Function to return subword of a camelcase string
up vote
6
down vote
favorite
Given a camelcase string and an index, return the subword of the string that includes that index, e.g.:
find_word('CamelCaseString', 6) -> 'Case'
find_word('ACamelCaseString', 0) -> 'A'
My code:
def find_word(s, index):
for i in range(index, 0, -1):
if s[i].isupper():
left = i
break
else:
left = 0
for i in range(index, len(s)-1):
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
right = i
break
else:
right = len(s) - 1
return s[left:right+1]
Can this be made more concise/efficient?
python strings interview-questions
asked 7 hours ago
Eugene Yarmash
26329
add a comment |
up vote
6
down vote
favorite
Given a camelcase string and an index, return the subword of the string that includes that index, e.g.:
find_word('CamelCaseString', 6) -> 'Case'
find_word('ACamelCaseString', 0) -> 'A'
My code:
def find_word(s, index):
for i in range(index, 0, -1):
if s[i].isupper():
left = i
break
else:
left = 0
for i in range(index, len(s)-1):
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
right = i
break
else:
right = len(s) - 1
return s[left:right+1]
Can this be made more concise/efficient?
python strings interview-questions
asked 7 hours ago
Eugene Yarmash
26329
Shouldn't the first example returnString?
– Heslacher
7 hours ago
1
No, the index 6 is character'a'in subword'Case'.
– Eugene Yarmash
7 hours ago
OK, now I got it.
– Heslacher
7 hours ago
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Given a camelcase string and an index, return the subword of the string that includes that index, e.g.:
find_word('CamelCaseString', 6) -> 'Case'
find_word('ACamelCaseString', 0) -> 'A'
My code:
def find_word(s, index):
for i in range(index, 0, -1):
if s[i].isupper():
left = i
break
else:
left = 0
for i in range(index, len(s)-1):
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
right = i
break
else:
right = len(s) - 1
return s[left:right+1]
Can this be made more concise/efficient?
python strings interview-questions
asked 7 hours ago
Eugene Yarmash
26329
Given a camelcase string and an index, return the subword of the string that includes that index, e.g.:
find_word('CamelCaseString', 6) -> 'Case'
find_word('ACamelCaseString', 0) -> 'A'
My code:
def find_word(s, index):
for i in range(index, 0, -1):
if s[i].isupper():
left = i
break
else:
left = 0
for i in range(index, len(s)-1):
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
right = i
break
else:
right = len(s) - 1
return s[left:right+1]
Can this be made more concise/efficient?
python strings interview-questions
python strings interview-questions
asked 7 hours ago
Eugene Yarmash
26329
asked 7 hours ago
Eugene Yarmash
26329
asked 7 hours ago
Eugene Yarmash
26329
asked 7 hours ago
Eugene Yarmash
26329
asked 7 hours ago
Eugene Yarmash
26329
26329
Shouldn't the first example returnString?
– Heslacher
7 hours ago
1
No, the index 6 is character'a'in subword'Case'.
– Eugene Yarmash
7 hours ago
OK, now I got it.
– Heslacher
7 hours ago
add a comment |
Shouldn't the first example returnString?
– Heslacher
7 hours ago
1
No, the index 6 is character'a'in subword'Case'.
– Eugene Yarmash
7 hours ago
OK, now I got it.
– Heslacher
7 hours ago
Shouldn't the first example return
String ?– Heslacher
7 hours ago
Shouldn't the first example return
String ?– Heslacher
7 hours ago
1
1
No, the index 6 is character
'a' in subword 'Case'.– Eugene Yarmash
7 hours ago
No, the index 6 is character
'a' in subword 'Case'.– Eugene Yarmash
7 hours ago
OK, now I got it.
– Heslacher
7 hours ago
OK, now I got it.
– Heslacher
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
7
down vote
Review
Add docstrings and tests... or both in the form of doctests!
def find_word(s, index):
"""
Finds the CamalCased word surrounding the givin index in the string
>>> find_word('CamelCaseString', 6)
'Case'
>>> find_word('ACamelCaseString', 0)
'A'
"""
...
Loop like a native.
Instead of going over the indexes we can loop over the item directly
range(index, 0, -1)
We can loop over the item and index at the same time using enumerate
for i, s in enumerate(string[index:0:-1])
However this would be slower since it will create a new string object with every slice.
If we can be sure that the givin string is a CamalCase string
Then we can drop some of your second if statement
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
Would be
if s[i+1].isupper():
Actually your code (from a performance aspect) is quite good
We could however use a while loop to increment both side at once, for a little performance gain.
(slower, yet more readable) Alternative
A different approach to finding CamalCase words can be done with regex,
We can find all CamalCase words with the following regex: r"([A-Z][a-z]*)"
And we can use re.finditer to create a generator for our matches and loop over them, and return when our index is in between the end and the start.
import re
def find_word_2(string, index):
for match in re.finditer(r"([A-Z][a-z]*)", string):
if match.start() <= index < match.end():
return match.group()
NOTE This yields more readable code, but it should be alot slower for large inputs.
answered 6 hours ago
Ludisposed
6,93921959
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
Review
Add docstrings and tests... or both in the form of doctests!
def find_word(s, index):
"""
Finds the CamalCased word surrounding the givin index in the string
>>> find_word('CamelCaseString', 6)
'Case'
>>> find_word('ACamelCaseString', 0)
'A'
"""
...
Loop like a native.
Instead of going over the indexes we can loop over the item directly
range(index, 0, -1)
We can loop over the item and index at the same time using enumerate
for i, s in enumerate(string[index:0:-1])
However this would be slower since it will create a new string object with every slice.
If we can be sure that the givin string is a CamalCase string
Then we can drop some of your second if statement
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
Would be
if s[i+1].isupper():
Actually your code (from a performance aspect) is quite good
We could however use a while loop to increment both side at once, for a little performance gain.
(slower, yet more readable) Alternative
A different approach to finding CamalCase words can be done with regex,
We can find all CamalCase words with the following regex: r"([A-Z][a-z]*)"
And we can use re.finditer to create a generator for our matches and loop over them, and return when our index is in between the end and the start.
import re
def find_word_2(string, index):
for match in re.finditer(r"([A-Z][a-z]*)", string):
if match.start() <= index < match.end():
return match.group()
NOTE This yields more readable code, but it should be alot slower for large inputs.
answered 6 hours ago
Ludisposed
6,93921959
add a comment |
up vote
7
down vote
Review
Add docstrings and tests... or both in the form of doctests!
def find_word(s, index):
"""
Finds the CamalCased word surrounding the givin index in the string
>>> find_word('CamelCaseString', 6)
'Case'
>>> find_word('ACamelCaseString', 0)
'A'
"""
...
Loop like a native.
Instead of going over the indexes we can loop over the item directly
range(index, 0, -1)
We can loop over the item and index at the same time using enumerate
for i, s in enumerate(string[index:0:-1])
However this would be slower since it will create a new string object with every slice.
If we can be sure that the givin string is a CamalCase string
Then we can drop some of your second if statement
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
Would be
if s[i+1].isupper():
Actually your code (from a performance aspect) is quite good
We could however use a while loop to increment both side at once, for a little performance gain.
(slower, yet more readable) Alternative
A different approach to finding CamalCase words can be done with regex,
We can find all CamalCase words with the following regex: r"([A-Z][a-z]*)"
And we can use re.finditer to create a generator for our matches and loop over them, and return when our index is in between the end and the start.
import re
def find_word_2(string, index):
for match in re.finditer(r"([A-Z][a-z]*)", string):
if match.start() <= index < match.end():
return match.group()
NOTE This yields more readable code, but it should be alot slower for large inputs.
answered 6 hours ago
Ludisposed
6,93921959
add a comment |
up vote
7
down vote
up vote
7
down vote
Review
Add docstrings and tests... or both in the form of doctests!
def find_word(s, index):
"""
Finds the CamalCased word surrounding the givin index in the string
>>> find_word('CamelCaseString', 6)
'Case'
>>> find_word('ACamelCaseString', 0)
'A'
"""
...
Loop like a native.
Instead of going over the indexes we can loop over the item directly
range(index, 0, -1)
We can loop over the item and index at the same time using enumerate
for i, s in enumerate(string[index:0:-1])
However this would be slower since it will create a new string object with every slice.
If we can be sure that the givin string is a CamalCase string
Then we can drop some of your second if statement
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
Would be
if s[i+1].isupper():
Actually your code (from a performance aspect) is quite good
We could however use a while loop to increment both side at once, for a little performance gain.
(slower, yet more readable) Alternative
A different approach to finding CamalCase words can be done with regex,
We can find all CamalCase words with the following regex: r"([A-Z][a-z]*)"
And we can use re.finditer to create a generator for our matches and loop over them, and return when our index is in between the end and the start.
import re
def find_word_2(string, index):
for match in re.finditer(r"([A-Z][a-z]*)", string):
if match.start() <= index < match.end():
return match.group()
NOTE This yields more readable code, but it should be alot slower for large inputs.
answered 6 hours ago
Ludisposed
6,93921959
Review
Add docstrings and tests... or both in the form of doctests!
def find_word(s, index):
"""
Finds the CamalCased word surrounding the givin index in the string
>>> find_word('CamelCaseString', 6)
'Case'
>>> find_word('ACamelCaseString', 0)
'A'
"""
...
Loop like a native.
Instead of going over the indexes we can loop over the item directly
range(index, 0, -1)
We can loop over the item and index at the same time using enumerate
for i, s in enumerate(string[index:0:-1])
However this would be slower since it will create a new string object with every slice.
If we can be sure that the givin string is a CamalCase string
Then we can drop some of your second if statement
if s[i].islower() and s[i+1].isupper() or s[i:i+2].isupper():
Would be
if s[i+1].isupper():
Actually your code (from a performance aspect) is quite good
We could however use a while loop to increment both side at once, for a little performance gain.
(slower, yet more readable) Alternative
A different approach to finding CamalCase words can be done with regex,
We can find all CamalCase words with the following regex: r"([A-Z][a-z]*)"
And we can use re.finditer to create a generator for our matches and loop over them, and return when our index is in between the end and the start.
import re
def find_word_2(string, index):
for match in re.finditer(r"([A-Z][a-z]*)", string):
if match.start() <= index < match.end():
return match.group()
NOTE This yields more readable code, but it should be alot slower for large inputs.
answered 6 hours ago
Ludisposed
6,93921959
answered 6 hours ago
Ludisposed
6,93921959
answered 6 hours ago
Ludisposed
6,93921959
answered 6 hours ago
Ludisposed
6,93921959
6,93921959
add a comment |
add a comment |
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Shouldn't the first example return
String?– Heslacher
7 hours ago
1
No, the index 6 is character
'a'in subword'Case'.– Eugene Yarmash
7 hours ago
OK, now I got it.
– Heslacher
7 hours ago