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?
Answers:
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.
I would use code like
path = "/Users/smcho/Desktop/bracket/[10,20]"
replacements = {"[": "[[]", "]": "[]]"}
new_path = "".join(replacements.get(c, c) for c in path)
import re
path2 = re.sub(r'([|])', r'[1]', path1)
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")
There is also this generic python multiple replace recipe: Single pass multiple replace
X = TE$%ST C@"DE
specialChars = "@#$%&"
for specialChar in specialChars:
X = X.replace(specialChar, '')
Y = appname1.replace(" ", "")
print(Y)
TESTCODE
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?
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.
I would use code like
path = "/Users/smcho/Desktop/bracket/[10,20]"
replacements = {"[": "[[]", "]": "[]]"}
new_path = "".join(replacements.get(c, c) for c in path)
import re
path2 = re.sub(r'([|])', r'[1]', path1)
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")
There is also this generic python multiple replace recipe: Single pass multiple replace
X = TE$%ST C@"DE
specialChars = "@#$%&"
for specialChar in specialChars:
X = X.replace(specialChar, '')
Y = appname1.replace(" ", "")
print(Y)
TESTCODE