[R] Query about calculating the monthly average of daily data columns
PIKAL Petr
petr@p|k@| @end|ng |rom prechez@@cz
Mon Sep 16 13:00:26 CEST 2019
Original email did not come through (some problems with formating).
Hi
No, on contrary. I **am** suggesting to change date column to real date asi it is easy to handle with appropriate functions.
Here are some fake data
> str(spdat)
'data.frame': 260 obs. of 3 variables:
$ dates : Date, format: "1995-01-01" "1995-01-02" "1995-01-03" "1995-01-04" ...
$ coutryA : num 0.188 0.405 -0.107 -0.596 -0.529 ...
$ countryB: num 9.4 10.76 11.24 8.26 10.71 ..
> head(spdat)
dates coutryA countryB
1 1995-01-01 0.1875060 9.402851
2 1995-01-02 0.4045193 10.755112
3 1995-01-03 -0.1073904 11.243663
4 1995-01-04 -0.5959683 8.256424
5 1995-01-05 -0.5293772 10.705431
6 1995-01-06 -0.2228029 10.171461
First I melt it
spdat.m <- melt(spdat, id.var="dates")
> head(spdat.m)
dates variable value
1 1995-01-01 coutryA 0.1875060
2 1995-01-02 coutryA 0.4045193
3 1995-01-03 coutryA -0.1073904
4 1995-01-04 coutryA -0.5959683
5 1995-01-05 coutryA -0.5293772
6 1995-01-06 coutryA -0.2228029
I do aggregation
> spdat.ag <- aggregate(spdat.m$value, list(format(spdat.m$dates, "%m.%Y"), spdat.m$variable), mean)
And now I use dcast to get required result.
> dcast(spdat.ag, Group.1~Group.2)
Using x as value column: use value.var to override.
Group.1 coutryA countryB
1 01.1995 0.098688137 10.177696
2 02.1995 0.352264682 9.609261
3 03.1995 0.155521876 10.043503
4 04.1995 -0.166092393 10.129844
5 05.1995 0.164665188 10.308275
6 06.1995 0.260633585 10.210129
7 07.1995 0.003671979 10.549016
8 08.1995 0.045295990 10.087435
9 09.1995 -0.145488206 9.689876
10 10.1995 -0.225645950 9.743744
11 11.1995 0.030273383 10.025435
12 12.1995 0.043557468 10.105626
Cheers
Petr
Here are the data.
> dput(spdat)
spdat <- structure(list(dates = structure(c(9131, 9132, 9133, 9134, 9135,
9136, 9137, 9138, 9139, 9140, 9141, 9142, 9143, 9144, 9145, 9146,
9147, 9148, 9149, 9150, 9151, 9152, 9162, 9163, 9164, 9165, 9166,
9167, 9168, 9169, 9170, 9171, 9172, 9173, 9174, 9175, 9176, 9177,
9178, 9179, 9180, 9181, 9190, 9191, 9192, 9193, 9194, 9195, 9196,
9197, 9198, 9199, 9200, 9201, 9202, 9203, 9204, 9205, 9206, 9207,
9208, 9209, 9210, 9211, 9212, 9221, 9222, 9223, 9224, 9225, 9226,
9227, 9228, 9229, 9230, 9231, 9232, 9233, 9234, 9235, 9236, 9237,
9238, 9239, 9240, 9251, 9252, 9253, 9254, 9255, 9256, 9257, 9258,
9259, 9260, 9261, 9262, 9263, 9264, 9265, 9266, 9267, 9268, 9269,
9270, 9271, 9272, 9273, 9282, 9283, 9284, 9285, 9286, 9287, 9288,
9289, 9290, 9291, 9292, 9293, 9294, 9295, 9296, 9297, 9298, 9299,
9300, 9301, 9302, 9303, 9312, 9313, 9314, 9315, 9316, 9317, 9318,
9319, 9320, 9321, 9322, 9323, 9324, 9325, 9326, 9327, 9328, 9329,
9330, 9331, 9332, 9343, 9344, 9345, 9346, 9347, 9348, 9349, 9350,
9351, 9352, 9353, 9354, 9355, 9356, 9357, 9358, 9359, 9360, 9361,
9362, 9363, 9364, 9365, 9374, 9375, 9376, 9377, 9378, 9379, 9380,
9381, 9382, 9383, 9384, 9385, 9386, 9387, 9388, 9389, 9390, 9391,
9392, 9393, 9394, 9404, 9405, 9406, 9407, 9408, 9409, 9410, 9411,
9412, 9413, 9414, 9415, 9416, 9417, 9418, 9419, 9420, 9421, 9422,
9423, 9424, 9456, 9435, 9436, 9437, 9438, 9439, 9440, 9441, 9442,
9443, 9444, 9445, 9446, 9447, 9448, 9449, 9450, 9451, 9452, 9453,
9454, 9455, 9486, 9465, 9466, 9467, 9468, 9469, 9470, 9471, 9472,
9473, 9474, 9475, 9476, 9477, 9478, 9479, 9480, 9481, 9482, 9483,
9484, 9151), class = "Date"), coutryA = c(0.187506004416315,
0.404519257417805, -0.107390371811605, -0.595968278805544, -0.529377240936012,
-0.222802921207767, 0.413182392872818, 0.689673026532298, -1.2768723266992,
-0.506308625809406, 0.113859233745174, -0.0963423819877653, 0.323987304768398,
1.63846917270538, 0.893233423250338, 0.297732439150487, 0.949323101836486,
-0.599518074708052, 0.366372319197032, -2.25734971953878, -0.190971733204918,
-0.0874143568874351, 1.46699645184047, 0.00702170238687361, 0.11221346278474,
-0.8060359607624, 0.340842350476532, 0.798838328074708, 0.449214745851041,
-0.664972890558734, 0.521830282184173, -1.35020467264521, -0.95240631225826,
1.25607320999849, 1.57018988549379, 0.99477900888445, -0.936218787378207,
1.48489932847779, 0.529222943794807, 0.0995675049147771, 0.477770516727839,
1.64567253670186, -0.0212651530684566, 0.558952796713992, 0.0409979382929057,
0.428675380654606, 0.0919422583362682, -0.819694497340459, 1.23998830450888,
0.607498144489643, -1.27724580163097, 1.41634774644371, -0.579094515769707,
2.02039606694223, 0.0740478208705996, -1.69826944583929, -0.321482399813063,
-0.489198601400924, 2.0066750686302, -1.90624857752322, 0.46762405849973,
1.31264724137396, -0.0473627194710677, 0.141362267796145, 0.329709761206515,
0.518454586458572, -1.39489985851779, -0.388303591187678, -0.668922704543522,
0.0735115674875065, 1.30737242978235, 0.198503397980751, 0.257831448122427,
-1.31173539205588, -1.45147941969116, 0.359725782295977, 0.612882118056585,
-0.0733768753346202, -0.508349204402508, 1.35776663767231, 0.997807735669086,
-1.41717534266382, -0.894170593324238, -0.68578120845151, -0.211509378018794,
0.436738904337909, -1.46932152770435, 0.0817388759874159, -0.0389350881653141,
0.709198476466861, -0.963669144724435, -0.548607422521798, -0.896886885575286,
0.322231150840934, 1.37327611339939, 0.0310213133870952, 0.796577750757324,
-0.2010067423637, -0.241723752424226, 1.37547329580654, -1.15382202538982,
0.101454200596915, 0.273663839664217, 1.8315140887841, 1.86096518756473,
-0.536393730924719, -0.45845011727266, 1.10226256157127, -0.385596991265563,
3.20218061566932, -1.25865250042183, -0.13613128784276, 0.483329357746514,
-0.597187329618306, 0.710977603908319, -1.07945708269043, -0.477626236401394,
1.51034914684104, 2.35886426985999, -0.0250526828683629, -0.29439443478131,
0.665774016744828, 0.464027472251246, 0.226658374792016, -0.802597030454373,
0.825517059805602, -1.11293193130819, -1.27677400513873, 1.60776237113347,
1.12490009531342, 0.95767047134623, 0.0475745549797055, -0.0591587460876868,
-0.690617365240342, -1.62111622938126, -1.3545210707469, 1.8607927043106,
0.764367674339969, 1.49261525602638, 0.549570728337346, -1.29658399741794,
-1.6289903797869, 0.00573336252135834, 0.0300702149640632, 0.440810830115721,
0.663568666361326, -0.126685900835146, -0.00221628368438927,
0.815321995886579, -0.499280888368945, -0.271814047751667, -0.071025546459042,
1.73165491816826, -0.0294770299043331, 0.833605607221529, -0.670108794857159,
-0.303323318026829, 1.29039844459134, -0.818806702120603, -0.445515595649677,
-0.0128796557666887, 0.320923705586147, 0.230597275812536, -1.54009153212366,
-0.294702981688559, 0.581209734391958, 0.121384768986639, 0.502914098451111,
-1.59018268505718, -0.635101104166451, 1.48005776676403, -0.25631761189957,
0.171947814411552, 0.444646195980014, 0.172655758440111, -0.00432159794094836,
-0.549321974240026, 0.585055026451421, -1.22813371480849, 0.846807540195381,
0.319629441352597, 0.393525732059709, -1.40275675444594, 1.11062585584811,
0.214809571213853, -0.636432711800391, -0.283087127251573, -1.46385553207618,
0.436928676930225, -1.34231945433777, 0.451281957595763, -0.523155001924496,
-2.69416779107545, 1.5513477373689, 0.989632029400036, 1.34636075948993,
0.346147428691405, -0.464527560160041, 0.337233933370495, 1.11331396366389,
-1.00060600083316, -0.734784444487169, 1.40476315358621, 1.01671092179193,
-0.0144306250829694, -0.923555930346906, -1.02275966525015, 0.619422010219383,
0.603484309754755, -0.774553813657576, 0.0932792545556387, -0.651884521428279,
-0.61965612647073, -1.22104834441579, -1.31439612639271, -2.87707752518163,
-0.0343801084491906, -0.640678302378492, -1.38653452986558, 0.884963139028743,
-0.657454283462004, 0.462842665244993, -0.20881674837534, 0.6345884135548,
0.707165108434729, -0.162090928425892, -0.998662309785188, 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,
0.179506648255554, 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,
10.6599997541876, 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, 11.2706666647314, 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,
10.3092266661814, 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, 9.25815555686085, 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.patra using gmail.com>
Sent: Friday, September 13, 2019 3:59 PM
To: PIKAL Petr <mailto:petr.pikal using precheza.cz>; r-help mailing list <mailto:r-help using 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
https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5& 09/13/19, 07:22:53 PM
On Fri, Sep 13, 2019 at 7:03 PM PIKAL Petr <mailto:petr.pikal using 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-bounces using r-project.org> On Behalf Of Subhamitra
> Patra
> Sent: Friday, September 13, 2019 3:20 PM
> To: Jim Lemon <mailto:drjimlemon using gmail.com>; r-help mailing list <r-help using r-
> 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 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 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.
>
>
>
>
>
>
>
> [image: Mailtrack]
> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_ca
> mpaign=signaturevirality5&>
> Sender
> notified by
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> 09/13/19,
> 06:41:41 PM
>
> On Fri, Sep 13, 2019 at 4:24 PM Jim Lemon <mailto:drjimlemon using gmail.com> 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 <
> > mailto:subhamitra.patra using 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 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,] 29.70000 30.43333
> > [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
> >
>
>
> --
> *Best Regards,*
> *Subhamitra Patra*
> *Phd. Research Scholar*
> *Department of Humanities and Social Sciences* *Indian Institute of
> Technology, Kharagpur*
> *INDIA*
>
> [[alternative HTML version deleted]]
>
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Best Regards,
Subhamitra Patra
Phd. Research Scholar
Department of Humanities and Social Sciences
Indian Institute of Technology, Kharagpur
INDIA
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