How do I convert a string to a valid variable name in Python?

Question

I need to convert an arbitrary string to a string that is a valid variable name in Python.

Here's a very basic example:

s1 = 'name/with/slashes'
s2 = 'name '

def clean(s):
    s = s.replace('/', '')
    s = s.strip()

    return s

# the _ is there so I can see the end of the string
print clean(s1) + '_'

That is a very naive approach. I need to check if the string contains invalid variable name characters and replace them with ''

What would be a pythonic way to do this?

What problem are you trying to solve using this method? There may be a better way. — Daenyth
– Daenyth, Commented Jul 21, 2010 at 20:06
print repr("a string ") shows the string in quotes - neater than appending _ to it. — Jochen Ritzel
– Jochen Ritzel, Commented Jul 21, 2010 at 20:38
I'm traversing a scene graph from cinema 4d and need to re-create it in blender. To keep things easy for me to understand I want to use the actual names of the cinema4d objects as variable names for Blender Python, so I need to adjust those first — George Profenza
– George Profenza, Commented Jul 21, 2010 at 20:40
BTW, the general conversion is called "slugifying" and searching stacoverflow or the internet for "python slugify" will find many solutions. — samwyse
– samwyse, Commented Jul 25, 2023 at 15:30
Thanks @samwyse ! if I only new this 13 years ago :))) Never too late to learn something new though. — George Profenza
– George Profenza, Commented Jul 25, 2023 at 21:03

Nas Banov · Accepted Answer · 2021-05-01 01:25:45Z

79

Well, I'd like to best Triptych's solution with ... a one-liner!

>>> def clean(varStr): return re.sub('\W|^(?=\d)','_', varStr)
...

>>> clean('32v2 g #Gmw845h$W b53wi ')
'_32v2_g__Gmw845h_W_b53wi_'

This substitution replaces any non-variable appropriate character with underscore and inserts underscore in front if the string starts with a digit. IMO, 'name/with/slashes' looks better as variable name name_with_slashes than as namewithslashes.

edited May 1, 2021 at 1:25

answered Jul 22, 2010 at 4:12

Nas Banov

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Kenan Banks · Accepted Answer · 2010-07-21 20:07:01Z

40

According to Python, an identifier is a letter or underscore, followed by an unlimited string of letters, numbers, and underscores:

import re

def clean(s):

   # Remove invalid characters
   s = re.sub('[^0-9a-zA-Z_]', '', s)

   # Remove leading characters until we find a letter or underscore
   s = re.sub('^[^a-zA-Z_]+', '', s)

   return s

Use like this:

>>> clean(' 32v2 g #Gmw845h$W b53wi ')
'v2gGmw845hWb53wi'

answered Jul 21, 2010 at 20:07

Kenan Banks

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ZXYNINE · Accepted Answer · 2022-04-14 01:21:16Z

5

You can use the built in func:str.isidentifier() in combination with filter(). This requires no imports such as re and works by iterating over each character and returning it if its an identifier. Then you just do a ''.join to convert the array to a string again.

s1 = 'name/with/slashes'
s2 = 'name '

def clean(s):
    s = ''.join(filter(str.isidentifier, s))
    return s

print f'{clean(s1)}_' #the _ is there so I can see the end of the string

EDIT:

If, like Hans Bouwmeester in the replies, want numeric values to be included as well, you can create a lambda which uses both the isIdentifier and the isdecimal functions to check the characters. Obviously this can be expanded as far as you want to take it. Code:

s1 = 'name/with/slashes'
s2 = 'name i2, i3    '
s3 = 'epng2 0-2g [ q4o 2-=2 t1  l32!@#$%*(vqv[r 0-34 2]] '

def clean(s):
    s = ''.join(filter( 
        lambda c: str.isidentifier(c) or str.isdecimal(c), s))
    return s
#the _ is there so I can see the end of the string
print(f'{ clean(s1) }_')
print(f'{ clean(s2) }_')
print(f'{ clean(s3) }_')

Gives :

namewithslashes_
namei2i3_
epng202gq4o22t1l32vqvr0342_

edited Apr 14, 2022 at 1:21

answered Dec 23, 2021 at 23:00

ZXYNINE

7824 silver badges6 bronze badges

Daenyth · Accepted Answer · 2010-07-21 20:01:15Z

3

You should build a regex that's a whitelist of permissible characters and replace everything that is not in that character class.

answered Jul 21, 2010 at 20:01

Daenyth

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S'pht'Kr · Accepted Answer · 2025-01-26 11:52:23Z

I wanted a solution that didn't import re and yet collapsed multiple underscores and any leading/trailing underscores. Combining most of the non-regex answers here with a quirk of the str.split() function when given no arguments (from this answer to another question), I came up with this "one-liner" which I think works well (and fast, but not really tested):

def clean(s):
   return '_'.join(''.join(map(
       lambda c: c if str.isidentifier(c) or str.isdecimal(c) else ' ',
       s
   )).split()).lstrip('0123456789')

This uses the lambda approach with map() instead of filter(), and replaces non-identifier characters with spaces... which is odd until the split()+'_'.join() trick comes in, because split() without arguments collapses multiple sequential whitespaces--which then get "replaced" with single underscores. The only thing novel here is the slightly kludgy .lstrip('01234567890') call which strips any leading numbers.

So, for...

>>> s3 = ' 42epng2 0-2g [ q4o 2-=2 t1  l32!@#$%*(vqv[r 0-34 2]] '
>>> clean(s3)
'epng2_0_2g_q4o_2_2_t1_l32_vqv_r_0_34_2'

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