如何在排除 NaN 的同时计算矩阵列表的元素平均值答案

【问题标题】：How to calculate element-wise mean of a list of matrices while excluding NaNs如何在排除 NaN 的同时计算矩阵列表的元素平均值
【发布时间】：2021-07-13 00:39:27
【问题描述】：

假设我有矩阵列表。您将如何计算另一个矩阵，其元素是矩阵列表的元素均值，考虑和排除 NaN？假设我们有一个矩阵列表： )

> A
     [,1] [,2] [,3]
[1,]   1    4    7
[2,]   2    5    8
[3,]   3    6    9

> B
     [,1] [,2] [,3]
[1,]   2    5    8
[2,]  NaN  NaN  NaN
[3,]   4    7   10

> C
     [,1] [,2] [,3]
[1,]   3    3    6
[2,]   2    3    7
[3,]  NaN  NaN  NaN

> my.list <- list(A, B, C)

我想要的输出是：

     [,1] [,2] [,3]
[1,]   2    4    7
[2,]   2    4   7.5
[3,]  3.5  6.5  9.5

如您所见，我希望计算列表的元素平均值，但不包括列表中矩阵的 NaN 值。因此，例如输出的元素 [1,1] 是 (1 + 2 + 3) / 3 = 2，但输出的元素 [3,1] 是 (3 + 4) / 2 = 3.5，因为我们排除了NaN 在矩阵 B 中的那个位置。知道如何计算吗？

【问题讨论】：

标签： r

【解决方案1】：

转换为array，然后使用apply循环，得到mean

 apply(simplify2array(my.list), c(1, 2), FUN = mean, na.rm = TRUE)

-输出

      [,1] [,2] [,3]
[1,]  2.0  4.0  7.0
[2,]  2.0  4.0  7.5
[3,]  3.5  6.5  9.5

或者可以循环一维并使用rowMeans

apply(simplify2array(my.list), 2, rowMeans, na.rm = TRUE)
     [,1] [,2] [,3]
[1,]  2.0  4.0  7.0
[2,]  2.0  4.0  7.5
[3,]  3.5  6.5  9.5

或者另一种选择是

library(dplyr)
library(purrr)
library(data.table)
 map_dfr(my.list, as.data.frame, .id = 'grp') %>% 
   group_by(grp = rowid(grp)) %>% 
   summarise(across(everything(), mean, na.rm = TRUE), 
       .groups = 'drop') %>%
   select(-grp)
# A tibble: 3 x 3
     V1    V2    V3
  <dbl> <dbl> <dbl>
1   2     4     7  
2   2     4     7.5
3   3.5   6.5   9.5

数据

my.list <- list(structure(1:9, .Dim = c(3L, 3L)), structure(c(2, NaN, 4, 
5, NaN, 7, 8, NaN, 10), .Dim = c(3L, 3L)), structure(c(3, 2, 
NaN, 3, 3, NaN, 6, 7, NaN), .Dim = c(3L, 3L)))

【讨论】：

【解决方案2】：

  # input prepare
  A= matrix(1:9,nrow=3,ncol=3)
  B = matrix(2:10,nrow=3,ncol=3)
  B[2,] = NaN
  C =matrix(c(3,2,NaN,3,3,NaN,6,7,NaN),nrow=3,ncol=3)

  # flatten every matrix as a column of rowMeans big matrix
  # so rowMeans do element-wise process
  mat.as.vec.list = lapply(list(A,B,C), as.vector)
  bind.by.col= do.call(cbind, mat.as.vec.list)

  
  mean.by.row = rowMeans(bind.by.col, na.rm = T)

  #convert back to matrix
  dim(mean.by.row) = dim(A)
  mean.by.row

     [,1] [,2] [,3]
[1,]  2.0  4.0  7.0
[2,]  2.0  4.0  7.5
[3,]  3.5  6.5  9.5

【讨论】：