Terry J. Reedy <[email protected]> added the comment:
Your proposal is underspecified. (And 'yours' is inconsistent with 'your';-).
If you want sequential replacememt in multiple scans, make sequential replace
calls. Use a loop if the number of replacement scans is variable or large. To
be sure of the order of replacements, a sequence is better than a dict, but
dict.values() instead would work in the following code.
s = 'this is my string'
for old, new in (('my', 'your'), ('string', 'car')):
s = s.replace(old, new)
print(s)
If you want parallel replacement in a single scan, a different scanner is
required. If the keys (search strings) are all single letters, one should use
str.translate. For a general string to string mapping, use re.sub with a
replacement function that does a lookup.
import re
s = 'this is my string'
subs = {'my': 'your', 'string': 'car'}
print(re.sub('|'.join(subs), lambda m: subs[m.group(0)], s))
In any case, the proposal to modify str.replace should be rejected.
However, the re code is not completely trivial (I did not work it out until
now), so I think it plausible to add the following to the re module.
def replace(string, map):
"""Return a string with map keys replaced by their values.
*string* is scanned once for non-overlapping occurrences of keys.
"""
return sub('|'.join(map), lambda m: map[m.group(0)], string)
I would reference this in the str.translate entry for when keys are not
restricted to letters.
If adding replace() is rejected, I would like an example added to the
sub(pattern, function, string) examples.
----------
nosy: +serhiy.storchaka, terry.reedy
stage: -> test needed
title: Make string.replace accept a dict instead of two arguments -> Add
re.replace(string, replacement_map)
_______________________________________
Python tracker <[email protected]>
<https://bugs.python.org/issue33647>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
