Transform an entire column using "date" command - linux

Question

Here is a dummy CSV file with 3 rows. The actual file has 7 million rows.
testdates.csv:
y_m_d
1997-01-01
1985-06-09
1943-07-14
The date tool can usually be formatted as such , to get the 'day' :
date -d "25 JUN 2011" +%A
=> output: Saturday
Query: How to provide an entire column as input for the date +%A transformation?
The resulting output should be appended to the end of the input file.
Intended output:
y_m_d, Day
1997-01-01, Thursday
1985-06-09, Sunday
1943-07-14, Tuesday

Answer 1

To read multiples dates from a file using GNU date, you can use the -f/--file option:
$ date -f testdates.csv '+%F, %A'
1997-01-01, Wednesday
1985-06-09, Sunday
1943-07-14, Wednesday
Since your file has a header row, we have to skip that, for example using process substitution and sed:
date -f <(sed '1d' testdates.csv) '+%F, %A'
To get your desired output, combine like this:
echo 'y_m_d, Day'
date -f <(sed '1d' testdates.csv) '+%F, %A'
or write to a new file:
{
echo 'y_m_d, Day'
date -f <(sed '1d' testdates.csv) '+%F, %A'
} > testdates.csv.tmp
and after inspection, you can rename with
mv testdates.csv.tmp testdates.csv

Answer 2

Hard to beat that date answer.
GNU awk would be OK too:
gawk -v OFS=', ' '
NR == 1 {$2 = "Day"}
NR > 1 {$2 = strftime("%A", mktime(gensub(/-/, " ", "g", $1) " 0 0 0"))}
1
' testdates.csv
y_m_d, Day
1997-01-01, Wednesday
1985-06-09, Sunday
1943-07-14, Wednesday
Or perl:
perl -MTime::Piece -lne '
print "$_, ", $. == 1
? "Day"
: Time::Piece->strptime($_, "%Y-%m-%d")->strftime("%A")
' testdates.csv

Answer 3

#/bin/bash
while read datespec; do
echo $datespec, $(date -d "$datespec" +%A)
done < testdates.csv
Output:
1997-01-01, Wednesday
1985-06-09, Sunday
1943-07-14, Wednesday