How to find substrings in a list of words
0
I'm trying to find if exists a substring in a list of strings,
FOR EXAMPLE:
I have the list of words ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
PINK is a substring of ASDEKNIP , because the reverse of PINK is KNIP
and the word BAR is in WORDRRAB because the reverse is RAB
How to find if substrip is exits? and if yes so put in reverse that string
so the new list should be:
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
I tried like this
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for word in d:
word = word[::-1]
if word in d:
print(word)
But it gives nothing
python python-3.x
edited Nov 21 '18 at 18:36
Austin
12.6k31031
asked Nov 21 '18 at 18:22
LordNordLordNord
707
add a comment |
0
I'm trying to find if exists a substring in a list of strings,
FOR EXAMPLE:
I have the list of words ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
PINK is a substring of ASDEKNIP , because the reverse of PINK is KNIP
and the word BAR is in WORDRRAB because the reverse is RAB
How to find if substrip is exits? and if yes so put in reverse that string
so the new list should be:
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
I tried like this
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for word in d:
word = word[::-1]
if word in d:
print(word)
But it gives nothing
python python-3.x
edited Nov 21 '18 at 18:36
Austin
12.6k31031
asked Nov 21 '18 at 18:22
LordNordLordNord
707
@balki d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
– LordNord
Nov 21 '18 at 18:41
add a comment |
0
0
0
I'm trying to find if exists a substring in a list of strings,
FOR EXAMPLE:
I have the list of words ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
PINK is a substring of ASDEKNIP , because the reverse of PINK is KNIP
and the word BAR is in WORDRRAB because the reverse is RAB
How to find if substrip is exits? and if yes so put in reverse that string
so the new list should be:
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
I tried like this
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for word in d:
word = word[::-1]
if word in d:
print(word)
But it gives nothing
python python-3.x
edited Nov 21 '18 at 18:36
Austin
12.6k31031
asked Nov 21 '18 at 18:22
LordNordLordNord
707
I'm trying to find if exists a substring in a list of strings,
FOR EXAMPLE:
I have the list of words ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
PINK is a substring of ASDEKNIP , because the reverse of PINK is KNIP
and the word BAR is in WORDRRAB because the reverse is RAB
How to find if substrip is exits? and if yes so put in reverse that string
so the new list should be:
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
I tried like this
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for word in d:
word = word[::-1]
if word in d:
print(word)
But it gives nothing
python python-3.x
python python-3.x
edited Nov 21 '18 at 18:36
Austin
12.6k31031
asked Nov 21 '18 at 18:22
LordNordLordNord
707
edited Nov 21 '18 at 18:36
Austin
12.6k31031
asked Nov 21 '18 at 18:22
LordNordLordNord
707
edited Nov 21 '18 at 18:36
Austin
12.6k31031
edited Nov 21 '18 at 18:36
Austin
12.6k31031
edited Nov 21 '18 at 18:36
Austin
12.6k31031
12.6k31031
asked Nov 21 '18 at 18:22
LordNordLordNord
707
asked Nov 21 '18 at 18:22
LordNordLordNord
707
asked Nov 21 '18 at 18:22
LordNordLordNord
707
707
@balki d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
– LordNord
Nov 21 '18 at 18:41
add a comment |
@balki d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
– LordNord
Nov 21 '18 at 18:41
@balki d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
– LordNord
Nov 21 '18 at 18:41
@balki d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
– LordNord
Nov 21 '18 at 18:41
add a comment |
1 Answer
1
active
oldest
votes
1
Use itertools.permutations:
from itertools import permutations
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for x, y in permutations(d, 2):
rev = y[::-1]
if rev in x:
d.append(rev)
print(d)
# ['GGBASDEPINK', 'ASDEKNIP', 'PINK', 'WORDRRAB', 'BAR', 'KNIP', 'RAB']
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
1
For long lists, you will save a lot of work usingcombinationsand checkshorter in longer.
– schwobaseggl
Nov 21 '18 at 18:40
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
1
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
|
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Use itertools.permutations:
from itertools import permutations
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for x, y in permutations(d, 2):
rev = y[::-1]
if rev in x:
d.append(rev)
print(d)
# ['GGBASDEPINK', 'ASDEKNIP', 'PINK', 'WORDRRAB', 'BAR', 'KNIP', 'RAB']
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
1
For long lists, you will save a lot of work usingcombinationsand checkshorter in longer.
– schwobaseggl
Nov 21 '18 at 18:40
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
1
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
|
show 1 more comment
1
Use itertools.permutations:
from itertools import permutations
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for x, y in permutations(d, 2):
rev = y[::-1]
if rev in x:
d.append(rev)
print(d)
# ['GGBASDEPINK', 'ASDEKNIP', 'PINK', 'WORDRRAB', 'BAR', 'KNIP', 'RAB']
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
1
For long lists, you will save a lot of work usingcombinationsand checkshorter in longer.
– schwobaseggl
Nov 21 '18 at 18:40
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
1
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
|
show 1 more comment
1
1
1
Use itertools.permutations:
from itertools import permutations
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for x, y in permutations(d, 2):
rev = y[::-1]
if rev in x:
d.append(rev)
print(d)
# ['GGBASDEPINK', 'ASDEKNIP', 'PINK', 'WORDRRAB', 'BAR', 'KNIP', 'RAB']
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
Use itertools.permutations:
from itertools import permutations
d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR']
for x, y in permutations(d, 2):
rev = y[::-1]
if rev in x:
d.append(rev)
print(d)
# ['GGBASDEPINK', 'ASDEKNIP', 'PINK', 'WORDRRAB', 'BAR', 'KNIP', 'RAB']
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
answered Nov 21 '18 at 18:35
AustinAustin
12.6k31031
12.6k31031
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
1
For long lists, you will save a lot of work usingcombinationsand checkshorter in longer.
– schwobaseggl
Nov 21 '18 at 18:40
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
1
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
|
show 1 more comment
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
1
For long lists, you will save a lot of work usingcombinationsand checkshorter in longer.
– schwobaseggl
Nov 21 '18 at 18:40
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
1
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
"for x, y in permutations(d, 2):" What does mean the number 2?
– LordNord
Nov 21 '18 at 18:40
1
1
For long lists, you will save a lot of work using
combinations and check shorter in longer.– schwobaseggl
Nov 21 '18 at 18:40
For long lists, you will save a lot of work using
combinations and check shorter in longer.– schwobaseggl
Nov 21 '18 at 18:40
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@schwobaseggl, I believe that doesn't test every words, since it's shorter like you say, ofcourse it works in this example.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
@LordNord, Taking 2 words at a time.
– Austin
Nov 21 '18 at 18:42
1
1
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
Combinations catches all pairs once. Permutations catches all pairs twice. But only the shorter of the two involved strings can be a substring of the other, so one comparison is spurious. But for pairs the gain is marginal, as the iterations will just half while you add a comparison of lengthes.
– schwobaseggl
Nov 21 '18 at 18:44
|
show 1 more comment
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@balki d = ['GGBASDEPINK','ASDEKNIP','PINK','WORDRRAB','BAR' ,'KNIP', 'RAB']
– LordNord
Nov 21 '18 at 18:41