Question

考虑到第一列即日期

，从行到列的过渡是强制性的

输入文件

72918,111000009,111000009,111000009,111000009,111000009,111000009,111000009,111000009,111000009
72918,2356,2357,2358,2359,2360,2361,2362,2363,2364
72918,0,0,0,0,0,0,0,0,0
72918,0,0,0,0,0,0,1,0,0
72918,1496,1502,1752,1752,1752,1752,1751,974,972
73018,111000009,111000009,111000009,111000009,111000009,111000009,111000009,111000009,111000009
73018,2349,2350,2351,2352,2353,2354,2355,2356,2357
73018,0,0,0,0,0,0,0,0,0
73018,0,0,0,0,0,0,0,0,0
73018,1524,1526,1752,1752,1752,1752,1752,256,250

所需的输出

72918,111000009,2356,0,0,1496
72918,111000009,2357,0,0,1502
72918,111000009,2358,0,0,1752
72918,111000009,2359,0,0,1752
72918,111000009,2360,0,0,1752
72918,111000009,2361,0,0,1752
72918,111000009,2362,0,1,1751
72918,111000009,2363,0,0,974
72918,111000009,2364,0,0,972
73018,111000009,2349,0,0,1524
73018,111000009,2350,0,0,1526
73018,111000009,2351,0,0,1752
73018,111000009,2352,0,0,1752
73018,111000009,2353,0,0,1752
73018,111000009,2354,0,0,1752
73018,111000009,2355,0,0,1752
73018,111000009,2356,0,0,256
73018,111000009,2357,0,0,250

请告知，谢谢。

Answer 1

此代码似乎完全符合您的要求：

awk -F, '
 func init_block() {ts=$1;delete a;cnt=0;nf0=NF}
 func dump_block() {for(f=2;f<=nf0;f+=1){printf("%s",ts);for(r=1;r<=cnt;r+=1){printf(",%s",a[r,f])};print ""}}
 BEGIN{ts=-1}
 ts<0{init_block()}
 ts!=$1{dump_block();init_block()}
 {cnt+=1;for(f=1; f<=NF; f++) a[cnt,f]=$f}
 END{dump_block()}' <input.txt >output.txt

它将收集行，直到时间戳更改，然后在保持相同时间戳的情况下打印块的转置。每个块中输入的字段数必须相同，以便此代码正确运行。

从行到列转置

1 个答案: