Sorting a csv object by dates in python

Question

I am trying to read and sort a csv file that has data that looks like

Date    Open    High    Low Close   Volume
27-Mar-12   8.25    8.35    8.17    8.19    9801989
26-Mar-12   8.16    8.25    8.12    8.24    8694416
23-Mar-12   8.05    8.12    7.95    8.09    8149170

I do this with

import csv
data = csv.reader(open('data.csv','r'))

To sort data by Date. I do:

sorteddata = sorted(data,key=operator.itemgetter(1),reverse=False)

The problem is, that it sorted the dates by reading them as String and not as dates. So the data is sorted like so,

['3-Aug-11', '7.06', '7.23', '6.84', '7.16', '31583617']
['3-Feb-12', '7.02', '7.12', '6.98', '7.08', '15318044']
['3-Jan-12', '5.53', '5.59', '5.44', '5.48', '12678923']
['3-Jun-11', '8.09', '8.17', '7.92', '7.97', '21273812']
['3-May-11', '9.00', '9.04', '8.63', '8.80', '17356005']

Does anybody know how to sort by dates?

agf · Accepted Answer · 2012-03-29 01:18:10Z

8

Use datetime.strptime to get a datetime from the date field:

from datetime import datetime

data = sorted(data, key = lambda row: datetime.strptime(row[0], "%d-%b-%y"))

answered Mar 29, 2012 at 1:18

agf

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1 Comment

Li-aung Yip Over a year ago

+1. Didn't know you could sort on datetime.datetime objects directly - looking at the docs, they do implement the < operator. Excellent.

Li-aung Yip · Accepted Answer · 2012-03-29 01:18:57Z

0

Use the time module for time format conversions, and convert your time strings (3-Aug-11) into numbers you can sort.

Here's some food for thought:

>>> t = time.strptime("3-Aug-11","%d-%b-%y")
>>> t
time.struct_time(tm_year=2011, tm_mon=8, tm_mday=3, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=2, tm_yday=215, tm_isdst=-1)
>>> time.mktime(t)
1312300800.0

Documentation for time module.

answered Mar 29, 2012 at 1:18

Li-aung Yip

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