数组中每12个矩阵的元素平均值，重复一个沿着12次，无for循环

dffbzjpn 于 2023-04-18 发布在其他

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我有一个维度数组[360，180，396]。这些维度是经度、纬度和33年月度数据的月-年。元素是该纬度/经度的百分比。
从这里，我想做一个汇总数组，我将在以后的分析中使用，而不默认使用for循环。我想得到所有33年的每个月数据的平均值，然后是所有年份的年平均值。这是我为包含数据而做的汇总数组的空白。

mca <- array(data = NA, 
             dim = c(360,180,13), 
             dimnames = list(lon, 
                             lat, 
                             c(month.abb, "Ann")))

下面是本例中较小的测试输入和输出数组

#input

set.seed(42)
smallin <- array(data = rnorm(n = 600, mean = 60, sd = 20),
               dim = c(5, 5, 24))

#output to fill
smallout <- array(data = NA, 
             dim = c(5,5,13), 
             dimnames = list(c("1", "2", "3", "4", "5"), 
                             c("-89.5", "-88.5", "-87.5", "-86.5", "-85.5"), 
                             c(month.abb, "Ann")))

基于我尝试的this question的第二个答案

jan <- apply(ca, c(seq(from = 1, to = 385, by = 12)), mean)

#also 

ind_jan <- c(seq(from = 1, to = 385, by = 12))
jan <- apply(ca, ind_jan, mean)

我认为这相当于

jan <- apply(smallin, c(seq(from = 1, to = 13, by = 12)), mean)

考虑到保证金，我需要把我想要的第三维平均，但收到了错误：
应用中的误差（ca，c（seq（from = 1，to = 385，by = 12）），mean）：'MARGIN'与dim（X）不匹配
我返回到上面的查询，realized margin = 1：2必须选择所有矩阵（维度1和2）。因此，使用它，我可以得到所有矩阵的平均值，它应该是我的输出数组[，，13]的百分比的年平均值。

smallout[,,13] <- apply(smallin, 1:2, mean)

但我还是不知道如何从1开始，然后从2开始，然后从3开始，平均每12个矩阵...
我已经阅读了apply文档，但发现它在这种情况下毫无帮助/令人费解。所有提出的建议问题似乎都是用Python（或其他语言）编写的。
我也不确定我是否可以一次完成所有这些，或者像上面一样通过索引将一个矩阵传递到输出数组。
我能想到的最接近的是

ind_jan <- c(seq(from = 1, to = 13, by = 12))
smallout[,,1] <- apply(smallin[,,c(ind_jan)], 1:2, mean)

对数组中的每个输出矩阵重复。有没有更少手动/更有效/更好的方法？

来源：https://stackoverflow.com/questions/75979986/element-wise-mean-of-every-12th-matrix-in-array-repeated-one-sequence-along-12

3条答案

按热度按时间

vbopmzt11#

考虑这个简化的数组A（参见下面的数据）。

str(A)
# int [1:2, 1:3, 1:6] 1 1 1 1 1 1 2 2 2 2 ...

我们可以使用sapply来“循环”这些年，使用simplify='array'选项来返回一个年平均值数组。

yrs <- seq_len(dim(A)[3]/nm)
sapply(yrs, \(i) apply(A[, , 1:nm + i - 1], 1:2, mean), simplify='array')
# , , 1
# 
#      [,1] [,2] [,3]
# [1,]    2    2    2
# [2,]    2    2    2
# 
# , , 2
# 
#      [,1] [,2] [,3]
# [1,]    2    2    2
# [2,]    2    2    2

以及相应的历年每月平均数：

mnt <- seq_len(nm)
sapply(mnt, \(i) apply(A[, , i], 1:2, mean), simplify='array')
# , , 1
# 
#      [,1] [,2] [,3]
# [1,]    1    1    1
# [2,]    1    1    1
# 
# , , 2
# 
#      [,1] [,2] [,3]
# [1,]    2    2    2
# [2,]    2    2    2
# 
# , , 3
# 
#      [,1] [,2] [,3]
# [1,]    3    3    3
# [2,]    3    3    3

数据：*

nm <- 3  ## no. "months"  ## actually 12 months in real years
ny <- 2  ## no. "years"  ## in your case 33
A <- array(rep(1:nm, each=nm*ny), c(2, 3, nm*ny))  ## think this is your `ca`

赞(0）回复(0）举报 2023-04-18

4nkexdtk2#

您可以通过拆分最后一个维度（包含月和年）来将月和年的维度分开，从而向数组中添加另一个维度。

i <- dim(smallin)
dim(smallin) <- c(i[1:2], 12L, i[3]/12L)

有了这个，你可以得到所有年份每个月的平均值：

apply(smallin, 1:3, mean)
#, , 1
#
#         [,1]     [,2]     [,3]     [,4]     [,5]
#[1,] 73.66338 58.35988 72.33907 62.19628 52.08766
#[2,] 61.95544 79.93891 75.27725 49.30859 44.07820
#[3,] 64.02119 68.98285 35.76780 35.06961 58.79089
#[4,] 73.67935 67.72028 50.90479 23.22819 72.14434
#[5,] 62.57796 59.03798 64.53486 83.65987 97.04576
#
#...
#
#, , 12
#
#         [,1]     [,2]     [,3]     [,4]     [,5]
#[1,] 83.55254 68.77645 48.88358 52.99573 56.82992
#[2,] 83.47723 39.02472 95.08051 65.97988 54.00097
#[3,] 47.59936 36.93396 38.35189 57.86126 83.99976
#[4,] 73.00906 53.71818 36.93229 80.85843 39.27094
#[5,] 81.67441 64.50031 62.71359 56.27758 54.01388

单年平均值：

apply(smallin, c(1,2,4), mean)
#, , 1
#
#         [,1]     [,2]     [,3]     [,4]     [,5]
#[1,] 60.77253 60.15417 54.71206 67.31820 62.05012
#[2,] 56.60298 59.14604 73.17469 57.66912 53.36540
#[3,] 56.52924 56.31096 58.73874 67.47850 59.06819
#[4,] 67.75999 56.45636 49.43743 55.14660 65.46497
#[5,] 60.28056 62.17656 55.08681 54.15788 60.05240
#
#, , 2
#
#         [,1]     [,2]     [,3]     [,4]     [,5]
#[1,] 60.55035 65.21223 59.92112 59.75500 69.77088
#[2,] 60.89782 54.59722 55.17699 59.06815 60.03906
#[3,] 58.85733 54.02893 47.31326 63.10434 59.56569
#[4,] 60.96362 61.82648 55.45109 54.50272 45.21176
#[5,] 59.94452 54.31497 60.64839 64.65777 80.86525

所有年份的年平均值：

apply(smallin, 1:2, mean)
#         [,1]     [,2]     [,3]     [,4]     [,5]
#[1,] 60.66144 62.68320 57.31659 63.53660 65.91050
#[2,] 58.75040 56.87163 64.17584 58.36864 56.70223
#[3,] 57.69329 55.16994 53.02600 65.29142 59.31694
#[4,] 64.36180 59.14142 52.44426 54.82466 55.33836
#[5,] 60.11254 58.24577 57.86760 59.40782 70.45883

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xmd2e60i3#

我相信有一个更好的方法，如果有人有它，我仍然热衷于学习，但下面的结果是工作，一旦我弄清楚如何做索引，以选择每个月的数据，以采取平均使用应用。

mca[,,1] <- apply(ca[,,c(seq(from = 1, to = 396, by = 12))], 1:2, mean)
mca[,,2] <- apply(ca[,,c(seq(from = 2, to = 396, by = 12))], 1:2, mean)
mca[,,3] <- apply(ca[,,c(seq(from = 3, to = 396, by = 12))], 1:2, mean)
mca[,,4] <- apply(ca[,,c(seq(from = 4, to = 396, by = 12))], 1:2, mean)
mca[,,5] <- apply(ca[,,c(seq(from = 5, to = 396, by = 12))], 1:2, mean)
mca[,,6] <- apply(ca[,,c(seq(from = 6, to = 396, by = 12))], 1:2, mean)
mca[,,7] <- apply(ca[,,c(seq(from = 7, to = 396, by = 12))], 1:2, mean)
mca[,,8] <- apply(ca[,,c(seq(from = 8, to = 396, by = 12))], 1:2, mean)
mca[,,9] <- apply(ca[,,c(seq(from = 9, to = 396, by = 12))], 1:2, mean)
mca[,,10] <- apply(ca[,,c(seq(from = 10, to = 396, by = 12))], 1:2, mean)
mca[,,11] <- apply(ca[,,c(seq(from = 11, to = 396, by = 12))], 1:2, mean)
mca[,,12] <- apply(ca[,,c(seq(from = 12, to = 396, by = 12))], 1:2, mean)
mca[,,13] <- apply(ca, 1:2, mean)

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数组中每12个矩阵的元素平均值，重复一个沿着12次，无for循环

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