按格式化为字符串的键对python字典进行排序

Question

I have a dictionary with strings as keys formatted as yyyy-mm-dd and want to sort the dictionary by keys with the earliest dates first:

I am currently using sorted(datesAndText.keys()) but this isn't reliably working because the month and day fields are not always zero padded.

I have looked at Sort python dictionary by date keys and How do I sort this list in Python, if my date is in a String? but I can't seem to adopt them to by specific case.

Answer 1

Are you sure your keys are exactly in the format yyyy-mm-dd ? For example:

>>> '2010-1-15' < '2010-02-15'
False

You may be forced to sort something like this:

sorted(d,key=lambda x: [int(y) for y in x.split('-')])

Another solution (assuming your years are all 4 digits):

sorted(d,key=lambda x: [y.zfill(2) for y in x.split('-')]) 

I'm not sure which would be faster. I suppose it's a candidate for timeit .

Answer 2

Dates in yyyy-mm-dd format sort the same way both alphabetically and chronologically, so you can use standard sorted :

for k, v in sorted(datesAndText.items()):
    # do something with key and value

Answer 3

Your format, yyyy-mm-dd , allows a lexicographic sort, so your code should work fine unless your values aren't zero padded (ex 2012-10-9 instead of 2012-10-09 ).

Fix this problem by relying on a comparison of dates rather than strings:

sorted(datesAndText, key=lambda x: datetime.strptime(x, '%Y-%m-%d'))

This utilizes the key parameter to sorted, which is a function which accepts one argument (an element of the list being compared during sort) and returns a value on which sorted can use to sort.

This has the ancillary benefit of allowing you to explicitly specify the string format of the date, should your data need to change.

Edit:

mgilson brought up an interesting point. str.split is probably more efficient. Let's see if he's correct:

strptime solution:

bburns@virgil:~$ python -mtimeit -s"from datetime import datetime;d={'2012-2-12':None, '2012-10-9':None, '1978-1-1':None, '1985-10-9':None}" 'sorted(d, key=lambda x: datetime.strptime(x,"%Y-%m-%d"))'
10000 loops, best of 3: 79.7 usec per loop

mgilson's original str.split solution:

bburns@virgil:~$ python -mtimeit -s"from datetime import datetime;d={'2012-2-12':None, '2012-10-9':None, '1978-1-1':None, '1985-10-9':None}" 'sorted(d,key=lambda x: [int(y) for y in x.split("-")])'
100000 loops, best of 3: 17.6 usec per loop

mgilson's zfill str.split solution:

bburns@virgil:~$ python -mtimeit -s"from datetime import datetime;d={'2012-2-12':None, '2012-10-9':None, '1978-1-1':None, '1985-10-9':None}" 'sorted(d,key=lambda x: [y.zfill(2) for y in x.split("-")])'
100000 loops, best of 3: 7.4 usec per loop

Looks like he's correct! mgilson's original answer is 4-5 times faster, and his final answer is 10-11 times faster! However, as we agreed in the comments, readability matters. Unless you're presently CPU-bound, I'd still advise going with datetime.strptime over str.split .