How to replace multiple substrings at the same time

Question

I have a string like

a = "X1+X2*X3*X1"
b = {"X1":"XX0","X2":"XX1","X0":"XX2"}

I want to replace the substring 'X1,X2,X3' using dict b.

However, when I replace using the below code,

for x in b:
    a = a.replace(x,b[x])
print(a)

'XXX2+XX1*X3'

Expected result is XX0 + XX1*X3*XX0

I know it is because the substring is replaced in a loop, but I don't know how to solve it.

you are updating the string too often. the easiest way to do this is with the `re` (regex) module. — Rick, Oct 05 '21 at 13:22
See _Split by multiple different delimiters_ at https://note.nkmk.me/en/python-split-rsplit-splitlines-re/ — Sharuzzaman Ahmat Raslan, Oct 05 '21 at 13:23
keys and values are "incested"... thats why you got a "strange" result — cards, Oct 05 '21 at 13:27
you should use `print(a)` inside `for`-loop to see what you get after every `replace`. And you should see that after first `replace` you have `XX0 + ....` but later `"X0":"XX2"` will treats `XX0 + ... ` as normal `X0` and it will put `XX2`. — furas, Oct 05 '21 at 13:48
problem it that old variables `X0`, `X1`, `X2` are similar to new variables `XX0`, `XX1`, `XX2` but if you would use litte different variables ie. `XX!0`, `XX!1`, `XX!2` and use your loop then you could get `XX!0 + XX!1 * X3 * XX!0` and you would need only `replace('!', '')` to get expected `XX0 + XX1 * X3 * XX0` — furas, Oct 05 '21 at 13:53
There is no pattern in original string, I just want to replace only once in the original string. — Jilong Yin, Oct 05 '21 at 13:53

score 8 · Accepted Answer · answered Oct 05 '21 at 13:24

You can create a pattern with '|' then search in dictionary transform like below.

Try this:

import re
a = "X1+X2*X3*X1"
b = {"X1":"XX0","X2":"XX1","X0":"XX2"}

pattern = re.compile("|".join(b.keys()))
out = pattern.sub(lambda x: b[re.escape(x.group(0))], a)

Output:

>>> out
'XX0+XX1*X3*XX0'

score 5 · Answer 2 · answered Oct 05 '21 at 13:21

5

You can use the repl parameter of re.sub:

import re
re.sub('X\d', lambda x: b.get(x.group(), x.group()), a)

output:

'XX0+XX1*X3*XX0'

answered Oct 05 '21 at 13:21

mozway

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jsofri · Answer 3 · 2022-06-13T13:07:19.067

The reason for this is beacuse you are replacing the same string multiple times, so behind the scenes (or between the iterations) there are a few more switches in the middle that you probably don't see (unless debugging this code). Please note that dictionary keys are not ordered, so you cannot assume what's replaced when. I suggest you use template

Edit: newer versions of python do preserve insertion order - Are dictionaries ordered in Python 3.6+?

score 1 · Answer 4 · answered Oct 05 '21 at 13:54

With just built-in functions. The given b dictionary contains cycles between keys and values, so used the f-string notation to perform the substitutions. To escape the {} one should double them {{}}, used for repeated substitution. The enumerate is needed to get unique keys in the new dictionary, so no more cycles.

a = "X1+X2*X3*X1"
b = {"X1":"XX0","X2":"XX1","X0":"XX2"}

new_dict = {}
for i, k in enumerate(b):
    sub_format =  f'{k}' + f'{i}'
    new_dict[sub_format] = b[k]
    a = a.replace(k, f'{{{sub_format}}}')

print(a.format(**new_dict))

Output

XX0+XX1*X3*XX0

How to replace multiple substrings at the same time

4 Answers4