column 4 has date of birth (dd/mm/yyyy). I would like this column to be sorted in an increasing order; and then output to a file.
Please assist
Asked
Active
Viewed 443 times
-3
-
What have you tried? What is the format of the DOB? Please post sample input – inspectorG4dget Feb 26 '14 at 21:48
-
You'll want to read all the date of births into a list, sort the list, and output the list. – Leah Sapan Feb 26 '14 at 21:49
-
I edited it and now it shows the format of DOB – ap306 Feb 26 '14 at 21:50
1 Answers
0
import csv
from datetime import datetime
with open('path/to/file') as infile, open('path/to/output', 'w') as outfile:
rows = [line for line in csv.reader(infile)]
rows.sort(key=lambda row:datetime.strptime(row[3], "%d/%m/%Y"))
writer = csv.writer(outfile)
writer.writerows(rows)
inspectorG4dget
- 110,290
- 27
- 149
- 241
-
-
@ap306: That lambda converts the 4th column into machine parseable date/time, for sorting – inspectorG4dget Feb 26 '14 at 22:44