find a substring in a selected portion of a list elements
up vote
1
down vote
favorite
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
edited Nov 22 at 15:37
jpp
87.8k195099
asked Nov 22 at 15:33
HT121
817
add a comment |
up vote
1
down vote
favorite
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
edited Nov 22 at 15:37
jpp
87.8k195099
asked Nov 22 at 15:33
HT121
817
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
edited Nov 22 at 15:37
jpp
87.8k195099
asked Nov 22 at 15:33
HT121
817
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
python string list list-comprehension python-3.5
edited Nov 22 at 15:37
jpp
87.8k195099
asked Nov 22 at 15:33
HT121
817
edited Nov 22 at 15:37
jpp
87.8k195099
asked Nov 22 at 15:33
HT121
817
edited Nov 22 at 15:37
jpp
87.8k195099
edited Nov 22 at 15:37
jpp
87.8k195099
edited Nov 22 at 15:37
jpp
87.8k195099
87.8k195099
asked Nov 22 at 15:33
HT121
817
asked Nov 22 at 15:33
HT121
817
asked Nov 22 at 15:33
HT121
817
817
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 at 15:35
jpp
87.8k195099
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 at 15:35
jpp
87.8k195099
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
add a comment |
up vote
2
down vote
accepted
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 at 15:35
jpp
87.8k195099
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 at 15:35
jpp
87.8k195099
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 at 15:35
jpp
87.8k195099
edited Nov 22 at 15:37
answered Nov 22 at 15:35
jpp
87.8k195099
answered Nov 22 at 15:35
jpp
87.8k195099
answered Nov 22 at 15:35
jpp
87.8k195099
87.8k195099
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
add a comment |
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 at 15:47
1
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.– jpp
Nov 22 at 15:59
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.– jpp
Nov 22 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 at 16:10
add a comment |
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