如果我们有三年的数据：
dat =（x1：x1096）
我想用这种方式计算平均值：
[x1+x366(the first day in the second year)+ x731(the first day in the third year)]/3
[x2+x367(the second day in the second year)+ x732(the second day in the third year)]/3
等到第365天：
[x365+x730(the last day in the second year)+ x1096(the last day in the third year)]/3
最后我会得到365 values。
dat= c(1:1096)
关于如何做到这一点的任何想法？
答案 0 :(得分：3)
data.table在这里非常方便:(即使基本R解决方案完全可行！）：
> set.seed(1)
> dat <- data.table(date=seq(as.Date("2010-01-01"), as.Date("2012-12-31"), "days"),
+ var=rnorm(1096))
> dat
date var
1: 2010-01-01 -0.626453811
2: 2010-01-02 0.183643324
3: 2010-01-03 -0.835628612
4: 2010-01-04 1.595280802
5: 2010-01-05 0.329507772
---
1092: 2012-12-27 0.711213964
1093: 2012-12-28 -0.337691156
1094: 2012-12-29 -0.009148952
1095: 2012-12-30 -0.125309208
1096: 2012-12-31 -2.090846097
> dat[, mean(var), by=list(month=month(date), mday(date))]
month mday V1
1: 1 1 -0.16755484
2: 1 2 0.59942582
3: 1 3 -0.44336168
4: 1 4 0.01297244
5: 1 5 -0.20317854
---
362: 12 28 -0.18076284
363: 12 29 0.07302903
364: 12 30 -0.01790655
365: 12 31 -0.87164859
366: 2 29 -0.78859794
2月29日结束，因为[.data.table当那天的小组是最后一个独特的组合（month(date)和mday(date)），导致它出现在2012年的第一次一旦你得到了你的结果，你可以分配钥匙，然后对表格进行排序：
> result <- dat[, mean(var), by=list(month=month(date), mday(date))]
> setkey(result, month, mday)
> result
month mday V1
1: 1 1 -0.16755484
2: 1 2 0.59942582
3: 1 3 -0.44336168
4: 1 4 0.01297244
5: 1 5 -0.20317854
---
362: 12 27 -0.60348463
363: 12 28 -0.18076284
364: 12 29 0.07302903
365: 12 30 -0.01790655
366: 12 31 -0.87164859
答案 1 :(得分：2)
也许是这样的？我尝试了一个比1:1096向量略小的例子 - 我每年使用5个值。
# the data, here 3 years with 5 values per year.
dat <- 1:15
# put your vector in a matrix
# by default, the matrix is filled column-wise
# thus, each column corresponds to a year, and each row to day of year
mm <- matrix(dat, ncol = 3)
# calculate row means
mm <- cbind(mm, rowMeans(mm))
mm
# [,1] [,2] [,3] [,4]
# [1,] 1 6 11 6
# [2,] 2 7 12 7
# [3,] 3 8 13 8
# [4,] 4 9 14 9
# [5,] 5 10 15 10
<强>更新
使用来自@ Michele的答案的相同（即base）'完整'数据来解释闰年的另一个set.seed(1)替代方案：
df2 <- aggregate(var ~ format(date, "%m-%d"), data = dat, FUN = mean)
head(df2)
# format(date, "%m-%d") var
# 1 01-01 -0.16755484
# 2 01-02 0.59942582
# 3 01-03 -0.44336168
# 4 01-04 0.01297244
# 5 01-05 -0.20317854
# 6 01-06 -0.55350137
答案 2 :(得分：2)
这是解决闰年问题的base解决方案：
# First your data
set.seed(1)
dat <- rnorm(1096) #Value for each day
day <- seq(as.Date("2010-01-01"), as.Date("2012-12-31"), "days") #Corresponding days
sapply(split(dat,format(day,"%m-%d")),mean)
01-01 01-02 01-03 01-04 01-05 01-06 01-07 01-08 01-09
-0.167554841 0.599425816 -0.443361675 0.012972442 -0.203178536 -0.553501370 0.563475994 -0.094459075 0.567263811
01-10 01-11 01-12 01-13 01-14 01-15 01-16 01-17 01-18
-0.325835336 -0.247226807 -0.272224241 0.171886332 -0.562604980 0.640473418 -0.209380261 0.709635402 -0.263715734
01-19 01-20 01-21 01-22 01-23 01-24 01-25 01-26 01-27
0.929096171 1.173422823 -0.197411808 -0.730959553 -0.277022971 -1.075673025 -0.494038031 -0.255709319 0.827062779
01-28 01-29 01-30 01-31 02-01 02-02 02-03 02-04 02-05
0.208963353 0.215192803 -0.118735162 0.141028516 0.703267761 -0.282852177 -0.297731589 -0.112031601 0.784073396
02-06 02-07 02-08 02-09 02-10 02-11 02-12 02-13 02-14
0.714499179 0.206640777 0.283234842 -0.255182989 -0.293285997 -0.761585755 0.443379228 1.138436815 -0.483004921
02-15 02-16 02-17 02-18 02-19 02-20 02-21 02-22 02-23
-0.692188333 0.701422889 0.677544133 -0.423576371 0.498868978 0.053960271 0.518228979 -0.250840385 -0.722647734
02-24 02-25 02-26 02-27 02-28 02-29 03-01 03-02 03-03
1.344507325 0.693403586 -0.226489715 -0.406929668 -0.171335064 -0.788597935 0.115894011 1.798749522 -0.502676829
03-04 03-05 03-06 03-07 03-08 03-09 03-10 03-11 03-12
0.244453933 -0.278023124 -0.817932086 -0.618472996 -0.842995408 -0.887451556 0.432459430 0.559562525 -0.516256302
03-13 03-14 03-15 03-16 03-17 03-18 03-19 03-20 03-21
0.392447923 0.191049834 -0.727128826 -0.261740657 -0.189455949 0.775326029 0.236835450 -0.266491426 -0.010319849
03-22 03-23 03-24 03-25 03-26 03-27 03-28 03-29 03-30
-0.949967889 -0.277676523 -0.556777524 -0.507373521 0.076952129 0.697147181 -0.416867359 -0.906909972 -0.231494410
03-31 04-01 04-02 04-03 04-04 04-05 04-06 04-07 04-08
-0.453616811 0.158367456 0.670354625 -0.285493660 -0.040162162 0.762953404 -0.388049908 1.079423205 -0.246508050
04-09 04-10 04-11 04-12 04-13 04-14 04-15 04-16 04-17
-0.215358691 -0.337611847 0.486368813 0.115883308 -0.282207017 0.614554509 0.531435739 1.063455284 -0.199968099
04-18 04-19 04-20 04-21 04-22 04-23 04-24 04-25 04-26
-0.080662691 -0.052822528 1.679629547 -1.341639141 0.986160744 0.468143827 0.029621883 -0.025910053 0.061093981
04-27 04-28 04-29 04-30 05-01 05-02 05-03 05-04 05-05
-0.387992910 -0.917561336 0.161867089 0.874549452 0.866708261 0.048304939 -1.209756576 -0.825689257 -0.176605953
05-06 05-07 05-08 05-09 05-10 05-11 05-12 05-13 05-14
-0.381265758 0.419105218 -0.440418731 -0.293923704 1.427366374 -0.020773738 -0.358619841 -0.294738750 -0.269765222
05-15 05-16 05-17 05-18 05-19 05-20 05-21 05-22 05-23
0.277361477 -0.505072373 -0.765572754 -0.493223200 -0.253297588 0.902399037 0.007676731 -0.273059247 -0.784701888
05-24 05-25 05-26 05-27 05-28 05-29 05-30 05-31 06-01
0.063532445 -0.681369105 -1.034300631 0.689037398 -0.209889037 -0.535166412 -0.994984541 0.438795387 -0.167806908
06-02 06-03 06-04 06-05 06-06 06-07 06-08 06-09 06-10
0.079629296 -0.063908968 0.484892252 -0.922112094 0.978258635 -0.790949931 -0.303356059 0.681310315 -0.512109593
06-11 06-12 06-13 06-14 06-15 06-16 06-17 06-18 06-19
0.337126461 0.526594905 0.742784618 -0.163083706 0.027435241 0.709630255 -1.144544436 -0.374108608 0.102721328
06-20 06-21 06-22 06-23 06-24 06-25 06-26 06-27 06-28
0.577569049 0.224528626 0.206667019 0.392007605 -0.557974448 0.068685789 0.460201512 1.101334023 0.035838933
06-29 06-30 07-01 07-02 07-03 07-04 07-05 07-06 07-07
0.873903793 -0.586658280 -0.395094221 0.303312480 -0.631756580 0.088308518 0.046129624 0.642985443 -0.615693218
07-08 07-09 07-10 07-11 07-12 07-13 07-14 07-15 07-16
0.372776652 0.453644860 0.466905164 -0.526930331 -0.351139797 0.250132593 -0.881175203 -1.090136940 0.409708249
07-17 07-18 07-19 07-20 07-21 07-22 07-23 07-24 07-25
0.206436178 0.056134229 -0.057927905 0.807127686 0.423170493 -0.325181464 -0.053593067 0.261438323 0.520617153
07-26 07-27 07-28 07-29 07-30 07-31 08-01 08-02 08-03
0.053800701 0.326492953 -0.471839346 0.438963172 0.499502012 0.620917026 0.619923442 -1.422177067 0.212056501
08-04 08-05 08-06 08-07 08-08 08-09 08-10 08-11 08-12
0.497181456 0.703607380 -0.054104370 0.931407619 0.545759743 -0.323646872 0.127371847 0.017697636 -0.033060879
08-13 08-14 08-15 08-16 08-17 08-18 08-19 08-20 08-21
-0.583034512 0.824859915 -0.019064796 -0.226035270 -1.026526076 -0.882074229 -0.079167867 -2.073168805 0.378121135
08-22 08-23 08-24 08-25 08-26 08-27 08-28 08-29 08-30
-0.004516521 -0.661187139 0.339497500 -0.042210229 0.026970585 0.431653210 0.104619786 0.149562359 -0.473661114
08-31 09-01 09-02 09-03 09-04 09-05 09-06 09-07 09-08
-0.235250025 -0.624645896 0.141205349 -0.485201261 0.097633486 0.462059099 -0.500082678 1.386621118 -0.070895288
09-09 09-10 09-11 09-12 09-13 09-14 09-15 09-16 09-17
-0.126090048 -0.371028573 -0.010479329 0.192555782 0.025085776 -1.410061589 1.046273116 0.938254501 -0.072773342
09-18 09-19 09-20 09-21 09-22 09-23 09-24 09-25 09-26
-0.272947102 0.279357832 0.172702983 0.219560592 0.922992902 -0.612832806 -0.450896711 -1.134353324 -0.336199724
09-27 09-28 09-29 09-30 10-01 10-02 10-03 10-04 10-05
-0.459242718 0.049888664 0.079844541 -0.058636867 0.581553407 -0.315806482 -0.163864166 -1.513984901 0.069093641
10-06 10-07 10-08 10-09 10-10 10-11 10-12 10-13 10-14
-0.325709367 0.114176104 -0.470510646 -0.393891025 -0.659031395 -0.224657523 -0.336803115 -0.510526475 -0.941899166
10-15 10-16 10-17 10-18 10-19 10-20 10-21 10-22 10-23
0.559205646 0.346629848 0.310935589 -0.851962382 0.387930834 0.505692192 -0.738722861 0.410302113 -0.181359914
10-24 10-25 10-26 10-27 10-28 10-29 10-30 10-31 11-01
0.831105889 -0.398852239 -0.164535170 -0.870295447 0.057609116 -1.058556114 0.809784093 0.188277796 1.432543613
11-02 11-03 11-04 11-05 11-06 11-07 11-08 11-09 11-10
0.040680316 0.711553107 0.565285429 -0.829181807 0.455487776 -0.037182199 -0.644669824 -0.704611643 0.491631958
11-11 11-12 11-13 11-14 11-15 11-16 11-17 11-18 11-19
-0.051188454 0.963031185 -0.511791970 0.193671830 -0.333065645 -0.176479500 0.367566807 -0.056534518 1.391773053
11-20 11-21 11-22 11-23 11-24 11-25 11-26 11-27 11-28
0.162741879 -0.269991630 0.866532461 -0.352034768 -0.028515790 -0.671437717 -0.393703641 0.394041604 -0.959721458
11-29 11-30 12-01 12-02 12-03 12-04 12-05 12-06 12-07
-0.187149463 0.203037321 -0.824439261 -0.081277243 0.361409692 -0.300022665 -0.067589145 -0.265877741 -0.474834675
12-08 12-09 12-10 12-11 12-12 12-13 12-14 12-15 12-16
-0.903405316 0.026396956 0.930117145 -0.489879346 -0.481598661 0.122388492 0.042287328 -0.160328704 0.777249363
12-17 12-18 12-19 12-20 12-21 12-22 12-23 12-24 12-25
-0.359802827 0.252189848 0.754686655 -0.012767780 0.683605939 0.782528149 -0.786087093 0.751560196 -0.610885984
12-26 12-27 12-28 12-29 12-30 12-31
0.203570612 -0.603484627 -0.180762839 0.073029026 -0.017906554 -0.871648586
这个想法是根据一年中的某一天（％d-％m）进行分割，并做每个小组的平均值。
编辑 - Michele （我认为最好将此答案完全改为base相关，而不是我的：
如果上面的矢量用于创建data.frame，那么这个解决方案是很好的选择：
dat <- data.frame(date=day, var=dat)
> ddply(dat, .(day=format(date,"%m-%d")), summarise, result=mean(var))
day result
1 01-01 -0.167554841
2 01-02 0.599425816
3 01-03 -0.443361675
4 01-04 0.012972442
5 01-05 -0.203178536
6 01-06 -0.553501370
注意：抱歉，它实际上使用的是plyr个包，但它仍然使用data.frame而ddply可能会被来自by个包的base替换。