awk 求和多列

如何计算所有单独列（115 列）的总和？

输入.txt

1st,2nd,3rd,4th,5th,Till-115thColumn
51,34,27,67,88,99
56,39,32,72,93,104
66,49,42,82,103,114

输出.txt

1st,2nd,3rd,4th,5th,Till-115thColumn
173,122,101,221,284,317

我尝试了这个命令：

awk -F"," 'BEGIN {sum=0; OFS=","} {for (i=1 i<=NF; i++) a[i]+=$i } END {for (i in a) print a[i]}' Input.txt

但我没有得到所需的输出。

这可以是一种方式：

awk 'BEGIN{FS=OFS=","}
     NR==1{print}
     NR>1{for (i=1;i<=NF;i++) a[i]+=$i}
     END{for (i=1;i<=NF;i++) printf a[i] OFS; printf "\n"}' file

它将逗号设置为输入/输出字段分隔符，然后将每列的总和存储在数组中a[]。最后，它循环遍历结果并打印它们。笔记NR==1是用户打印标题。

对于您给定的输入，它返回：

$ awk 'BEGIN{FS=OFS=","} NR==1{print} NR>1{for (i=1;i<=NF;i++) a[i]+=$i} END{for (i=1;i<=NF;i++) printf a[i] OFS; printf "\n"}' file
1st,2nd,3rd,4th,5th,Till-115thColumn
173,122,101,221,284,317,

为什么你的脚本不起作用？

因为你错过了一个; in the for宣言：

awk -F"," 'BEGIN {sum=0; OFS=","} {for (i=1 i<=NF; i++) a[i]+=$i } END {for (i in a) print a[i]}' a
awk: line 1: syntax error at or near )

所以这使得：

$ awk -F"," 'BEGIN {sum=0; OFS=","} {for (i=1; i<=NF; i++) a[i]+=$i } END {for (i in a) print a[i]}' file
                                             ^
174
124
104
225
289
317