劈裂CSV字符串列“日期时间”到“日期”，“时间”文件

Question

我有以下形式的csv文件：

$ head purchases.csv 
id,userID,itemID,price,platform,day 
1,9132,id_005,3600,2,2014-10-30 17:29:46 
2,67894,id_005,3000,1,2015-04-23 21:22:55 
3,272780,id_004,1000,1,2014-11-27 16:58:30 
4,302396,id_001,100,1,2014-12-11 08:35:07 

现在，我想改变CSV的最后一列。目前，它的格式为day列2014-10-30 17:29:46即日期和时间之间有空格。但我想这列分成两列day和time使变更后的csv文件就变成了：

$ head purchases.csv 
id,userID,itemID,price,platform,day,time 
1,9132,id_005,3600,2,2014-10-30,17:29:46 
2,67894,id_005,3000,1,2015-04-23,21:22:55 
3,272780,id_004,1000,1,2014-11-27,16:58:30 

如何从终端做呢？

Answer 1

在$ 6使用split：

$ awk -v OFS=\, -F\, 'NR==1{print $0,"time";next} {split($6,a," "); print $1,$2,$3,$4,$5,a[1],a[2]}' test.in 
id,userID,itemID,price,platform,day,time 
1,9132,id_005,3600,2,2014-10-30,17:29:46 
2,67894,id_005,3000,1,2015-04-23,21:22:55 
3,272780,id_004,1000,1,2014-11-27,16:58:30 
4,302396,id_001,100,1,2014-12-11,08:35:07 

或者你可以使用gsub，只是用逗号代替空间：

$ awk -v OFS=\, -F\, 'NR==1{print $0,"time";next} {gsub(/ /,",",$6); print $0}' test.in 

Answer 2

James Brown's answer是有帮助的，但是硬编码来修改，而列还假设它是最后。

一些简单的调整概括解决方案：

awk -v ndx=6 -F, 'NR==1 {sub(/$/, ",time", $ndx); print; next} sub(" ", ",", $ndx)' \ 
    purchases.csv