[R] Query about calculating the monthly average of daily data columns
Subhamitra Patra
@ubh@m|tr@@p@tr@ @end|ng |rom gm@||@com
Sun Oct 20 14:09:00 CEST 2019
Dear Sir,
Thank you very much for your suggestions.
Due to certain inconveniences, I was unable to work on your suggestions.
Today I worked on both suggestions and got the result that I really wanted
that monthly averages for each country.
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.
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 0.066888865
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 0.000233338
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
112 19940328 0.000228713
112 19940329 0.000228894
112 19940330 0.000229255
112 19940331 0.000231003
112 19940401 0.000796567
112 19940404 0.000790668
112 19940405 0.00078195
112 19940406 0.000780475
112 19940407 0.000228355
112 19940408 0.000781723
112 19940411 0.000775741
112 19940412 0.000226647
112 19940413 0.000778876
112 19940414 0.000777336
112 19940415 0.000775253
112 19940418 0.000226362
112 19940419 0.000779554
112 19940420 0.000774824
112 19940421 0.000225582
112 19940422 0.000225724
112 19940425 0.000773361
112 19940426 0.0002256
112 19940427 0.000776416
113 19940428 0.000280542
113 19940429 0.000964148
113 19940103 0.000962654
113 19940104 0.000281768
113 19940105 0.000962219
113 19940106 0.000961965
113 19940107 0.000958602
113 19940110 0.000280056
113 19940111 0.000956348
113 19940112 0.000952171
113 19940113 0.000948176
113 19940114 0.000275607
113 19940117 0.000275773
113 19940118 0.000276738
113 19940119 0.000947068
113 19940120 0.000940959
113 19940121 0.000275224
113 19940124 0.000948489
113 19940125 0.000940076
113 19940126 0.0009309
113 19940127 0.000269955
113 19940128 0.000270328
113 19940131 0.000924234
113 19940201 0.000924038
113 19940202 0.000269088
113 19940203 0.000270247
113 19940204 0.000270562
113 19940207 0.00092656
113 19940208 0.000921819
113 19940209 0.000920361
113 19940210 0.000268958
113 19940211 0.000924758
113 19940214 0.000266768
113 19940215 0.000911325
113 19940216 0.000909294
113 19940217 0.000905887
113 19940218 0.000262919
113 19940221 0.000262978
113 19940222 0.000263189
113 19940223 0.000904439
113 19940224 0.000263512
113 19940225 0.000906184
113 19940228 0.000265198
113 19940301 0.000906126
113 19940302 0.000264357
113 19940303 0.000265392
113 19940304 0.000912495
113 19940307 0.000910641
113 19940308 0.000266143
113 19940309 0.000910113
113 19940310 0.000909277
113 19940311 0.000905056
113 19940314 0.00090285
113 19940315 0.000898831
113 19940316 0.000896118
113 19940317 0.000261294
113 19940318 0.000892563
113 19940321 0.000890852
113 19940322 0.00088639
113 19940323 0.000258509
113 19940324 0.000260286
113 19940325 0.000889354
113 19940328 0.000888373
113 19940329 0.000885049
113 19940330 0.000259116
113 19940331 0.000259474
113 19940401 0.000260316
113 19940404 0.000897493
113 19940405 0.000894592
113 19940406 0.000260435
113 19940407 0.000260989
113 19940408 0.000262061
113 19940411 0.000262262
113 19940412 0.000263604
113 19940413 0.000908682
113 19940414 0.000265348
113 19940415 0.000265637
113 19940418 0.00026862
113 19940419 0.000918882
113 19940420 0.000909904
113 19940421 0.000901725
113 19940422 0.000900062
113 19940425 0.000893547
113 19940426 0.000260899
Here, the 1st column is the name of the countries panel which I identified
in number, 2nd is the date column arranged in yyyy-mm-dd format, and 3rd is
the daily variable for each country for 4 months of 1994. I need to take
the monthly average of spread variable for each country (i.e. noted as 111,
112, and 113) in the above example. In short, my monthly spread variable
should be sorted on the basis of both PERMNO (i.e. country identifier), and
months of the particular year.
Please educate me that in this data format, how the average code can be
written?
Thank you very much.
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10/20/19,
05:31:23 PM
On Mon, Sep 16, 2019 at 4:31 PM PIKAL Petr <petr.pikal using precheza.cz> wrote:
> 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,
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> 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,
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> 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.
>
>
>
>
>
>
> https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&
> 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
> > Mailtrack
> > <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_ca
> > mpaign=signaturevirality5&>
> > 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]]
> >
> > ______________________________________________
> > mailto:R-help using 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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>
>
> --
> Best Regards,
> Subhamitra Patra
> Phd. Research Scholar
> Department of Humanities and Social Sciences
> Indian Institute of Technology, Kharagpur
> INDIA
> ______________________________________________
> R-help using 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.
>
--
*Best Regards,*
*Subhamitra Patra*
*Phd. Research Scholar*
*Department of Humanities and Social Sciences*
*Indian Institute of Technology, Kharagpur*
*INDIA*
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