Regular Expressions: Search in list
44
I want to filter strings in a list based on a regular expression.
Is there something better than [x for x in list if r.match(x)] ?
python regex
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
add a comment |
44
I want to filter strings in a list based on a regular expression.
Is there something better than [x for x in list if r.match(x)] ?
python regex
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
add a comment |
44
44
44
18
I want to filter strings in a list based on a regular expression.
Is there something better than [x for x in list if r.match(x)] ?
python regex
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
I want to filter strings in a list based on a regular expression.
Is there something better than [x for x in list if r.match(x)] ?
python regex
python regex
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
edited Nov 6 '14 at 19:46
RockPaperLizard
2,13252236
2,13252236
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
asked Sep 4 '10 at 0:13
leolukleoluk
9,00843445
9,00843445
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
79
You can create an iterator in Python 3.x or a list in Python 2.x by using:
filter(r.match, list)
To convert the Python 3.x iterator to a list, simply cast it; list(filter(..)).
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
27
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, thefilterversion is perfectly clear and has much less noise.
– sepp2k
Sep 4 '10 at 0:47
what isr.matchhere?
– rbatt
Oct 12 '18 at 11:48
@rbattr.matchis a method that, when applied to a given string, finds whether the regexrmatches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)
– sepp2k
Oct 12 '18 at 21:33
add a comment |
71
Full Example (Python 3):
For Python 2.x look into Note below
import re
mylist = ["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
r = re.compile(".*cat")
newlist = list(filter(r.match, mylist)) # Read Note
print(newlist)
Prints:
['cat', 'wildcat', 'thundercat']
Note:
For Python 2.x users, filter returns a list already. In Python 3.x filter was changed to return an iterator so it has to be converted to list (in order to see it printed out nicely).
Python 3 code example
Python 2.x code example
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
4
Hi there, When I run the above code, I get<filter object at 0x1057acda0>What am I doing wrong?
– Joshua
Oct 13 '16 at 11:08
1
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
1
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed thestrsincefilterreturns a list anyway, to no avail...
– Joshua
Oct 14 '16 at 12:25
4
@joshua you've probably figured this out by now but tryprint(list(newlist))orprint([i for i in newlist])
– James Draper
Jan 10 '17 at 18:42
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
add a comment |
3
Just in case someone comes here in future, there is another pythonic way to do it. First you need to create the regex and then then filter
import re
inilist =["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
regex = re.compile(r'.*cat')
selectobj = filter(regex.search, inilist)
selectobj
results:
['cat', 'wildcat', 'thundercat']
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
79
You can create an iterator in Python 3.x or a list in Python 2.x by using:
filter(r.match, list)
To convert the Python 3.x iterator to a list, simply cast it; list(filter(..)).
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
27
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, thefilterversion is perfectly clear and has much less noise.
– sepp2k
Sep 4 '10 at 0:47
what isr.matchhere?
– rbatt
Oct 12 '18 at 11:48
@rbattr.matchis a method that, when applied to a given string, finds whether the regexrmatches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)
– sepp2k
Oct 12 '18 at 21:33
add a comment |
79
You can create an iterator in Python 3.x or a list in Python 2.x by using:
filter(r.match, list)
To convert the Python 3.x iterator to a list, simply cast it; list(filter(..)).
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
27
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, thefilterversion is perfectly clear and has much less noise.
– sepp2k
Sep 4 '10 at 0:47
what isr.matchhere?
– rbatt
Oct 12 '18 at 11:48
@rbattr.matchis a method that, when applied to a given string, finds whether the regexrmatches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)
– sepp2k
Oct 12 '18 at 21:33
add a comment |
79
79
79
You can create an iterator in Python 3.x or a list in Python 2.x by using:
filter(r.match, list)
To convert the Python 3.x iterator to a list, simply cast it; list(filter(..)).
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
You can create an iterator in Python 3.x or a list in Python 2.x by using:
filter(r.match, list)
To convert the Python 3.x iterator to a list, simply cast it; list(filter(..)).
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
edited Jun 4 '18 at 6:29
Ev. Kounis
10.8k21546
10.8k21546
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
answered Sep 4 '10 at 0:17
sepp2ksepp2k
294k38595610
294k38595610
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
27
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, thefilterversion is perfectly clear and has much less noise.
– sepp2k
Sep 4 '10 at 0:47
what isr.matchhere?
– rbatt
Oct 12 '18 at 11:48
@rbattr.matchis a method that, when applied to a given string, finds whether the regexrmatches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)
– sepp2k
Oct 12 '18 at 21:33
add a comment |
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
27
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, thefilterversion is perfectly clear and has much less noise.
– sepp2k
Sep 4 '10 at 0:47
what isr.matchhere?
– rbatt
Oct 12 '18 at 11:48
@rbattr.matchis a method that, when applied to a given string, finds whether the regexrmatches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)
– sepp2k
Oct 12 '18 at 21:33
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
Actually, list comprehensions are usually prefered over functional constructs such as filter, reduce, lambda, etc.
– Ivo van der Wijk
Sep 4 '10 at 0:41
27
27
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, the
filter version is perfectly clear and has much less noise.– sepp2k
Sep 4 '10 at 0:47
@Ivo: They are usually preferred because they're usually clearer and often more succinct. However in this case, the
filter version is perfectly clear and has much less noise.– sepp2k
Sep 4 '10 at 0:47
what is
r.match here?– rbatt
Oct 12 '18 at 11:48
what is
r.match here?– rbatt
Oct 12 '18 at 11:48
@rbatt
r.match is a method that, when applied to a given string, finds whether the regex r matches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)– sepp2k
Oct 12 '18 at 21:33
@rbatt
r.match is a method that, when applied to a given string, finds whether the regex r matches that string (and returns a corresponding match object if so, but that doesn't matter in this case as we just care whether the result is truthy)– sepp2k
Oct 12 '18 at 21:33
add a comment |
71
Full Example (Python 3):
For Python 2.x look into Note below
import re
mylist = ["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
r = re.compile(".*cat")
newlist = list(filter(r.match, mylist)) # Read Note
print(newlist)
Prints:
['cat', 'wildcat', 'thundercat']
Note:
For Python 2.x users, filter returns a list already. In Python 3.x filter was changed to return an iterator so it has to be converted to list (in order to see it printed out nicely).
Python 3 code example
Python 2.x code example
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
4
Hi there, When I run the above code, I get<filter object at 0x1057acda0>What am I doing wrong?
– Joshua
Oct 13 '16 at 11:08
1
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
1
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed thestrsincefilterreturns a list anyway, to no avail...
– Joshua
Oct 14 '16 at 12:25
4
@joshua you've probably figured this out by now but tryprint(list(newlist))orprint([i for i in newlist])
– James Draper
Jan 10 '17 at 18:42
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
add a comment |
71
Full Example (Python 3):
For Python 2.x look into Note below
import re
mylist = ["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
r = re.compile(".*cat")
newlist = list(filter(r.match, mylist)) # Read Note
print(newlist)
Prints:
['cat', 'wildcat', 'thundercat']
Note:
For Python 2.x users, filter returns a list already. In Python 3.x filter was changed to return an iterator so it has to be converted to list (in order to see it printed out nicely).
Python 3 code example
Python 2.x code example
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
4
Hi there, When I run the above code, I get<filter object at 0x1057acda0>What am I doing wrong?
– Joshua
Oct 13 '16 at 11:08
1
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
1
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed thestrsincefilterreturns a list anyway, to no avail...
– Joshua
Oct 14 '16 at 12:25
4
@joshua you've probably figured this out by now but tryprint(list(newlist))orprint([i for i in newlist])
– James Draper
Jan 10 '17 at 18:42
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
add a comment |
71
71
71
Full Example (Python 3):
For Python 2.x look into Note below
import re
mylist = ["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
r = re.compile(".*cat")
newlist = list(filter(r.match, mylist)) # Read Note
print(newlist)
Prints:
['cat', 'wildcat', 'thundercat']
Note:
For Python 2.x users, filter returns a list already. In Python 3.x filter was changed to return an iterator so it has to be converted to list (in order to see it printed out nicely).
Python 3 code example
Python 2.x code example
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
Full Example (Python 3):
For Python 2.x look into Note below
import re
mylist = ["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
r = re.compile(".*cat")
newlist = list(filter(r.match, mylist)) # Read Note
print(newlist)
Prints:
['cat', 'wildcat', 'thundercat']
Note:
For Python 2.x users, filter returns a list already. In Python 3.x filter was changed to return an iterator so it has to be converted to list (in order to see it printed out nicely).
Python 3 code example
Python 2.x code example
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
edited Oct 14 '18 at 7:00
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
answered Sep 20 '16 at 11:41
MercuryMercury
2,6832027
2,6832027
4
Hi there, When I run the above code, I get<filter object at 0x1057acda0>What am I doing wrong?
– Joshua
Oct 13 '16 at 11:08
1
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
1
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed thestrsincefilterreturns a list anyway, to no avail...
– Joshua
Oct 14 '16 at 12:25
4
@joshua you've probably figured this out by now but tryprint(list(newlist))orprint([i for i in newlist])
– James Draper
Jan 10 '17 at 18:42
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
add a comment |
4
Hi there, When I run the above code, I get<filter object at 0x1057acda0>What am I doing wrong?
– Joshua
Oct 13 '16 at 11:08
1
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
1
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed thestrsincefilterreturns a list anyway, to no avail...
– Joshua
Oct 14 '16 at 12:25
4
@joshua you've probably figured this out by now but tryprint(list(newlist))orprint([i for i in newlist])
– James Draper
Jan 10 '17 at 18:42
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
4
4
Hi there, When I run the above code, I get
<filter object at 0x1057acda0> What am I doing wrong?– Joshua
Oct 13 '16 at 11:08
Hi there, When I run the above code, I get
<filter object at 0x1057acda0> What am I doing wrong?– Joshua
Oct 13 '16 at 11:08
1
1
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
According to python docs (python 2.7.12): docs.python.org/2/library/functions.html#filter filter returns a list not an object. You can also check that code: repl.it/X3G/5786 (just hit run)
– Mercury
Oct 13 '16 at 13:24
1
1
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed the
str since filter returns a list anyway, to no avail...– Joshua
Oct 14 '16 at 12:25
Thank you. I am using Python 3.5.2 on a Mac. I tried your link. Of course it works, though not sure why I get that msg. I even removed the
str since filter returns a list anyway, to no avail...– Joshua
Oct 14 '16 at 12:25
4
4
@joshua you've probably figured this out by now but try
print(list(newlist)) or print([i for i in newlist])– James Draper
Jan 10 '17 at 18:42
@joshua you've probably figured this out by now but try
print(list(newlist)) or print([i for i in newlist])– James Draper
Jan 10 '17 at 18:42
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
@James Draper Thank you James. It was a good reminder.
– Joshua
Jan 10 '17 at 18:48
add a comment |
3
Just in case someone comes here in future, there is another pythonic way to do it. First you need to create the regex and then then filter
import re
inilist =["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
regex = re.compile(r'.*cat')
selectobj = filter(regex.search, inilist)
selectobj
results:
['cat', 'wildcat', 'thundercat']
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
add a comment |
3
Just in case someone comes here in future, there is another pythonic way to do it. First you need to create the regex and then then filter
import re
inilist =["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
regex = re.compile(r'.*cat')
selectobj = filter(regex.search, inilist)
selectobj
results:
['cat', 'wildcat', 'thundercat']
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
add a comment |
3
3
3
Just in case someone comes here in future, there is another pythonic way to do it. First you need to create the regex and then then filter
import re
inilist =["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
regex = re.compile(r'.*cat')
selectobj = filter(regex.search, inilist)
selectobj
results:
['cat', 'wildcat', 'thundercat']
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
Just in case someone comes here in future, there is another pythonic way to do it. First you need to create the regex and then then filter
import re
inilist =["dog", "cat", "wildcat", "thundercat", "cow", "hooo"]
regex = re.compile(r'.*cat')
selectobj = filter(regex.search, inilist)
selectobj
results:
['cat', 'wildcat', 'thundercat']
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
answered Nov 13 '18 at 9:51
MEdwinMEdwin
964114
964114
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
add a comment |
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
That's literally what the accepted answer suggests
– leoluk
Nov 14 '18 at 12:31
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
@leoluk Yes, but it shows greater detail and is probably more useful to those who ask questions vs answer questions on SO. You have 9K reputation! Of course you don't need any other details! :)
– enter_display_name_here
Jan 15 at 15:42
add a comment |
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