How can I do the multiple replace in python?

Question

As asked and answered in this post, I need to replace '[' with '[[]', and ']' with '[]]'.

I tried to use s.replace(), but as it's not in place change, I ran as follows to get a wrong anwser.

path1 = "/Users/smcho/Desktop/bracket/[10,20]"
path2 = path1.replace('[','[[]')
path3 = path2.replace(']','[]]')
pathName = os.path.join(path3, "*.txt")
print pathName
-->
/Users/smcho/Desktop/bracket/[[[]]10,20[]]/*.txt

How can I do the multiple replace in python?
Or how can I replace '[' and ']' at the same time?

Use regex to split on '[' or ']', then replace individual '[' and ']' with what you want, then join back. — Hamish Grubijan
– Hamish Grubijan, Commented Apr 12, 2010 at 16:16

interjay · Accepted Answer · 2010-04-12 16:15:04Z

12

import re
path2 = re.sub(r'(\[|])', r'[\1]', path)

Explanation:

\[|] will match a bracket (opening or closing). Placing it in the parentheses will make it capture into a group. Then in the replacement string, \1 will be substituted with the content of the group.

answered Apr 12, 2010 at 16:15

interjay

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1 Comment

SamB Over a year ago

I was hoping to see this solution here :-). It is wise to avoid pointlessly splitting and rejoining strings, due to the potentially large number of unnecessary allocations this would cause.

Mike Graham · Accepted Answer · 2010-04-12 16:15:14Z

3

I would use code like

path = "/Users/smcho/Desktop/bracket/[10,20]"
replacements = {"[": "[[]", "]": "[]]"}
new_path = "".join(replacements.get(c, c) for c in path)

answered Apr 12, 2010 at 16:15

Mike Graham

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Comments

vikramsjn · Accepted Answer · 2012-05-09 17:30:58Z

1

There is also this generic python multiple replace recipe: Single pass multiple replace

answered May 9, 2012 at 17:30

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vikramsjn

Comments

Will McCutchen · Accepted Answer · 2010-04-12 16:18:06Z

0

import re
path2 = re.sub(r'(\[|\])', r'[\1]', path1)

answered Apr 12, 2010 at 16:18

Will McCutchen

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neil · Accepted Answer · 2010-04-12 16:32:54Z

0

Or, to avoid regex, I would replace the opening bracket with a unique string, then replace the closing bracket and then replace the unique string - maybe a round about way, but to my mind it looks simpler - only a test would say if it is faster. Also, I'd tend to reuse the same name.

i.e.

path1 = "/Users/smcho/Desktop/bracket/[10,20]"
path1 = path1.replace('[','*UNIQUE*')
path1 = path1.replace(']','[]]')
path1 = path1.replace('*UNIQUE*','[[]')

pathName = os.path.join(path1, "*.txt")

answered Apr 12, 2010 at 16:32

neil

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Comments

Sandeep Reddy Kesava Reddy · Accepted Answer · 2022-09-23 07:27:28Z

0

X = TE$%ST C@"DE

specialChars = "@#$%&"

for specialChar in specialChars:

X = X.replace(specialChar, '')

Y = appname1.replace(" ", "")

print(Y)

TESTCODE

edited Sep 23, 2022 at 7:27

answered Sep 23, 2022 at 7:25

Sandeep Reddy Kesava Reddy

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