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Calculate Levenshtein distance between two strings in Python
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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I need a function that checks how different are two different strings. I chose the Levenshtein distance as a quick approach, and implemented this function:
from difflib import ndiff
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference between two sequences.
It is calculated as the minimum number of single-character edits necessary to transform one string into another
"""
distance = 0
buffer_removed = buffer_added = 0
for x in ndiff(str_1, str_2):
code = x[0]
# Code ? is ignored as it does not translate to any modification
if code == ' ':
distance += max(buffer_removed, buffer_added)
buffer_removed = buffer_added = 0
elif code == '-':
buffer_removed += 1
elif code == '+':
buffer_added += 1
distance += max(buffer_removed, buffer_added)
return distance
Then calling it as:
similarity = 1 - calculate_levenshtein_distance(str_1, str_2) / max(len(str_1), len(str_2))
How sloppy/prone to errors is this code? How can it be improved?
python edit-distance
New contributor
$endgroup$
add a comment |
$begingroup$
I need a function that checks how different are two different strings. I chose the Levenshtein distance as a quick approach, and implemented this function:
from difflib import ndiff
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference between two sequences.
It is calculated as the minimum number of single-character edits necessary to transform one string into another
"""
distance = 0
buffer_removed = buffer_added = 0
for x in ndiff(str_1, str_2):
code = x[0]
# Code ? is ignored as it does not translate to any modification
if code == ' ':
distance += max(buffer_removed, buffer_added)
buffer_removed = buffer_added = 0
elif code == '-':
buffer_removed += 1
elif code == '+':
buffer_added += 1
distance += max(buffer_removed, buffer_added)
return distance
Then calling it as:
similarity = 1 - calculate_levenshtein_distance(str_1, str_2) / max(len(str_1), len(str_2))
How sloppy/prone to errors is this code? How can it be improved?
python edit-distance
New contributor
$endgroup$
add a comment |
$begingroup$
I need a function that checks how different are two different strings. I chose the Levenshtein distance as a quick approach, and implemented this function:
from difflib import ndiff
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference between two sequences.
It is calculated as the minimum number of single-character edits necessary to transform one string into another
"""
distance = 0
buffer_removed = buffer_added = 0
for x in ndiff(str_1, str_2):
code = x[0]
# Code ? is ignored as it does not translate to any modification
if code == ' ':
distance += max(buffer_removed, buffer_added)
buffer_removed = buffer_added = 0
elif code == '-':
buffer_removed += 1
elif code == '+':
buffer_added += 1
distance += max(buffer_removed, buffer_added)
return distance
Then calling it as:
similarity = 1 - calculate_levenshtein_distance(str_1, str_2) / max(len(str_1), len(str_2))
How sloppy/prone to errors is this code? How can it be improved?
python edit-distance
New contributor
$endgroup$
I need a function that checks how different are two different strings. I chose the Levenshtein distance as a quick approach, and implemented this function:
from difflib import ndiff
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference between two sequences.
It is calculated as the minimum number of single-character edits necessary to transform one string into another
"""
distance = 0
buffer_removed = buffer_added = 0
for x in ndiff(str_1, str_2):
code = x[0]
# Code ? is ignored as it does not translate to any modification
if code == ' ':
distance += max(buffer_removed, buffer_added)
buffer_removed = buffer_added = 0
elif code == '-':
buffer_removed += 1
elif code == '+':
buffer_added += 1
distance += max(buffer_removed, buffer_added)
return distance
Then calling it as:
similarity = 1 - calculate_levenshtein_distance(str_1, str_2) / max(len(str_1), len(str_2))
How sloppy/prone to errors is this code? How can it be improved?
python edit-distance
python edit-distance
New contributor
New contributor
edited 10 hours ago
Reinderien
5,340927
5,340927
New contributor
asked 19 hours ago
Kyra_WKyra_W
462
462
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is a module available for exactly that calculation, python-Levenshtein
. You can install it with pip install python-Levenshtein
.
It is implemented in C, so is probably faster than anything you can come up with yourself.
from Levenshtein import distance as levenshtein_distance
According to the docstring
conventions, your docstring should look like this, i.e. with the indentation aligned to the """
and the line length curtailed to 80 characters.
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference
between two sequences.
It is calculated as the minimum number of single-character edits necessary to
transform one string into another.
"""
...
$endgroup$
7
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
add a comment |
$begingroup$
The code itself is rather clear. There are some smaller changes I would make
tuple unpacking
You can use tuple unpacking to do:
for code, *_ in ndiff(str1, str2):
instead of:
for x in ndiff(str_1, str_2):
code = x[0]
dict results:
Instead of a counter for the additions and removals, I would keep it in 1 dict: counter = ({"+": 0, "-": 0})
def levenshtein_distance(str1, str2, ):
counter = {"+": 0, "-": 0}
distance = 0
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
distance += max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
distance += max(counter.values())
return distance
generators
A smaller, less useful variation, is to let this method be a generator, and use the builtin sum
to do the summary. this saves 1 variable inside the function:
def levenshtein_distance_gen(str1, str2, ):
counter = {"+": 0, "-": 0}
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
yield max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
yield max(counter.values())
sum(levenshtein_distance_gen(str1, str2))
timings
The differences in timings between the original and both these variations are minimal, and within the variation of results. This is rather logical, since for simple strings (aaabbbc
and abcabcabc
) 90% of the time is spent in ndiff
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a module available for exactly that calculation, python-Levenshtein
. You can install it with pip install python-Levenshtein
.
It is implemented in C, so is probably faster than anything you can come up with yourself.
from Levenshtein import distance as levenshtein_distance
According to the docstring
conventions, your docstring should look like this, i.e. with the indentation aligned to the """
and the line length curtailed to 80 characters.
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference
between two sequences.
It is calculated as the minimum number of single-character edits necessary to
transform one string into another.
"""
...
$endgroup$
7
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
add a comment |
$begingroup$
There is a module available for exactly that calculation, python-Levenshtein
. You can install it with pip install python-Levenshtein
.
It is implemented in C, so is probably faster than anything you can come up with yourself.
from Levenshtein import distance as levenshtein_distance
According to the docstring
conventions, your docstring should look like this, i.e. with the indentation aligned to the """
and the line length curtailed to 80 characters.
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference
between two sequences.
It is calculated as the minimum number of single-character edits necessary to
transform one string into another.
"""
...
$endgroup$
7
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
add a comment |
$begingroup$
There is a module available for exactly that calculation, python-Levenshtein
. You can install it with pip install python-Levenshtein
.
It is implemented in C, so is probably faster than anything you can come up with yourself.
from Levenshtein import distance as levenshtein_distance
According to the docstring
conventions, your docstring should look like this, i.e. with the indentation aligned to the """
and the line length curtailed to 80 characters.
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference
between two sequences.
It is calculated as the minimum number of single-character edits necessary to
transform one string into another.
"""
...
$endgroup$
There is a module available for exactly that calculation, python-Levenshtein
. You can install it with pip install python-Levenshtein
.
It is implemented in C, so is probably faster than anything you can come up with yourself.
from Levenshtein import distance as levenshtein_distance
According to the docstring
conventions, your docstring should look like this, i.e. with the indentation aligned to the """
and the line length curtailed to 80 characters.
def calculate_levenshtein_distance(str_1, str_2):
"""
The Levenshtein distance is a string metric for measuring the difference
between two sequences.
It is calculated as the minimum number of single-character edits necessary to
transform one string into another.
"""
...
edited 18 hours ago
answered 19 hours ago
GraipherGraipher
26.9k54397
26.9k54397
7
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
add a comment |
7
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
7
7
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
16 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
$begingroup$
Just to point out a small nitpick to other people who may stumble upon this answer, as per help center: "Every answer must make at least one insightful observation about the code in the question. Answers that merely provide an alternate solution with no explanation or justification do not constitute valid Code Review answers and may be deleted." While this answer does provide alternative and existing module suggestion, it also goes into some suggestions about improving code quality. So it's an example of a decent answer
$endgroup$
– Sergiy Kolodyazhnyy
5 hours ago
add a comment |
$begingroup$
The code itself is rather clear. There are some smaller changes I would make
tuple unpacking
You can use tuple unpacking to do:
for code, *_ in ndiff(str1, str2):
instead of:
for x in ndiff(str_1, str_2):
code = x[0]
dict results:
Instead of a counter for the additions and removals, I would keep it in 1 dict: counter = ({"+": 0, "-": 0})
def levenshtein_distance(str1, str2, ):
counter = {"+": 0, "-": 0}
distance = 0
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
distance += max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
distance += max(counter.values())
return distance
generators
A smaller, less useful variation, is to let this method be a generator, and use the builtin sum
to do the summary. this saves 1 variable inside the function:
def levenshtein_distance_gen(str1, str2, ):
counter = {"+": 0, "-": 0}
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
yield max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
yield max(counter.values())
sum(levenshtein_distance_gen(str1, str2))
timings
The differences in timings between the original and both these variations are minimal, and within the variation of results. This is rather logical, since for simple strings (aaabbbc
and abcabcabc
) 90% of the time is spent in ndiff
$endgroup$
add a comment |
$begingroup$
The code itself is rather clear. There are some smaller changes I would make
tuple unpacking
You can use tuple unpacking to do:
for code, *_ in ndiff(str1, str2):
instead of:
for x in ndiff(str_1, str_2):
code = x[0]
dict results:
Instead of a counter for the additions and removals, I would keep it in 1 dict: counter = ({"+": 0, "-": 0})
def levenshtein_distance(str1, str2, ):
counter = {"+": 0, "-": 0}
distance = 0
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
distance += max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
distance += max(counter.values())
return distance
generators
A smaller, less useful variation, is to let this method be a generator, and use the builtin sum
to do the summary. this saves 1 variable inside the function:
def levenshtein_distance_gen(str1, str2, ):
counter = {"+": 0, "-": 0}
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
yield max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
yield max(counter.values())
sum(levenshtein_distance_gen(str1, str2))
timings
The differences in timings between the original and both these variations are minimal, and within the variation of results. This is rather logical, since for simple strings (aaabbbc
and abcabcabc
) 90% of the time is spent in ndiff
$endgroup$
add a comment |
$begingroup$
The code itself is rather clear. There are some smaller changes I would make
tuple unpacking
You can use tuple unpacking to do:
for code, *_ in ndiff(str1, str2):
instead of:
for x in ndiff(str_1, str_2):
code = x[0]
dict results:
Instead of a counter for the additions and removals, I would keep it in 1 dict: counter = ({"+": 0, "-": 0})
def levenshtein_distance(str1, str2, ):
counter = {"+": 0, "-": 0}
distance = 0
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
distance += max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
distance += max(counter.values())
return distance
generators
A smaller, less useful variation, is to let this method be a generator, and use the builtin sum
to do the summary. this saves 1 variable inside the function:
def levenshtein_distance_gen(str1, str2, ):
counter = {"+": 0, "-": 0}
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
yield max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
yield max(counter.values())
sum(levenshtein_distance_gen(str1, str2))
timings
The differences in timings between the original and both these variations are minimal, and within the variation of results. This is rather logical, since for simple strings (aaabbbc
and abcabcabc
) 90% of the time is spent in ndiff
$endgroup$
The code itself is rather clear. There are some smaller changes I would make
tuple unpacking
You can use tuple unpacking to do:
for code, *_ in ndiff(str1, str2):
instead of:
for x in ndiff(str_1, str_2):
code = x[0]
dict results:
Instead of a counter for the additions and removals, I would keep it in 1 dict: counter = ({"+": 0, "-": 0})
def levenshtein_distance(str1, str2, ):
counter = {"+": 0, "-": 0}
distance = 0
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
distance += max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
distance += max(counter.values())
return distance
generators
A smaller, less useful variation, is to let this method be a generator, and use the builtin sum
to do the summary. this saves 1 variable inside the function:
def levenshtein_distance_gen(str1, str2, ):
counter = {"+": 0, "-": 0}
for edit_code, *_ in ndiff(str1, str2):
if edit_code == " ":
yield max(counter.values())
counter = {"+": 0, "-": 0}
else:
counter[edit_code] += 1
yield max(counter.values())
sum(levenshtein_distance_gen(str1, str2))
timings
The differences in timings between the original and both these variations are minimal, and within the variation of results. This is rather logical, since for simple strings (aaabbbc
and abcabcabc
) 90% of the time is spent in ndiff
answered 15 hours ago
Maarten FabréMaarten Fabré
5,139517
5,139517
add a comment |
add a comment |
Kyra_W is a new contributor. Be nice, and check out our Code of Conduct.
Kyra_W is a new contributor. Be nice, and check out our Code of Conduct.
Kyra_W is a new contributor. Be nice, and check out our Code of Conduct.
Kyra_W is a new contributor. Be nice, and check out our Code of Conduct.
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