sort a list using lambda and regex in python
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy' ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x: ........)
I want to sort the lists in an order starting with the earliest date. How should I go about it using lambda and regex?
python regex lambda
add a comment |
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy' ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x: ........)
I want to sort the lists in an order starting with the earliest date. How should I go about it using lambda and regex?
python regex lambda
Why do you have these weird strings? What is the expected output for the given list?
– timgeb
Nov 14 '18 at 16:33
sorry the original string had '<' characters in it which interfered with the way it was displayed. I have edited the question as you can see now
– dratoms
Nov 14 '18 at 16:34
1
Avoidlist
as a variable name, there's already the builtinlist
.
– timgeb
Nov 14 '18 at 16:51
yes list should not have been used as a variable name. thnx
– dratoms
Nov 14 '18 at 17:07
add a comment |
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy' ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x: ........)
I want to sort the lists in an order starting with the earliest date. How should I go about it using lambda and regex?
python regex lambda
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy' ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x: ........)
I want to sort the lists in an order starting with the earliest date. How should I go about it using lambda and regex?
python regex lambda
python regex lambda
edited Nov 14 '18 at 16:59
timgeb
50.9k116593
50.9k116593
asked Nov 14 '18 at 16:31
dratomsdratoms
195
195
Why do you have these weird strings? What is the expected output for the given list?
– timgeb
Nov 14 '18 at 16:33
sorry the original string had '<' characters in it which interfered with the way it was displayed. I have edited the question as you can see now
– dratoms
Nov 14 '18 at 16:34
1
Avoidlist
as a variable name, there's already the builtinlist
.
– timgeb
Nov 14 '18 at 16:51
yes list should not have been used as a variable name. thnx
– dratoms
Nov 14 '18 at 17:07
add a comment |
Why do you have these weird strings? What is the expected output for the given list?
– timgeb
Nov 14 '18 at 16:33
sorry the original string had '<' characters in it which interfered with the way it was displayed. I have edited the question as you can see now
– dratoms
Nov 14 '18 at 16:34
1
Avoidlist
as a variable name, there's already the builtinlist
.
– timgeb
Nov 14 '18 at 16:51
yes list should not have been used as a variable name. thnx
– dratoms
Nov 14 '18 at 17:07
Why do you have these weird strings? What is the expected output for the given list?
– timgeb
Nov 14 '18 at 16:33
Why do you have these weird strings? What is the expected output for the given list?
– timgeb
Nov 14 '18 at 16:33
sorry the original string had '<' characters in it which interfered with the way it was displayed. I have edited the question as you can see now
– dratoms
Nov 14 '18 at 16:34
sorry the original string had '<' characters in it which interfered with the way it was displayed. I have edited the question as you can see now
– dratoms
Nov 14 '18 at 16:34
1
1
Avoid
list
as a variable name, there's already the builtin list
.– timgeb
Nov 14 '18 at 16:51
Avoid
list
as a variable name, there's already the builtin list
.– timgeb
Nov 14 '18 at 16:51
yes list should not have been used as a variable name. thnx
– dratoms
Nov 14 '18 at 17:07
yes list should not have been used as a variable name. thnx
– dratoms
Nov 14 '18 at 17:07
add a comment |
3 Answers
3
active
oldest
votes
With the code you have there it is sufficient to do:
list.sort(key=lambda x: datet.search(x).group(1))
(but please, don't use list
as a variable name).
There is no need to convert the extracted string to a datetime
as it is already in a format that will sort naturally.
Note however that if any string does not match the regex this will generate an error, so you may be better to split the key out into a named multi-line function and test for a successful match before returning the matched group.
def sort_key(line):
match = datet.search(line)
if match:
return match.group(1)
return ''
data = [
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime'
]
data.sort(key=sort_key)
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
add a comment |
You can use dateutil.parser.parse
(see this answer: Parse date strings?) to parse the date and re.findall
to get it from a string
import re
from dateutil.parser import parse
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x : parse(re.findall(datet, x)[0]))
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
add a comment |
I think the simplest solution without any imports would be:
data = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
sorted_data = sorted(data, key=lambda x: x[20:36])
print(sorted_data)
Output:
['xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
add a comment |
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3 Answers
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3 Answers
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votes
With the code you have there it is sufficient to do:
list.sort(key=lambda x: datet.search(x).group(1))
(but please, don't use list
as a variable name).
There is no need to convert the extracted string to a datetime
as it is already in a format that will sort naturally.
Note however that if any string does not match the regex this will generate an error, so you may be better to split the key out into a named multi-line function and test for a successful match before returning the matched group.
def sort_key(line):
match = datet.search(line)
if match:
return match.group(1)
return ''
data = [
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime'
]
data.sort(key=sort_key)
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
add a comment |
With the code you have there it is sufficient to do:
list.sort(key=lambda x: datet.search(x).group(1))
(but please, don't use list
as a variable name).
There is no need to convert the extracted string to a datetime
as it is already in a format that will sort naturally.
Note however that if any string does not match the regex this will generate an error, so you may be better to split the key out into a named multi-line function and test for a successful match before returning the matched group.
def sort_key(line):
match = datet.search(line)
if match:
return match.group(1)
return ''
data = [
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime'
]
data.sort(key=sort_key)
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
add a comment |
With the code you have there it is sufficient to do:
list.sort(key=lambda x: datet.search(x).group(1))
(but please, don't use list
as a variable name).
There is no need to convert the extracted string to a datetime
as it is already in a format that will sort naturally.
Note however that if any string does not match the regex this will generate an error, so you may be better to split the key out into a named multi-line function and test for a successful match before returning the matched group.
def sort_key(line):
match = datet.search(line)
if match:
return match.group(1)
return ''
data = [
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime'
]
data.sort(key=sort_key)
With the code you have there it is sufficient to do:
list.sort(key=lambda x: datet.search(x).group(1))
(but please, don't use list
as a variable name).
There is no need to convert the extracted string to a datetime
as it is already in a format that will sort naturally.
Note however that if any string does not match the regex this will generate an error, so you may be better to split the key out into a named multi-line function and test for a successful match before returning the matched group.
def sort_key(line):
match = datet.search(line)
if match:
return match.group(1)
return ''
data = [
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime'
]
data.sort(key=sort_key)
answered Nov 14 '18 at 16:54
DuncanDuncan
63k679132
63k679132
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
add a comment |
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
Ah, good observation about the lexicographical order.
– timgeb
Nov 14 '18 at 16:58
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
thanks for that syntax. that was elusive to me. and thanks for that neat little function there. Although the list element part is autogenerated and is unlikely that there will be missing values, your function is going to help me a lot in future, a newbie to python (and programming in general) that I am.
– dratoms
Nov 14 '18 at 17:13
add a comment |
You can use dateutil.parser.parse
(see this answer: Parse date strings?) to parse the date and re.findall
to get it from a string
import re
from dateutil.parser import parse
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x : parse(re.findall(datet, x)[0]))
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
add a comment |
You can use dateutil.parser.parse
(see this answer: Parse date strings?) to parse the date and re.findall
to get it from a string
import re
from dateutil.parser import parse
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x : parse(re.findall(datet, x)[0]))
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
add a comment |
You can use dateutil.parser.parse
(see this answer: Parse date strings?) to parse the date and re.findall
to get it from a string
import re
from dateutil.parser import parse
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x : parse(re.findall(datet, x)[0]))
You can use dateutil.parser.parse
(see this answer: Parse date strings?) to parse the date and re.findall
to get it from a string
import re
from dateutil.parser import parse
list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]
datet = re.compile(r'ResultDatetime:(d4-d2-d2 d2:d2)')
list.sort(key = lambda x : parse(re.findall(datet, x)[0]))
answered Nov 14 '18 at 16:58
mrzasamrzasa
10.6k104078
10.6k104078
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
add a comment |
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
I haven't used dateutil so far. But it seems promising. Will keep this in mind.
– dratoms
Nov 14 '18 at 17:15
add a comment |
I think the simplest solution without any imports would be:
data = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
sorted_data = sorted(data, key=lambda x: x[20:36])
print(sorted_data)
Output:
['xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
add a comment |
I think the simplest solution without any imports would be:
data = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
sorted_data = sorted(data, key=lambda x: x[20:36])
print(sorted_data)
Output:
['xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
add a comment |
I think the simplest solution without any imports would be:
data = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
sorted_data = sorted(data, key=lambda x: x[20:36])
print(sorted_data)
Output:
['xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
I think the simplest solution without any imports would be:
data = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
sorted_data = sorted(data, key=lambda x: x[20:36])
print(sorted_data)
Output:
['xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime']
answered Nov 14 '18 at 17:06
NickNick
79911231
79911231
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
add a comment |
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
The last string has the date at a slightly different offset. I think the OP's intention is that xxxx and yyyyy could be any arbitrarily long strings.
– Duncan
Nov 14 '18 at 17:08
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
exactly. and there could be other string numbers before the regex pattern that would impede in natural sorting here.
– dratoms
Nov 14 '18 at 17:20
add a comment |
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Why do you have these weird strings? What is the expected output for the given list?
– timgeb
Nov 14 '18 at 16:33
sorry the original string had '<' characters in it which interfered with the way it was displayed. I have edited the question as you can see now
– dratoms
Nov 14 '18 at 16:34
1
Avoid
list
as a variable name, there's already the builtinlist
.– timgeb
Nov 14 '18 at 16:51
yes list should not have been used as a variable name. thnx
– dratoms
Nov 14 '18 at 17:07