code, I have a query
> > that I need to define  spdat$returnA, and  spdat$returnB separately
> > before writing this code, right? Like this, I need to define for each
> > 84 countries separately with their respective number of months, and
> > years before writing this code, right?
> >
>
> I don't think so. Because I don't know what your data looks like, I am
> guessing that for each row, it has columns for each of the 84 countries. I
> don't know what these columns are named, either. Maybe:
>
> date Australia   Belarus   ...Zambia
> 01/01/1994   20 21 22
> ...
>
> Here, due to my misunderstanding about the code, I was wrong. But, what
> data structure you guessed, it is absolutely right that for each row, I have
> columns for each of the 84 countries. So, I think, I need to define the date
> column with no. of months, and years once for all the countries.
> Therefore, I got my answer to the first and third question in the previous
> email (what you suggested) that I no need to define the column of each
> country, as the date, and no. of observations are same for all countries.
> But, the no. of days are different for each month, and similarly, for each
> year. So, I think I need to define date for each year separately.  Hence, I 
> have
> given an example of 12 months, for 2 years (i.e. 1994, and 1995), and have
> written the following code. Please correct me in case I am wrong.
>
>  spdat<-data.frame(
>
> dates=paste(c(1:21,1:20,1:23,1:21,1:22,1:22,1:21,1:23,1:22,1:21,1:22,1:22),c(r
> ep(1,21),rep(2,20),
> rep(3,23), rep(4,21),
> rep(5,22),rep(6,22),rep(7,21),rep(8,23),rep(9,22),rep(10,21),rep(11,22),rep(12
> ,22)
> ),rep(1994,260)
>  dates1=
> paste(c(1:22,1:20,1:23,1:20,1:23,1:22,1:21,1:23,1:21,1:22,1:22,1:21),c(rep(1,2
> 2),rep(2,20),
> rep(3,23), rep(4,20),
> rep(5,23),rep(6,22),rep(7,21),rep(8,23),rep(9,21),rep(10,21),rep(11,22),rep(12
> ,21)
> ),rep(1995,259) ,sep="-")
>
> Concerning the exporting of structure of the dataset to excel, I will have
> 12*84 matrix. But, please suggest me the way to proceed for the large
> sample. I have mentioned below what I understood from your code. Please
> correct me if I am wrong.
> 1. I need to define the date for each year as the no. of days in each month
> are different for each year (as mentioned in my above code). For instance, in
> my data file, Jan 1994 has 21 days wh

Re: [R] Query about calculating the monthly average of daily data columns

[Subhamitra Patra]

Dear Sir,

Yes, I understood the logic. But, still, I have a few queries that I
mentioned below your answers.

"# if you only have to get the monthly averages, it can be done this way
> spdat$month<-sapply(strsplit(spdat$dates,"-"),"["*,2*)
> spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",*3*)"
>
> B. Here, I need to define the no. of months, and years separately, right?
> or else what 2, and 3 (in bold) indicates?
>

To get the grouping variable of sequential months that you want, you only
need the month and year values of the dates in the first column. First I
used the "strsplit" function to split the date field at the hyphens, then
used "sapply" to extract ("[") the second (month) and *third (year)* parts
as two new columns. Because you have more than one year of data, you need
the year values or you will group all Januarys, all Februarys and so on.
Notice how I pass both of the new columns as a list (a data frame is a type
of list) in the call to get the mean of each month.

1. Here, as per my understanding, the "3" indicates the 3rd year, right?
But, you showed an average for 2 months of the same year. Then, what "3" in
the  spdat$year object indicate?


C. From this part, I got the exact average values of both January and
> February of 1994 for country A, and B. But, in code, I have a query that I
> need to define  spdat$returnA, and  spdat$returnB separately before writing
> this code, right? Like this, I need to define for each 84 countries
> separately with their respective number of months, and years before writing
> this code, right?
>

I don't think so. Because I don't know what your data looks like, I am
guessing that for each row, it has columns for each of the 84 countries. I
don't know what these columns are named, either. Maybe:

date Australia   Belarus   ...Zambia
01/01/1994   20 21 22
...

Here, due to my misunderstanding about the code, I was wrong. But, what
data structure you guessed, it is absolutely right that for each row, I
have columns for each of the 84 countries. So, I think, I need to define
the date column with no. of months, and years once for all the countries.
Therefore, I got my answer to the first and third question in the previous
email (what you suggested) that I no need to define the column of each
country, as the date, and no. of observations are same for all countries.
But, the no. of days are different for each month, and similarly, for each
year. So, I think I need to define date for each year separately.  Hence, I
have given an example of 12 months, for 2 years (i.e. 1994, and 1995), and
have written the following code. Please correct me in case I am wrong.

 spdat<-data.frame(

dates=paste(c(1:21,1:20,1:23,1:21,1:22,1:22,1:21,1:23,1:22,1:21,1:22,1:22),c(rep(1,21),rep(2,20),
rep(3,23), rep(4,21),
rep(5,22),rep(6,22),rep(7,21),rep(8,23),rep(9,22),rep(10,21),rep(11,22),rep(12,22)
),rep(1994,260)
 dates1=
paste(c(1:22,1:20,1:23,1:20,1:23,1:22,1:21,1:23,1:21,1:22,1:22,1:21),c(rep(1,22),rep(2,20),
rep(3,23), rep(4,20),
rep(5,23),rep(6,22),rep(7,21),rep(8,23),rep(9,21),rep(10,21),rep(11,22),rep(12,21)
),rep(1995,259) ,sep="-")

Concerning the exporting of structure of the dataset to excel, I will have
12*84 matrix. But, please suggest me the way to proceed for the large
sample. I have mentioned below what I understood from your code. Please
correct me if I am wrong.
1. I need to define the date for each year as the no. of days in each month
are different for each year (as mentioned in my above code). For instance,
in my data file, Jan 1994 has 21 days while Jan 1995 has 22 days.
2. Need to define the date column as character.
3. Need to define the monthly average for each month, and year. So, now
code will be as follows.
spdat$month<-sapply(strsplit(spdat$dates,"-"),"[",2,3,4,5,6,7,8,9,10,11,12)
  As I need all months average sequentially.
spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",3)

Here, this meaning of "3", I am really unable to get.

4. Need to define each country with each month and year as mentioned in the
last part of your code.

Please suggest me in this regard.

Thank you.







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On Fri, Sep 13, 2019 at 4:24 PM Jim Lemon  wrote:

> Hi Subhamitra,
> I'll try to write my answers adjacent to your questions below.
>
> On Fri, Sep 13, 2019 at 6:08 PM Subhamitra Patra <
> subhamitra.pa...@gmail.com> wrote:
>
>> Dear Sir,
>>
>> Thank you very much for your suggestion.
>>
>> Yes, your suggested code worked. But, actually, I have data from 3rd
>> January 1994 to 3rd August 2017 for very large (i.e. for 84 countries)
>> sample. From this, I have given the example of the years up to 2000. Before
>> applying the same code for the long 24 years, I want to learn the logic
>> 

Re: [R] Query about calculating the monthly average of daily data columns

[Jim Lemon]

Hi Subhamitra,
I'll try to write my answers adjacent to your questions below.

On Fri, Sep 13, 2019 at 6:08 PM Subhamitra Patra 
wrote:

> Dear Sir,
>
> Thank you very much for your suggestion.
>
> Yes, your suggested code worked. But, actually, I have data from 3rd
> January 1994 to 3rd August 2017 for very large (i.e. for 84 countries)
> sample. From this, I have given the example of the years up to 2000. Before
> applying the same code for the long 24 years, I want to learn the logic
> behind the code. Actually, some part of the code is not understandable to
> me which I mentioned in the bold letter as follows.
>
> "spdat<-data.frame(
>   dates=paste(c(1:30,1:28),c(rep(1,30),rep(2,28)),rep(1994,58),sep="-"),
>   returnA=sample(*15:50*,58,TRUE),returnB=sample(*10:45*,58,TRUE))"
>
> A. Here, I need to define the no. of days in a month, and the no. of
> countries name separately, right? But, what is meant by 15:50, and 10:45 in
> return A, and B respectively?
>

To paraphrase Donald Trump, this is FAKE DATA! I have no idea what the real
values of return are, so I made them up using the "sample" function.
However, this is not meant to mislead anyone, just to show how whatever
numbers are in your data can be used in calculations. The colon (":")
operator creates a sequence of numbers starting with the one to the left
and ending with the one to the right.

>
> "# if you only have to get the monthly averages, it can be done this way
> spdat$month<-sapply(strsplit(spdat$dates,"-"),"["*,2*)
> spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",*3*)"
>
> B. Here, I need to define the no. of months, and years separately, right?
> or else what 2, and 3 (in bold) indicates?
>

To get the grouping variable of sequential months that you want, you only
need the month and year values of the dates in the first column. First I
used the "strsplit" function to split the date field at the hyphens, then
used "sapply" to extract ("[") the second (month) and third (year) parts as
two new columns. Because you have more than one year of data, you need the
year values or you will group all Januarys, all Februarys and so on. Notice
how I pass both of the new columns as a list (a data frame is a type of
list) in the call to get the mean of each month.

>
> "# get the averages by month and year - is this correct?
> monthlyA<-by(*spdat$returnA*,spdat[,c("month","year")],mean)
> monthlyB<-by(*spdat$returnB*,spdat[,c("month","year")],mean)"
>
> C. From this part, I got the exact average values of both January and
> February of 1994 for country A, and B. But, in code, I have a query that I
> need to define  spdat$returnA, and  spdat$returnB separately before writing
> this code, right? Like this, I need to define for each 84 countries
> separately with their respective number of months, and years before writing
> this code, right?
>

I don't think so. Because I don't know what your data looks like, I am
guessing that for each row, it has columns for each of the 84 countries. I
don't know what these columns are named, either. Maybe:

date Australia   Belarus   ...Zambia
01/01/1994   20 21 22
...


> Yes, after obtaining the monthly average for each country's data, I need
> to use them for further calculations. So, I want to export the result to
> excel. But, until understanding the code, I think I willn't able to apply
> for the entire sample, and cannot be able to discuss the format of the
> resulted column to export to excel.
>

Say that we perform the grouped mean calculation for the first two country
columns like this:
monmeans<-sapply(spdat[,2:3],by,spdat[,c("month","year")],mean)
monmeans
Australia  Belarus
[1,]  29.7 30.4
[2,]  34.17857 27.39286

We are presented with a 2x2 matrix of monthly means in just the format
someone might use for importing into Excel. The first row is January 1994,
the second February 1994 and so on. By expanding the columns to include all
the countries in your data, You should have the result you want.

Jim

[[alternative HTML version deleted]]

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Re: [R] Query about calculating the monthly average of daily data columns

[PIKAL Petr]

Hi

I may be completely wrong but reshape/aggregate should by what you want
spdat
   dates returnA returnB
1   1-1-1994  16  13
2   2-1-1994  44  10
3   3-1-1994  24  32
.
> library(reshape2)
> spdat.m <- melt(spdat)
Using dates as id variables
> str(spdat.m)
'data.frame':   116 obs. of  3 variables:
 $ dates   : Factor w/ 58 levels "1-1-1994","1-2-1994",..: 1 23 44 47 49 51 53 
55 57 3 ...
 $ variable: Factor w/ 2 levels "returnA","returnB": 1 1 1 1 1 1 1 1 1 1 ...
 $ value   : int  16 44 24 47 16 35 34 34 26 36 ...
> spdat.m$realdate <- as.Date(spdat.m[,1], format="%d-%m-%Y")
> aggregate(spdat.m$value, list(format(spdat.m$realdate, "%m.%Y"), 
> spdat.m$variable), mean)
  Group.1 Group.2x
1 01.1994 returnA 31.9
2 02.1994 returnA 32.39286
3 01.1994 returnB 24.26667
4 02.1994 returnB 30.03571

Cheers
Petr

> -Original Message-
> From: R-help  On Behalf Of Subhamitra
> Patra
> Sent: Friday, September 13, 2019 10:08 AM
> To: Jim Lemon 
> Cc: r-help mailing list 
> Subject: Re: [R] Query about calculating the monthly average of daily data
> columns
>
> Dear Sir,
>
> Thank you very much for your suggestion.
>
> Yes, your suggested code worked. But, actually, I have data from 3rd January
> 1994 to 3rd August 2017 for very large (i.e. for 84 countries) sample. From
> this, I have given the example of the years up to 2000. Before applying the
> same code for the long 24 years, I want to learn the logic behind the code.
> Actually, some part of the code is not understandable to me which I
> mentioned in the bold letter as follows.
>
> "spdat<-data.frame(
>   dates=paste(c(1:30,1:28),c(rep(1,30),rep(2,28)),rep(1994,58),sep="-"),
>   returnA=sample(*15:50*,58,TRUE),returnB=sample(*10:45*,58,TRUE))"
>
> A. Here, I need to define the no. of days in a month, and the no. of countries
> name separately, right? But, what is meant by 15:50, and 10:45 in return A,
> and B respectively?
>
> "# if you only have to get the monthly averages, it can be done this way
> spdat$month<-sapply(strsplit(spdat$dates,"-"),"["*,2*)
> spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",*3*)"
>
> B. Here, I need to define the no. of months, and years separately, right?
> or else what 2, and 3 (in bold) indicates?
>
> "# get the averages by month and year - is this correct?
> monthlyA<-by(*spdat$returnA*,spdat[,c("month","year")],mean)
> monthlyB<-by(*spdat$returnB*,spdat[,c("month","year")],mean)"
>
> C. From this part, I got the exact average values of both January and
> February of 1994 for country A, and B. But, in code, I have a query that I
> need to define  spdat$returnA, and  spdat$returnB separately before writing
> this code, right? Like this, I need to define for each 84 countries separately
> with their respective number of months, and years before writing this code,
> right?
>
> Yes, after obtaining the monthly average for each country's data, I need to
> use them for further calculations. So, I want to export the result to excel. 
> But,
> until understanding the code, I think I willn't able to apply for the entire
> sample, and cannot be able to discuss the format of the resulted column to
> export to excel.
>
> Therefore, kindly help me to understand the code.
>
> Thank you very much, Sir, and thanks to this R forum for helping the R-
> beginners.
>
>
>
> [image: Mailtrack]
> <https://mailtrack.io?utm_source=gmail_medium=signature_ca
> mpaign=signaturevirality5&>
> Sender
> notified by
> Mailtrack
> <https://mailtrack.io?utm_source=gmail_medium=signature_ca
> mpaign=signaturevirality5&>
> 09/13/19,
> 12:57:58 PM
>
> On Fri, Sep 13, 2019 at 3:15 AM Jim Lemon  wrote:
>
> > Hi Subhamitra,
> > Your data didn't make it through, so I guess the first thing is to
> > guess what it looks like. Here's a try at just January and February of
> > 1994 so that we can see the result on the screen. The logic will work
> > just as well for the whole seven years.
> >
> > # create fake data for the first two months spdat<-data.frame(
> > dates=paste(c(1:30,1:28),c(rep(1,30),rep(2,28)),rep(1994,58),sep="-"),
> >  returnA=sample(15:50,58,TRUE),returnB=sample(10:45,58,TRUE))
> > # I'll assume that the dates in your file are character, not factor
> > spdat$dates<-as.character(spdat$dates)
> > # if you only have to get the monthly averages, it can be done this
> > way
> > spdat$month<-sapply(strsplit(spdat$dates,"-"),"[",2)
> >

Re: [R] Query about calculating the monthly average of daily data columns

[Subhamitra Patra]

Dear Sir,

Thank you very much for your suggestion.

Yes, your suggested code worked. But, actually, I have data from 3rd
January 1994 to 3rd August 2017 for very large (i.e. for 84 countries)
sample. From this, I have given the example of the years up to 2000. Before
applying the same code for the long 24 years, I want to learn the logic
behind the code. Actually, some part of the code is not understandable to
me which I mentioned in the bold letter as follows.

"spdat<-data.frame(
  dates=paste(c(1:30,1:28),c(rep(1,30),rep(2,28)),rep(1994,58),sep="-"),
  returnA=sample(*15:50*,58,TRUE),returnB=sample(*10:45*,58,TRUE))"

A. Here, I need to define the no. of days in a month, and the no. of
countries name separately, right? But, what is meant by 15:50, and 10:45 in
return A, and B respectively?

"# if you only have to get the monthly averages, it can be done this way
spdat$month<-sapply(strsplit(spdat$dates,"-"),"["*,2*)
spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",*3*)"

B. Here, I need to define the no. of months, and years separately, right?
or else what 2, and 3 (in bold) indicates?

"# get the averages by month and year - is this correct?
monthlyA<-by(*spdat$returnA*,spdat[,c("month","year")],mean)
monthlyB<-by(*spdat$returnB*,spdat[,c("month","year")],mean)"

C. From this part, I got the exact average values of both January and
February of 1994 for country A, and B. But, in code, I have a query that I
need to define  spdat$returnA, and  spdat$returnB separately before writing
this code, right? Like this, I need to define for each 84 countries
separately with their respective number of months, and years before writing
this code, right?

Yes, after obtaining the monthly average for each country's data, I need to
use them for further calculations. So, I want to export the result to
excel. But, until understanding the code, I think I willn't able to apply
for the entire sample, and cannot be able to discuss the format of the
resulted column to export to excel.

Therefore, kindly help me to understand the code.

Thank you very much, Sir, and thanks to this R forum for helping the
R-beginners.



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09/13/19,
12:57:58 PM

On Fri, Sep 13, 2019 at 3:15 AM Jim Lemon  wrote:

> Hi Subhamitra,
> Your data didn't make it through, so I guess the first thing is to
> guess what it looks like. Here's a try at just January and February of
> 1994 so that we can see the result on the screen. The logic will work
> just as well for the whole seven years.
>
> # create fake data for the first two months
> spdat<-data.frame(
>  dates=paste(c(1:30,1:28),c(rep(1,30),rep(2,28)),rep(1994,58),sep="-"),
>  returnA=sample(15:50,58,TRUE),returnB=sample(10:45,58,TRUE))
> # I'll assume that the dates in your file are character, not factor
> spdat$dates<-as.character(spdat$dates)
> # if you only have to get the monthly averages, it can be done this way
> spdat$month<-sapply(strsplit(spdat$dates,"-"),"[",2)
> spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",3)
> # get the averages by month and year - is this correct?
> monthlyA<-by(spdat$returnA,spdat[,c("month","year")],mean)
> monthlyB<-by(spdat$returnB,spdat[,c("month","year")],mean)
>
> Now you have what you say you want:
>
> monthlyA
> month: 1
> year: 1994
> [1] 34.1
> 
> month: 2
> year: 1994
> [1] 33.32143
>
> monthlyB
> month: 1
> year: 1994
> [1] 29.7
> 
> month: 2
> year: 1994
> [1] 27.28571
>
> Sorry I didn't use a loop (for(month in 1:12) ... for (year in
> 1994:2000) ...), too lazy.
> Now you have to let us know how this information is to be formatted to
> go into Excel. Excel will import the text as above, but I think you
> want something that you can use for further calculations.
>
> Jim
>
> On Fri, Sep 13, 2019 at 12:54 AM Subhamitra Patra
>  wrote:
> >
> > Dear R-users,
> >
> > I have daily data from 03-01-1994 to 29-12-2000. In my datafile, he first
> > column is date and the second and third columns are the returns of the
> > country A, and B. Here, the date column is same for both countries. I
> want
> > to calculate the monthly average of both country's returns by using a
> loop,
> > and then, I want to export the results into excel.
> >
> > Please help me in this regard.
> >
> > Please find the attached datasheet.
> >
> > Thank you.
> >
> > --
> > *Best Regards,*
> > *Subhamitra Patra*
> > *Phd. Research Scholar*
> > *Department of Humanities and Social Sciences*
> > *Indian Institute of Technology, Kharagpur*
> > *INDIA*
> >
> > [image: Mailtrack]
> > <
> https://mailtrack.io?utm_source=gmail_medium=signature_campaign=signaturevirality5;
> >
> > Sender
> > notified by
> > Mailtrack
> > <
> 

Re: [R] Query about calculating the monthly average of daily data columns

[Jim Lemon]

Hi Subhamitra,
Your data didn't make it through, so I guess the first thing is to
guess what it looks like. Here's a try at just January and February of
1994 so that we can see the result on the screen. The logic will work
just as well for the whole seven years.

# create fake data for the first two months
spdat<-data.frame(
 dates=paste(c(1:30,1:28),c(rep(1,30),rep(2,28)),rep(1994,58),sep="-"),
 returnA=sample(15:50,58,TRUE),returnB=sample(10:45,58,TRUE))
# I'll assume that the dates in your file are character, not factor
spdat$dates<-as.character(spdat$dates)
# if you only have to get the monthly averages, it can be done this way
spdat$month<-sapply(strsplit(spdat$dates,"-"),"[",2)
spdat$year<-sapply(strsplit(spdat$dates,"-"),"[",3)
# get the averages by month and year - is this correct?
monthlyA<-by(spdat$returnA,spdat[,c("month","year")],mean)
monthlyB<-by(spdat$returnB,spdat[,c("month","year")],mean)

Now you have what you say you want:

monthlyA
month: 1
year: 1994
[1] 34.1

month: 2
year: 1994
[1] 33.32143

monthlyB
month: 1
year: 1994
[1] 29.7

month: 2
year: 1994
[1] 27.28571

Sorry I didn't use a loop (for(month in 1:12) ... for (year in
1994:2000) ...), too lazy.
Now you have to let us know how this information is to be formatted to
go into Excel. Excel will import the text as above, but I think you
want something that you can use for further calculations.

Jim

On Fri, Sep 13, 2019 at 12:54 AM Subhamitra Patra
 wrote:
>
> Dear R-users,
>
> I have daily data from 03-01-1994 to 29-12-2000. In my datafile, he first
> column is date and the second and third columns are the returns of the
> country A, and B. Here, the date column is same for both countries. I want
> to calculate the monthly average of both country's returns by using a loop,
> and then, I want to export the results into excel.
>
> Please help me in this regard.
>
> Please find the attached datasheet.
>
> Thank you.
>
> --
> *Best Regards,*
> *Subhamitra Patra*
> *Phd. Research Scholar*
> *Department of Humanities and Social Sciences*
> *Indian Institute of Technology, Kharagpur*
> *INDIA*
>
> [image: Mailtrack]
> 
> Sender
> notified by
> Mailtrack
> 
> 09/12/19,
> 08:23:07 PM
> __
> 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.

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Re: [R] Query about calculating the monthly average of daily data columns

[Rui Barradas]

Hello,

Inline.

Às 17:33 de 12/09/19, Bert Gunter escreveu:
But she wants *monthly* averages, Rui. 


Thanks, my mistake.

Ergo ave() or tidyData

equivalent, right?


Maybe. But ave() returns as many values as the input length, this seems 
more suited for tapply or aggregate.



I will first create an example data set.

set.seed(1234)
start <- as.Date("03-01-1994", "%d-%m-%Y")
end <- as.Date("29-12-2000", "%d-%m-%Y")
date <- seq(start, end, by = "day")
date <- date[as.integer(format(date, "%u")) %in% 1:5]
df1 <- data.frame(date,
  CountryA = rnorm(length(date)),
  CountryB = rnorm(length(date)))


Now the averages by month

month <- zoo::as.yearmon(df1[[1]])
aggA <- aggregate(CountryA ~ month, df1, mean)
aggB <- aggregate(CountryB ~ month, df1, mean)
MonthReturns <- merge(aggA, aggB)
head(MonthReturns)


Final clean up.

rm(date, month, aggA, aggB)


Hope this helps,

Rui Barradas


-- Bert

On Thu, Sep 12, 2019 at 8:41 AM Rui Barradas > wrote:


Hello,

Please include data, say

dput(head(data, 20))  # post the output of this


But, is the problem as simple as

rowMeans(data[2:3], na.rm = TRUE)

?

Hope this helps,

Rui Barradas


Às 15:53 de 12/09/19, Subhamitra Patra escreveu:
 > Dear R-users,
 >
 > I have daily data from 03-01-1994 to 29-12-2000. In my datafile,
he first
 > column is date and the second and third columns are the returns
of the
 > country A, and B. Here, the date column is same for both
countries. I want
 > to calculate the monthly average of both country's returns by
using a loop,
 > and then, I want to export the results into excel.
 >
 > Please help me in this regard.
 >
 > Please find the attached datasheet.
 >
 > Thank you.
 >

__
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Re: [R] Query about calculating the monthly average of daily data columns

[Rui Barradas]

Hello,

Please include data, say

dput(head(data, 20))  # post the output of this


But, is the problem as simple as

rowMeans(data[2:3], na.rm = TRUE)

?

Hope this helps,

Rui Barradas


Às 15:53 de 12/09/19, Subhamitra Patra escreveu:

Dear R-users,

I have daily data from 03-01-1994 to 29-12-2000. In my datafile, he first
column is date and the second and third columns are the returns of the
country A, and B. Here, the date column is same for both countries. I want
to calculate the monthly average of both country's returns by using a loop,
and then, I want to export the results into excel.

Please help me in this regard.

Please find the attached datasheet.

Thank you.



__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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] Query about calculating the monthly average of daily data columns

[Bert Gunter]

No reproducible example, so hard to say. What class is your "date" column?
-- factor, character, Date?  See ?Date
Once you have an object of appropriate class -- see ?format.Date -- ?months
can extract the month and ?ave can do your averaging. No explicit looping
is needed.

The tidydata alternative universe can also do all these things if that's
where you prefer to live.

As usual, any attached data was stripped. See ?dput for one way to include
data in your post.

Cheers,
Bert


On Thu, Sep 12, 2019 at 7:54 AM Subhamitra Patra 
wrote:

> Dear R-users,
>
> I have daily data from 03-01-1994 to 29-12-2000. In my datafile, he first
> column is date and the second and third columns are the returns of the
> country A, and B. Here, the date column is same for both countries. I want
> to calculate the monthly average of both country's returns by using a loop,
> and then, I want to export the results into excel.
>
> Please help me in this regard.
>
> Please find the attached datasheet.
>
> Thank you.
>
> --
> *Best Regards,*
> *Subhamitra Patra*
> *Phd. Research Scholar*
> *Department of Humanities and Social Sciences*
> *Indian Institute of Technology, Kharagpur*
> *INDIA*
>
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[R] Query about calculating the monthly average of daily data columns

[Subhamitra Patra]

Dear R-users,

I have daily data from 03-01-1994 to 29-12-2000. In my datafile, he first
column is date and the second and third columns are the returns of the
country A, and B. Here, the date column is same for both countries. I want
to calculate the monthly average of both country's returns by using a loop,
and then, I want to export the results into excel.

Please help me in this regard.

Please find the attached datasheet.

Thank you.

-- 
*Best Regards,*
*Subhamitra Patra*
*Phd. Research Scholar*
*Department of Humanities and Social Sciences*
*Indian Institute of Technology, Kharagpur*
*INDIA*

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