Calculate Levenshtein distance between two strings in PythonEdit Distance Between Two StringsString Matching and ClusteringSorting movie search results by similarityEdit distance between 2 stringsMaking the Levenshtein distance code cleanerEdit distance (Optimal Alignment) - follow upGet Levenshtein DistanceMessage classification with Levenshtein DistanceCode to implement the Jaro similarity for fuzzy matching stringsFinding differences in strings with Levenshtein distance and soundex
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Calculate Levenshtein distance between two strings in Python
Edit Distance Between Two StringsString Matching and ClusteringSorting movie search results by similarityEdit distance between 2 stringsMaking the Levenshtein distance code cleanerEdit distance (Optimal Alignment) - follow upGet Levenshtein DistanceMessage classification with Levenshtein DistanceCode to implement the Jaro similarity for fuzzy matching stringsFinding differences in strings with Levenshtein distance and soundex
.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
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to Code Review! Good job on your first question here. I wish you good reviews!
$endgroup$
– Alex
8 hours ago
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
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 36 mins ago
Reinderien
5,335926
5,335926
New contributor
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Kyra_WKyra_W
311
311
New contributor
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kyra_W is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Welcome to Code Review! Good job on your first question here. I wish you good reviews!
$endgroup$
– Alex
8 hours ago
add a comment |
1
$begingroup$
Welcome to Code Review! Good job on your first question here. I wish you good reviews!
$endgroup$
– Alex
8 hours ago
1
1
$begingroup$
Welcome to Code Review! Good job on your first question here. I wish you good reviews!
$endgroup$
– Alex
8 hours ago
$begingroup$
Welcome to Code Review! Good job on your first question here. I wish you good reviews!
$endgroup$
– Alex
8 hours ago
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.
"""
...
$endgroup$
5
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
6 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 |
Your Answer
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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.
"""
...
$endgroup$
5
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
6 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.
"""
...
$endgroup$
5
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
6 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 9 hours ago
answered 9 hours ago
GraipherGraipher
26.9k54396
26.9k54396
5
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
6 hours ago
add a comment |
5
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
6 hours ago
5
5
$begingroup$
Just to note the module is licensed under GPL 2.0 so watch out if you're using it for work.
$endgroup$
– lucasgcb
6 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
6 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 6 hours ago
Maarten FabréMaarten Fabré
5,119517
5,119517
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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1
$begingroup$
Welcome to Code Review! Good job on your first question here. I wish you good reviews!
$endgroup$
– Alex
8 hours ago