Transpose from row to column - bash

Question

Traspose from line to column is the objetive, taking in consideration the first column, which is the date
Input file
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
Output desired
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
Please advise, thanks in advance.

Answer 1

This code seems to do exactly what you need:
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
It collects rows until the timestamp changes, then prints the transpose of the block with keeping the same timestamp. The number of fields in the input must be the same within each block so that this code behaves correctly.