CODEWITHSUNDEEP
pythonstring
thithien
thithien

Reputation: 235

Replacing characters of a string by a given string

Given this string 'www__ww_www_'

I need to replace all the '_' characters with characters from the following string '1234'. The result should be 'www12ww3www4'.

TEXT = 'aio__oo_ecc_'
INSERT = '1234'

insert = list(INSERT)
ret = ''

for char in TEXT:
    if char == '_':
        ret += insert[0]
        insert.pop(0)
    else:
        ret += char

print (ret)
>> aio12oo3ecc4

What is the right way to do this? Because this seems like the most inefficient way.

Upvotes: 2

Views: 93

Answers (4)

ggorlen
ggorlen

Reputation: 56935

You can loop over the TEXT using a list comprehension that uses a ternary to select from an INSERT iterator or from the current element in TEXT:

>>> TEXT = 'aio__oo_ecc_'
>>> INSERT = '1234'
>>> it = iter(INSERT)
>>> "".join([next(it) if x == "_" else x for x in TEXT])
'aio12oo3ecc4'

The benefits include avoiding Shlemiel the Painter's Algorithm with ret += char. Also, pop(0) requires the whole list to be shifted forward, so it's linear (better would be reversing INSERT and using pop()).

In response to some of the comments here, list comprehensions tend to be faster than generators when the whole iterable will be consumed on the spot.

Upvotes: 4

goutnet
goutnet

Reputation: 442

As pointed in the comments, you can use the str.replace directly:

for c in INSERT:
    TEXT = TEXT.replace('_', c, 1)

You can use also the regex replace for that:

import re
for c in INSERT:
    TEXT = re.sub('_', c, TEXT, 1)

see here: https://docs.python.org/3/library/re.html

Upvotes: 2

DYZ
DYZ

Reputation: 57033

Consider splitting the pattern string by the underscore and zipping it with the string of inserts:

TEXT = 'aio__oo_ecc_a' # '_a' added to illustrate the need for zip_longest
from itertools import zip_longest, chain
''.join(chain.from_iterable(zip_longest(TEXT.split('_'), INSERT, fillvalue='')))
#'aio12oo3ecc4a'

zip_longest is used instead of the "normal" zip to make sure the last fragment of the pattern, if any, is not lost.

A step-by-step exploration:

pieces = TEXT.split('_')
# ['aio', '', 'oo', 'ecc', 'a']
mix = zip_longest(pieces, INSERT, fillvalue='')
# [('aio', '1'), ('', '2'), ('oo', '3'), ('ecc', '4'), ('a', '')]
flat_mix = chain.from_iterable(mix)
# ['aio', '1', '', '2', 'oo', '3', 'ecc', '4', 'a', '']
result = ''.join(flat_mix)

Speed comparison:

  1. This solution: 1.32 µs ± 9.08 ns per loop
  2. Iterator + ternary + list comprehension: 1.77 µs ± 20.8 ns per loop
  3. Original solution: 2 µs ± 13.2 ns per loop
  4. The loop + regex solution: 3.66 µs ± 103 ns per loop

Upvotes: 1

blhsing
blhsing

Reputation: 106553

You can use an iterator in a replacement function for re.sub:

import re
TEXT = 'aio__oo_ecc_'
INSERT = '1234'
i = iter(INSERT)
print(re.sub('_', lambda _: next(i), TEXT))

This outputs:

aio12oo3ecc4

Upvotes: 1

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