Replacing digits with str.replace()

问题

I want to make a new string by replacing digits with %d for example:

Name.replace( "_u1_v1" , "_u%d_v%d") 

...but the number 1 can be any digit for example "_u2_v2.tx"

Can I give replace() a wildcard to expect any digit? Like "_u"%d"_v"%d".tx"

Or do I have to make a regular expression?

回答1:

Using regular expressions:

>>> import re
>>> s = "_u1_v1"
>>> print re.sub('\d', '%d', s)
_u%d_v%d

\d matches any number 0-9. re.sub replaces the number(s) with %d

回答2:

You cannot; str.replace() works with literal text only.

To replace patterns, use regular expressions:

re.sub(r'_u\d_v\d', '_u%d_v%d', inputtext)

Demo:

>>> import re
>>> inputtext = '42_u2_v3.txt'
>>> re.sub(r'_u\d_v\d', '_u%d_v%d', inputtext)
'42_u%d_v%d.txt'

回答3:

Just for variety, some non-regex approaches:

>>> s = "_u1_v1"
>>> ''.join("%d" if c.isdigit() else c for c in s)
'_u%d_v%d'

Or if you need to group multiple digits:

>>> from itertools import groupby, chain
>>> s = "_u1_v13"
>>> grouped = groupby(s, str.isdigit)
>>> ''.join(chain.from_iterable("%d" if k else g for k,g in grouped))
'_u%d_v%d'

(To be honest, though, while I'm generally anti-regex, this case is simple enough I'd probably use them.)

回答4:

A solution using translate (source):

remove_digits = str.maketrans('0123456789', '%%%%%%%%%%')
'_u1_v1'.translate(remove_digits)  # '_u%_v%'

来源：https://stackoverflow.com/questions/19084443/replacing-digits-with-str-replace