CODEWITHSUNDEEP
python
Android Beginner
Android Beginner

Reputation: 486

sorting a list containing dates as a substring

I have a list of strings something like this As you can make out there is a date embedded at the start of the string and also at the end of the string.

 a = ["08/19/2014100%ABC10/02/2014F","02/12/2012100%ABC10/02/2014F",
      "08/29/2014100%ABC10/02/2012F"]

I wanted to sort this list based on the date substring.

The way I wanted to implement is by extract the date, sort the date and join the date with the substring, but it's becoming too complicated.

Upvotes: 3

Views: 752

Answers (2)

Tim Biegeleisen
Tim Biegeleisen

Reputation: 521249

The following code defines a custom sorting function compare which takes two elements from your list, and sorts them based on the date which appears at the start of each string. If you wanted to use the dates embedded at the end of the strings, you can modify the code accordingly.

def compare(item1, item2):
    format = '%m/%d/%Y' #changed the date format
    # convert string dates to Date type
    date1 = datetime.datetime.strptime(item1[:10], format).date() #removed comma from item1[:,10]
    date2 = datetime.datetime.strptime(item2[:10], format).date() #same as above

    if date1 < date2:
        return -1
    elif date1 > date2:
        return 1
    else:
        return 0

a = ["08/19/2014100%ABC10/02/2014F","02/12/2012100%ABC10/02/2014F",
  "08/29/2014100%ABC10/02/2012F"]

a.sort(compare)

Upvotes: 0

mhawke
mhawke

Reputation: 87084

Just call sorted() with the key argument set to a function that extracts and converts the date from the string into a datetime.datetime object.

>>> from datetime import datetime
>>> a = ["08/19/2014100%ABC10/02/2014F","02/12/2012100%ABC10/02/2014F", "08/29/2014100%ABC10/02/2012F"]
>>> sorted_a = sorted(a, key=lambda s: datetime.strptime(s[:10], '%m/%d/%Y'))
>>> sorted_a
['02/12/2012100%ABC10/02/2014F', '08/19/2014100%ABC10/02/2014F', '08/29/2014100%ABC10/02/2012F']

Or, if you want to sort in place:

>>> a.sort(key=lambda s: datetime.strptime(s[:10], '%m/%d/%Y'))
>>> a
['02/12/2012100%ABC10/02/2014F', '08/19/2014100%ABC10/02/2014F', '08/29/2014100%ABC10/02/2012F']

If you actually mean to sort on the last date in the string just change the key function to:

lambda s: datetime.strptime(s[-11:-1], '%m/%d/%Y')

Upvotes: 3

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