矩阵的行/列的R部分和 [英] R partial sum of rows/columns of a matrix

问题描述

有没有办法从K NxK N方阵中获得K NxN矩阵和NxK N?

Is there a way to obtain a KNxN matrix and a NxKN from a KNxKN square matrix?

K:数据中的行业数. N:数据中的国家/地区数.

K: number of industries in the data. N: number of countries in the data.

每个行/名称对3个代码国家/地区名称进行编码，例如美国，分隔符.c和行业编号，例如USA.c1.

Each row/colname encodes the 3 code country name e.g. USA, the seperator .c and the number of the industry, e.g. USA.c1 .

我尝试使用colSums和rowSums，但是函数仅返回一个数字，而不是N个数字.

I tried to use colSums and rowSums but the functions return only one number instead of N numbers.

由2个行业和2个国家组成的矩阵的最小工作示例

Minimal working example for a matrix of 2 industries and 2 countries

           BEL.c30 BEL.c31 CAN.c25 CAN.c26
   BEL.c30   11844      14       1       0
   BEL.c31      85     227       0       0
   CAN.c25       0       0    1037       1
   CAN.c26       0       0      43    1113

第一个矩阵应如下所示(两个国家/地区的行总和):

The first matrix should like this (row sum across each of the two country ) :

               BEL     CAN
   BEL.c30    11858      1       
   BEL.c31      312     227              
   CAN.c25       0      1038  
   CAN.c26       0      1156

第二个矩阵应如下所示(每个国家/地区的列总和):

The second matrix should look like this (column sum for each country ) :

              BEL.c30 BEL.c31 CAN.c25 CAN.c26
      BEL    11929      241     1     0
      CAN       0         0     1080  1114

推荐答案

这里是一个选择:

do.call(rbind, tapply(as.data.frame(m), sub("\\.c.*", "", colnames(m)), colSums))
#    BEL.c30 BEL.c31 CAN.c25 CAN.c26
#BEL   11929     241       1       0
#CAN       0       0    1080    1114

do.call(cbind, tapply(as.data.frame(t(m)), sub("\\.c.*", "", colnames(m)), colSums))
#          BEL  CAN
#BEL.c30 11858    1
#BEL.c31   312    0
#CAN.c25     0 1038
#CAN.c26     0 1156

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