Sort list of urls by datetime path component
2
I have a list of urls that have a multiple datetime components.
urls = [ www.google.com/2018-10-26/2018-10-26T11:07:10.000Z, www.google.com//2018-10-26/2018-10-26T11:09:18.000Z]
I want to sort on it and then return the list back like this:
urls = [ www.google.com/2018-10-26/2018-10-26T11:09:18.000Z, www.google.com//2018-10-26/2018-10-26T11:07:10.000Z]
Because the first element has 2018-10-26T11:09:18.000Z it is greater than 2018-10-26T11:07:10.000Z. Is there a way to do this in a list comprehension? I am assuming this is the right path:
for url in urls:
date = url.split('/')[2]
But not sure what to do from there.
python list sorting datetime
edited Nov 14 '18 at 16:19
benvc
5,3341625
asked Nov 14 '18 at 16:01
skim8201skim8201
373
add a comment |
2
I have a list of urls that have a multiple datetime components.
urls = [ www.google.com/2018-10-26/2018-10-26T11:07:10.000Z, www.google.com//2018-10-26/2018-10-26T11:09:18.000Z]
I want to sort on it and then return the list back like this:
urls = [ www.google.com/2018-10-26/2018-10-26T11:09:18.000Z, www.google.com//2018-10-26/2018-10-26T11:07:10.000Z]
Because the first element has 2018-10-26T11:09:18.000Z it is greater than 2018-10-26T11:07:10.000Z. Is there a way to do this in a list comprehension? I am assuming this is the right path:
for url in urls:
date = url.split('/')[2]
But not sure what to do from there.
python list sorting datetime
edited Nov 14 '18 at 16:19
benvc
5,3341625
asked Nov 14 '18 at 16:01
skim8201skim8201
373
sorted() would work?
– kon_u
Nov 14 '18 at 16:04
add a comment |
2
2
2
I have a list of urls that have a multiple datetime components.
urls = [ www.google.com/2018-10-26/2018-10-26T11:07:10.000Z, www.google.com//2018-10-26/2018-10-26T11:09:18.000Z]
I want to sort on it and then return the list back like this:
urls = [ www.google.com/2018-10-26/2018-10-26T11:09:18.000Z, www.google.com//2018-10-26/2018-10-26T11:07:10.000Z]
Because the first element has 2018-10-26T11:09:18.000Z it is greater than 2018-10-26T11:07:10.000Z. Is there a way to do this in a list comprehension? I am assuming this is the right path:
for url in urls:
date = url.split('/')[2]
But not sure what to do from there.
python list sorting datetime
edited Nov 14 '18 at 16:19
benvc
5,3341625
asked Nov 14 '18 at 16:01
skim8201skim8201
373
I have a list of urls that have a multiple datetime components.
urls = [ www.google.com/2018-10-26/2018-10-26T11:07:10.000Z, www.google.com//2018-10-26/2018-10-26T11:09:18.000Z]
I want to sort on it and then return the list back like this:
urls = [ www.google.com/2018-10-26/2018-10-26T11:09:18.000Z, www.google.com//2018-10-26/2018-10-26T11:07:10.000Z]
Because the first element has 2018-10-26T11:09:18.000Z it is greater than 2018-10-26T11:07:10.000Z. Is there a way to do this in a list comprehension? I am assuming this is the right path:
for url in urls:
date = url.split('/')[2]
But not sure what to do from there.
python list sorting datetime
python list sorting datetime
edited Nov 14 '18 at 16:19
benvc
5,3341625
asked Nov 14 '18 at 16:01
skim8201skim8201
373
edited Nov 14 '18 at 16:19
benvc
5,3341625
asked Nov 14 '18 at 16:01
skim8201skim8201
373
edited Nov 14 '18 at 16:19
benvc
5,3341625
edited Nov 14 '18 at 16:19
benvc
5,3341625
edited Nov 14 '18 at 16:19
benvc
5,3341625
5,3341625
asked Nov 14 '18 at 16:01
skim8201skim8201
373
asked Nov 14 '18 at 16:01
skim8201skim8201
373
asked Nov 14 '18 at 16:01
skim8201skim8201
373
373
sorted() would work?
– kon_u
Nov 14 '18 at 16:04
add a comment |
sorted() would work?
– kon_u
Nov 14 '18 at 16:04
sorted() would work?
– kon_u
Nov 14 '18 at 16:04
sorted() would work?
– kon_u
Nov 14 '18 at 16:04
add a comment |
2 Answers
2
active
oldest
votes
3
You can sort your urls inplace based on the timestamp (i.e. the last element after the slash) in reverse order (most recent first) via:
urls.sort(key=lambda url: url.split('/')[-1], reverse=True)
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
add a comment |
1
You can sort the list using a function to define the sort key. For example:
urls = ['www.google.com/2018-10-26/2018-10-26T11:07:10.000Z', 'www.google.com//2018-10-26/2018-10-26T11:09:18.000Z']
sorted_urls = sorted(urls, key=lambda x: x.split('/')[-1], reverse=True)
print(sorted_urls)
# OUTPUT
# ['www.google.com//2018-10-26/2018-10-26T11:09:18.000Z', 'www.google.com/2018-10-26/2018-10-26T11:07:10.000Z']
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer useslist.sort()rather thansorted()which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.
– benvc
Nov 14 '18 at 16:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
You can sort your urls inplace based on the timestamp (i.e. the last element after the slash) in reverse order (most recent first) via:
urls.sort(key=lambda url: url.split('/')[-1], reverse=True)
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
add a comment |
3
You can sort your urls inplace based on the timestamp (i.e. the last element after the slash) in reverse order (most recent first) via:
urls.sort(key=lambda url: url.split('/')[-1], reverse=True)
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
add a comment |
3
3
3
You can sort your urls inplace based on the timestamp (i.e. the last element after the slash) in reverse order (most recent first) via:
urls.sort(key=lambda url: url.split('/')[-1], reverse=True)
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
You can sort your urls inplace based on the timestamp (i.e. the last element after the slash) in reverse order (most recent first) via:
urls.sort(key=lambda url: url.split('/')[-1], reverse=True)
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
answered Nov 14 '18 at 16:08
AlexanderAlexander
54.4k1487123
54.4k1487123
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
add a comment |
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
great this worked perfectly. im surprised it didnt need to be converted because of the T in the string. Thanks again!
– skim8201
Nov 14 '18 at 17:58
add a comment |
1
You can sort the list using a function to define the sort key. For example:
urls = ['www.google.com/2018-10-26/2018-10-26T11:07:10.000Z', 'www.google.com//2018-10-26/2018-10-26T11:09:18.000Z']
sorted_urls = sorted(urls, key=lambda x: x.split('/')[-1], reverse=True)
print(sorted_urls)
# OUTPUT
# ['www.google.com//2018-10-26/2018-10-26T11:09:18.000Z', 'www.google.com/2018-10-26/2018-10-26T11:07:10.000Z']
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer useslist.sort()rather thansorted()which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.
– benvc
Nov 14 '18 at 16:11
add a comment |
1
You can sort the list using a function to define the sort key. For example:
urls = ['www.google.com/2018-10-26/2018-10-26T11:07:10.000Z', 'www.google.com//2018-10-26/2018-10-26T11:09:18.000Z']
sorted_urls = sorted(urls, key=lambda x: x.split('/')[-1], reverse=True)
print(sorted_urls)
# OUTPUT
# ['www.google.com//2018-10-26/2018-10-26T11:09:18.000Z', 'www.google.com/2018-10-26/2018-10-26T11:07:10.000Z']
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer useslist.sort()rather thansorted()which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.
– benvc
Nov 14 '18 at 16:11
add a comment |
1
1
1
You can sort the list using a function to define the sort key. For example:
urls = ['www.google.com/2018-10-26/2018-10-26T11:07:10.000Z', 'www.google.com//2018-10-26/2018-10-26T11:09:18.000Z']
sorted_urls = sorted(urls, key=lambda x: x.split('/')[-1], reverse=True)
print(sorted_urls)
# OUTPUT
# ['www.google.com//2018-10-26/2018-10-26T11:09:18.000Z', 'www.google.com/2018-10-26/2018-10-26T11:07:10.000Z']
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
You can sort the list using a function to define the sort key. For example:
urls = ['www.google.com/2018-10-26/2018-10-26T11:07:10.000Z', 'www.google.com//2018-10-26/2018-10-26T11:09:18.000Z']
sorted_urls = sorted(urls, key=lambda x: x.split('/')[-1], reverse=True)
print(sorted_urls)
# OUTPUT
# ['www.google.com//2018-10-26/2018-10-26T11:09:18.000Z', 'www.google.com/2018-10-26/2018-10-26T11:07:10.000Z']
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
answered Nov 14 '18 at 16:08
benvcbenvc
5,3341625
5,3341625
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer useslist.sort()rather thansorted()which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.
– benvc
Nov 14 '18 at 16:11
add a comment |
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer useslist.sort()rather thansorted()which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.
– benvc
Nov 14 '18 at 16:11
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer uses
list.sort() rather than sorted() which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.– benvc
Nov 14 '18 at 16:11
@Alexander beat me to the punch by a few seconds (which is why our answers are near dups) and that answer uses
list.sort() rather than sorted() which avoids creating a probably unnecessary extra variable, so I recommend that you accept Alexander's answer.– benvc
Nov 14 '18 at 16:11
add a comment |
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sorted() would work?
– kon_u
Nov 14 '18 at 16:04