Subtract fixed number of days from date column using awk and add it to new column - csv

Question

Let's assume that we have a file with the values as seen bellow:
% head test.csv
20220601,A,B,1
20220530,A,B,1
And we want to add two new columns, one with the date minus 1 day and one with minus 7 days, resulting the following:
% head new_test.csv
20220601,A,B,20220525,20220531,1
20220530,A,B,20220523,20220529,1
The awk that was used to produce the above is:
% awk 'BEGIN{FS=OFS=","} { a="date -d \"$(date -d \""$1"\") -7 days\" +'%Y%m%d'"; a | getline st ; close(a) ;b="date -d \"$(date -d \""$1"\") -1 days\" +'%Y%m%d'"; b | getline cb ; close(b) ;print $1","$2","$3","st","cb","$4}' test.csv > new_test.csv
But after applying the above in a large file with more than 100K lines it runs for 20 minutes, is there any way to optimize the awk?

Answer 1

One GNU awk approach:
awk '
BEGIN { FS=OFS=","
secs_in_day = 60 * 60 * 24
}
{ dt = mktime( substr($1,1,4) " " substr($1,5,2) " " substr($1,7,2) " 12 0 0" )
dt1 = strftime("%Y%m%d",dt - secs_in_day )
dt7 = strftime("%Y%m%d",dt - (secs_in_day * 7) )
print $1,$2,$3,dt7,dt1,$4
}
' test.csv
This generates:
20220601,A,B,20220525,20220531,1
20220530,A,B,20220523,20220529,1
NOTES:
requires GNU awk for the mktime() and strftime() functions; see GNU awk time functions for more details
other flavors of awk may have similar functions, ymmv

Answer 2

You can try using function calls, it is faster than calling the .
awk -F, '
function cmd1(date){
a="date -d \"$(date -d \""date"\") -1days\" +'%Y%m%d'"
a | getline st
return st
close(a)
}
function cmd2(date){
b="date -d \"$(date -d \""date"\") -7days\" +'%Y%m%d'"
b | getline cm
return cm
close(b)
}
{
$5=cmd1($1)
$6=cmd2($1)
print $1","$2","$3","$5","$6","$4
}' OFS=, test > newFileTest
I executed this against a file with 20000 records in seconds, compared to the original awk which took around 5 minutes.