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Print a random anagram of a given string
Simple anagram or permutation generator in PythonGuessing a random number between a range of numbersRandom Distribution to fit an Average PythonRandom number generation seeding in C++Testing a Random number generatorGenerate cryptographically secure random numbers in a specific rangePython - Faster random business date generationTesting random number generator functionFrom a set of words, output all words that fit a string of random letters (like Scrabble)Split a given number so that their sum adds to another given numberPseudo-truly random number generator
$begingroup$
I am given a task to create a function that prints a random anagram of a given string:
def anagram(value):
'''Prints random anagram of given value'''
import random
newWord = ''
for i in range(len(value)):
pos = random.randint(0, len(value)-1)
newWord += value[pos]
value = value[:pos] + value[pos+1:]
print newWord
anagram('12345')
anagram('should')
What all edge cases should this program cover?
python python-2.x random
$endgroup$
add a comment |
$begingroup$
I am given a task to create a function that prints a random anagram of a given string:
def anagram(value):
'''Prints random anagram of given value'''
import random
newWord = ''
for i in range(len(value)):
pos = random.randint(0, len(value)-1)
newWord += value[pos]
value = value[:pos] + value[pos+1:]
print newWord
anagram('12345')
anagram('should')
What all edge cases should this program cover?
python python-2.x random
$endgroup$
$begingroup$
Possible duplicate of Simple anagram or permutation generator in Python
$endgroup$
– hjpotter92
Jan 17 '18 at 16:47
3
$begingroup$
@hjpotter92 Asking for a code review of code that solves the same problem as in some other question is fine here. Just having copy & pasted the other persons code would be a dupe (and also off-topic), or re-posting your own question (without having changed anything), instead of editing it.
$endgroup$
– Graipher
Jan 17 '18 at 16:54
1
$begingroup$
Is is acceptable to return the input string?
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:37
add a comment |
$begingroup$
I am given a task to create a function that prints a random anagram of a given string:
def anagram(value):
'''Prints random anagram of given value'''
import random
newWord = ''
for i in range(len(value)):
pos = random.randint(0, len(value)-1)
newWord += value[pos]
value = value[:pos] + value[pos+1:]
print newWord
anagram('12345')
anagram('should')
What all edge cases should this program cover?
python python-2.x random
$endgroup$
I am given a task to create a function that prints a random anagram of a given string:
def anagram(value):
'''Prints random anagram of given value'''
import random
newWord = ''
for i in range(len(value)):
pos = random.randint(0, len(value)-1)
newWord += value[pos]
value = value[:pos] + value[pos+1:]
print newWord
anagram('12345')
anagram('should')
What all edge cases should this program cover?
python python-2.x random
python python-2.x random
edited Jan 21 '18 at 0:51
Jamal♦
30.4k11121227
30.4k11121227
asked Jan 17 '18 at 16:41
Latika AgarwalLatika Agarwal
881426
881426
$begingroup$
Possible duplicate of Simple anagram or permutation generator in Python
$endgroup$
– hjpotter92
Jan 17 '18 at 16:47
3
$begingroup$
@hjpotter92 Asking for a code review of code that solves the same problem as in some other question is fine here. Just having copy & pasted the other persons code would be a dupe (and also off-topic), or re-posting your own question (without having changed anything), instead of editing it.
$endgroup$
– Graipher
Jan 17 '18 at 16:54
1
$begingroup$
Is is acceptable to return the input string?
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:37
add a comment |
$begingroup$
Possible duplicate of Simple anagram or permutation generator in Python
$endgroup$
– hjpotter92
Jan 17 '18 at 16:47
3
$begingroup$
@hjpotter92 Asking for a code review of code that solves the same problem as in some other question is fine here. Just having copy & pasted the other persons code would be a dupe (and also off-topic), or re-posting your own question (without having changed anything), instead of editing it.
$endgroup$
– Graipher
Jan 17 '18 at 16:54
1
$begingroup$
Is is acceptable to return the input string?
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:37
$begingroup$
Possible duplicate of Simple anagram or permutation generator in Python
$endgroup$
– hjpotter92
Jan 17 '18 at 16:47
$begingroup$
Possible duplicate of Simple anagram or permutation generator in Python
$endgroup$
– hjpotter92
Jan 17 '18 at 16:47
3
3
$begingroup$
@hjpotter92 Asking for a code review of code that solves the same problem as in some other question is fine here. Just having copy & pasted the other persons code would be a dupe (and also off-topic), or re-posting your own question (without having changed anything), instead of editing it.
$endgroup$
– Graipher
Jan 17 '18 at 16:54
$begingroup$
@hjpotter92 Asking for a code review of code that solves the same problem as in some other question is fine here. Just having copy & pasted the other persons code would be a dupe (and also off-topic), or re-posting your own question (without having changed anything), instead of editing it.
$endgroup$
– Graipher
Jan 17 '18 at 16:54
1
1
$begingroup$
Is is acceptable to return the input string?
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:37
$begingroup$
Is is acceptable to return the input string?
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You should usually not put the import
s into your functions. While Python does not re-import a module it has already imported, it still needs to run the check for this. So this introduces an unnecessary overhead, when running the function multiple times. There are a few use cases for doing this, though, like not wanting to pollute the global namespace if that module does some dirty hacks during its initialization or it takes a very long time to import and you only want to do this sometimes (and only run the function once). But random
is no such module.
Your code makes this also too complicated. The biggest part of your algorithm is making sure that you don't re-use a letter you already used. You can use random.sample
for this, instead. It randomly samples (duh) from an iterable (well it needs to be iterable and indexable, actually), without replacement:
import random
def anagram(value):
'''Returns random anagram of given value'''
return ''.join(random.sample(value, len(value)))
An alternative would be random.shuffle
, which shuffles a list in place, but this has the overhead of casting to a list first. The documentation actually recommends using the first one (which is also a lot clearer IMO).
def anagram(value):
'''Returns random anagram of given value'''
value_list = list(value)
random.shuffle(value_list)
return ''.join(value_list)
As for corner cases to test? The obvious one is the empty string ''
. Then maybe a string containing only one distinct letter, like 'aaa'
(to catch some very weird algorithm that only looks at the set of letters, which would be quite wrong, of course). And then finally maybe the scaling behavior of the algorithm so strings of increasing length.
All should be tested for example with this:
from collections import Counter
def test_anagram_function(s):
assert Counter(s) == Counter(anagram(s))
$endgroup$
add a comment |
$begingroup$
- When building a string you should build a list, and then use
''.join()
. This is as strings are immutable, and so generatingnewWord
takes $O(n^2)$ time. - You are mannually poping
pos
fromvalue
. If you changevalue
to a list, you can just uselist.pop
. - You should
import random
at the top of your code. Never in a function. - Common Python style is to use
snake_case
for variables. - Common Python style is to use
_
as a throw away variable. - You can use
random.randrange
, rather thanrandint
- It's best if you
return
rather thanprint
your anagram.
And so you could use:
def anagram(value):
new_word = []
value = list(value)
for _ in range(len(value)):
pos = random.randrange(len(value))
new_word.append(value.pop(pos))
return ''.join(new_word)
This however runs in $O(n^2)$ time. If you use a Fisher–Yates shuffle, you can do this in $O(n)$ time.
def anagram(value):
value = list(value)
for i in range(len(value)):
pos = random.randrange(i, len(value))
value[i], value[pos] = value[pos], value[i]
return ''.join(value)
You can also use random.shuffle
, which likely also uses the above algorithm. However you won't have to maintain it. Allowing:
def anagram(value):
value = list(value)
random.shuffle(value)
return ''.join(value)
$endgroup$
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
4
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
Just curious : why "never in a function"? It could make sense for aheavy_computation()
function, which uses many libraries but whose result gets cached for later use.
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
2
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you doimport module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.
$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
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
$begingroup$
You should usually not put the import
s into your functions. While Python does not re-import a module it has already imported, it still needs to run the check for this. So this introduces an unnecessary overhead, when running the function multiple times. There are a few use cases for doing this, though, like not wanting to pollute the global namespace if that module does some dirty hacks during its initialization or it takes a very long time to import and you only want to do this sometimes (and only run the function once). But random
is no such module.
Your code makes this also too complicated. The biggest part of your algorithm is making sure that you don't re-use a letter you already used. You can use random.sample
for this, instead. It randomly samples (duh) from an iterable (well it needs to be iterable and indexable, actually), without replacement:
import random
def anagram(value):
'''Returns random anagram of given value'''
return ''.join(random.sample(value, len(value)))
An alternative would be random.shuffle
, which shuffles a list in place, but this has the overhead of casting to a list first. The documentation actually recommends using the first one (which is also a lot clearer IMO).
def anagram(value):
'''Returns random anagram of given value'''
value_list = list(value)
random.shuffle(value_list)
return ''.join(value_list)
As for corner cases to test? The obvious one is the empty string ''
. Then maybe a string containing only one distinct letter, like 'aaa'
(to catch some very weird algorithm that only looks at the set of letters, which would be quite wrong, of course). And then finally maybe the scaling behavior of the algorithm so strings of increasing length.
All should be tested for example with this:
from collections import Counter
def test_anagram_function(s):
assert Counter(s) == Counter(anagram(s))
$endgroup$
add a comment |
$begingroup$
You should usually not put the import
s into your functions. While Python does not re-import a module it has already imported, it still needs to run the check for this. So this introduces an unnecessary overhead, when running the function multiple times. There are a few use cases for doing this, though, like not wanting to pollute the global namespace if that module does some dirty hacks during its initialization or it takes a very long time to import and you only want to do this sometimes (and only run the function once). But random
is no such module.
Your code makes this also too complicated. The biggest part of your algorithm is making sure that you don't re-use a letter you already used. You can use random.sample
for this, instead. It randomly samples (duh) from an iterable (well it needs to be iterable and indexable, actually), without replacement:
import random
def anagram(value):
'''Returns random anagram of given value'''
return ''.join(random.sample(value, len(value)))
An alternative would be random.shuffle
, which shuffles a list in place, but this has the overhead of casting to a list first. The documentation actually recommends using the first one (which is also a lot clearer IMO).
def anagram(value):
'''Returns random anagram of given value'''
value_list = list(value)
random.shuffle(value_list)
return ''.join(value_list)
As for corner cases to test? The obvious one is the empty string ''
. Then maybe a string containing only one distinct letter, like 'aaa'
(to catch some very weird algorithm that only looks at the set of letters, which would be quite wrong, of course). And then finally maybe the scaling behavior of the algorithm so strings of increasing length.
All should be tested for example with this:
from collections import Counter
def test_anagram_function(s):
assert Counter(s) == Counter(anagram(s))
$endgroup$
add a comment |
$begingroup$
You should usually not put the import
s into your functions. While Python does not re-import a module it has already imported, it still needs to run the check for this. So this introduces an unnecessary overhead, when running the function multiple times. There are a few use cases for doing this, though, like not wanting to pollute the global namespace if that module does some dirty hacks during its initialization or it takes a very long time to import and you only want to do this sometimes (and only run the function once). But random
is no such module.
Your code makes this also too complicated. The biggest part of your algorithm is making sure that you don't re-use a letter you already used. You can use random.sample
for this, instead. It randomly samples (duh) from an iterable (well it needs to be iterable and indexable, actually), without replacement:
import random
def anagram(value):
'''Returns random anagram of given value'''
return ''.join(random.sample(value, len(value)))
An alternative would be random.shuffle
, which shuffles a list in place, but this has the overhead of casting to a list first. The documentation actually recommends using the first one (which is also a lot clearer IMO).
def anagram(value):
'''Returns random anagram of given value'''
value_list = list(value)
random.shuffle(value_list)
return ''.join(value_list)
As for corner cases to test? The obvious one is the empty string ''
. Then maybe a string containing only one distinct letter, like 'aaa'
(to catch some very weird algorithm that only looks at the set of letters, which would be quite wrong, of course). And then finally maybe the scaling behavior of the algorithm so strings of increasing length.
All should be tested for example with this:
from collections import Counter
def test_anagram_function(s):
assert Counter(s) == Counter(anagram(s))
$endgroup$
You should usually not put the import
s into your functions. While Python does not re-import a module it has already imported, it still needs to run the check for this. So this introduces an unnecessary overhead, when running the function multiple times. There are a few use cases for doing this, though, like not wanting to pollute the global namespace if that module does some dirty hacks during its initialization or it takes a very long time to import and you only want to do this sometimes (and only run the function once). But random
is no such module.
Your code makes this also too complicated. The biggest part of your algorithm is making sure that you don't re-use a letter you already used. You can use random.sample
for this, instead. It randomly samples (duh) from an iterable (well it needs to be iterable and indexable, actually), without replacement:
import random
def anagram(value):
'''Returns random anagram of given value'''
return ''.join(random.sample(value, len(value)))
An alternative would be random.shuffle
, which shuffles a list in place, but this has the overhead of casting to a list first. The documentation actually recommends using the first one (which is also a lot clearer IMO).
def anagram(value):
'''Returns random anagram of given value'''
value_list = list(value)
random.shuffle(value_list)
return ''.join(value_list)
As for corner cases to test? The obvious one is the empty string ''
. Then maybe a string containing only one distinct letter, like 'aaa'
(to catch some very weird algorithm that only looks at the set of letters, which would be quite wrong, of course). And then finally maybe the scaling behavior of the algorithm so strings of increasing length.
All should be tested for example with this:
from collections import Counter
def test_anagram_function(s):
assert Counter(s) == Counter(anagram(s))
edited Jan 17 '18 at 21:01
answered Jan 17 '18 at 16:59
GraipherGraipher
26.1k54091
26.1k54091
add a comment |
add a comment |
$begingroup$
- When building a string you should build a list, and then use
''.join()
. This is as strings are immutable, and so generatingnewWord
takes $O(n^2)$ time. - You are mannually poping
pos
fromvalue
. If you changevalue
to a list, you can just uselist.pop
. - You should
import random
at the top of your code. Never in a function. - Common Python style is to use
snake_case
for variables. - Common Python style is to use
_
as a throw away variable. - You can use
random.randrange
, rather thanrandint
- It's best if you
return
rather thanprint
your anagram.
And so you could use:
def anagram(value):
new_word = []
value = list(value)
for _ in range(len(value)):
pos = random.randrange(len(value))
new_word.append(value.pop(pos))
return ''.join(new_word)
This however runs in $O(n^2)$ time. If you use a Fisher–Yates shuffle, you can do this in $O(n)$ time.
def anagram(value):
value = list(value)
for i in range(len(value)):
pos = random.randrange(i, len(value))
value[i], value[pos] = value[pos], value[i]
return ''.join(value)
You can also use random.shuffle
, which likely also uses the above algorithm. However you won't have to maintain it. Allowing:
def anagram(value):
value = list(value)
random.shuffle(value)
return ''.join(value)
$endgroup$
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
4
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
Just curious : why "never in a function"? It could make sense for aheavy_computation()
function, which uses many libraries but whose result gets cached for later use.
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
2
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you doimport module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.
$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
add a comment |
$begingroup$
- When building a string you should build a list, and then use
''.join()
. This is as strings are immutable, and so generatingnewWord
takes $O(n^2)$ time. - You are mannually poping
pos
fromvalue
. If you changevalue
to a list, you can just uselist.pop
. - You should
import random
at the top of your code. Never in a function. - Common Python style is to use
snake_case
for variables. - Common Python style is to use
_
as a throw away variable. - You can use
random.randrange
, rather thanrandint
- It's best if you
return
rather thanprint
your anagram.
And so you could use:
def anagram(value):
new_word = []
value = list(value)
for _ in range(len(value)):
pos = random.randrange(len(value))
new_word.append(value.pop(pos))
return ''.join(new_word)
This however runs in $O(n^2)$ time. If you use a Fisher–Yates shuffle, you can do this in $O(n)$ time.
def anagram(value):
value = list(value)
for i in range(len(value)):
pos = random.randrange(i, len(value))
value[i], value[pos] = value[pos], value[i]
return ''.join(value)
You can also use random.shuffle
, which likely also uses the above algorithm. However you won't have to maintain it. Allowing:
def anagram(value):
value = list(value)
random.shuffle(value)
return ''.join(value)
$endgroup$
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
4
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
Just curious : why "never in a function"? It could make sense for aheavy_computation()
function, which uses many libraries but whose result gets cached for later use.
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
2
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you doimport module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.
$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
add a comment |
$begingroup$
- When building a string you should build a list, and then use
''.join()
. This is as strings are immutable, and so generatingnewWord
takes $O(n^2)$ time. - You are mannually poping
pos
fromvalue
. If you changevalue
to a list, you can just uselist.pop
. - You should
import random
at the top of your code. Never in a function. - Common Python style is to use
snake_case
for variables. - Common Python style is to use
_
as a throw away variable. - You can use
random.randrange
, rather thanrandint
- It's best if you
return
rather thanprint
your anagram.
And so you could use:
def anagram(value):
new_word = []
value = list(value)
for _ in range(len(value)):
pos = random.randrange(len(value))
new_word.append(value.pop(pos))
return ''.join(new_word)
This however runs in $O(n^2)$ time. If you use a Fisher–Yates shuffle, you can do this in $O(n)$ time.
def anagram(value):
value = list(value)
for i in range(len(value)):
pos = random.randrange(i, len(value))
value[i], value[pos] = value[pos], value[i]
return ''.join(value)
You can also use random.shuffle
, which likely also uses the above algorithm. However you won't have to maintain it. Allowing:
def anagram(value):
value = list(value)
random.shuffle(value)
return ''.join(value)
$endgroup$
- When building a string you should build a list, and then use
''.join()
. This is as strings are immutable, and so generatingnewWord
takes $O(n^2)$ time. - You are mannually poping
pos
fromvalue
. If you changevalue
to a list, you can just uselist.pop
. - You should
import random
at the top of your code. Never in a function. - Common Python style is to use
snake_case
for variables. - Common Python style is to use
_
as a throw away variable. - You can use
random.randrange
, rather thanrandint
- It's best if you
return
rather thanprint
your anagram.
And so you could use:
def anagram(value):
new_word = []
value = list(value)
for _ in range(len(value)):
pos = random.randrange(len(value))
new_word.append(value.pop(pos))
return ''.join(new_word)
This however runs in $O(n^2)$ time. If you use a Fisher–Yates shuffle, you can do this in $O(n)$ time.
def anagram(value):
value = list(value)
for i in range(len(value)):
pos = random.randrange(i, len(value))
value[i], value[pos] = value[pos], value[i]
return ''.join(value)
You can also use random.shuffle
, which likely also uses the above algorithm. However you won't have to maintain it. Allowing:
def anagram(value):
value = list(value)
random.shuffle(value)
return ''.join(value)
edited Jan 17 '18 at 17:11
answered Jan 17 '18 at 17:00
PeilonrayzPeilonrayz
25.9k338109
25.9k338109
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
4
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
Just curious : why "never in a function"? It could make sense for aheavy_computation()
function, which uses many libraries but whose result gets cached for later use.
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
2
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you doimport module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.
$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
add a comment |
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
4
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
Just curious : why "never in a function"? It could make sense for aheavy_computation()
function, which uses many libraries but whose result gets cached for later use.
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
2
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you doimport module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.
$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
$begingroup$
“never” is a strong word. Sometimes, due to circular imports, you don’t have any other alternative:P (but of course it doesn’t apply in this scenario)
$endgroup$
– яүυк
Jan 17 '18 at 17:53
4
4
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
@MrGrj if you have circular imports you have much bigger problems than where your imports go.
$endgroup$
– Peilonrayz
Jan 17 '18 at 17:55
$begingroup$
Just curious : why "never in a function"? It could make sense for a
heavy_computation()
function, which uses many libraries but whose result gets cached for later use.$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
$begingroup$
Just curious : why "never in a function"? It could make sense for a
heavy_computation()
function, which uses many libraries but whose result gets cached for later use.$endgroup$
– Eric Duminil
Jan 17 '18 at 20:36
2
2
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you do
import module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
$begingroup$
@EricDuminil The simple answer is PEP 8 says so. However, why wouldn't you make it a module? You can have it so it only loads when you do
import module.submodule
, just like a lot of the math and game libraries do. This has the benefit that it's cached without you having to do anything too.$endgroup$
– Peilonrayz
Jan 17 '18 at 20:41
add a comment |
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$begingroup$
Possible duplicate of Simple anagram or permutation generator in Python
$endgroup$
– hjpotter92
Jan 17 '18 at 16:47
3
$begingroup$
@hjpotter92 Asking for a code review of code that solves the same problem as in some other question is fine here. Just having copy & pasted the other persons code would be a dupe (and also off-topic), or re-posting your own question (without having changed anything), instead of editing it.
$endgroup$
– Graipher
Jan 17 '18 at 16:54
1
$begingroup$
Is is acceptable to return the input string?
$endgroup$
– Eric Duminil
Jan 17 '18 at 20:37