Replacing All Non-Alphanumeric Characters with Empty Strings

Replacing all non-alphanumeric characters with empty strings

Use [^A-Za-z0-9].

Note: removed the space since that is not typically considered alphanumeric.

Replace all non alphanumeric characters, new lines, and multiple white space with one space

Be aware, that \W leaves the underscore. A short equivalent for [^a-zA-Z0-9] would be [\W_]

text.replace(/[\W_]+/g," ");

\W is the negation of shorthand \w for [A-Za-z0-9_] word characters (including the underscore)

Example at regex101.com

How do I remove all non alphanumeric characters from a string except dash?

Replace [^a-zA-Z0-9 -] with an empty string.

Regex rgx = new Regex("[^a-zA-Z0-9 -]");
str = rgx.Replace(str, "");

Replace all non-alphanumeric characters in a string

Regex to the rescue!

import re

s = re.sub('[^0-9a-zA-Z]+', '*', s)

Example:

>>> re.sub('[^0-9a-zA-Z]+', '*', 'h^&ell`.,|o w]{+orld')
'h*ell*o*w*orld'

Regular Expression to replace all non alphanumeric characters except / to empty() character

Use regex [^a-zA-Z0-9//] insted of [^a-zA-Z0-9] to escape /

 LibString.replaceAll(token, "[^a-zA-Z0-9//]", "");

Replacing all non-alphanumeric characters except some characters

Use explicit white list instead:

replaceAll("[^a-zA-Z0-9ÆØÅæøå]","_")

Look at the similar question

Replace all Non-Alphanumeric Characters except one particular pattern using RegEx in Python

You can use re.sub and a pattern with a capture group

(\d+(?:\.\d+))|\W+

The pattern matches:

• (\d+(?:\.\d+)) Capture digits with an optional decimal part with a dot in group 1
• | OR
• \W+ Match 1+ non word characters to replace with a single space (or use a negated character class [^a-zA-Z0-9]+ to keep matching an underscore)

In the replacement keep the capture group, and replace the match for 1+ non word characters with a space.

See a regex demo and a Python demo

import re

s = "xyz - aabc 123.56 cancer s15.2 date 12/03/2021 @ dd hospital www.someurl ocr.5rror 123.sometext"
pattern = r"(\d+(?:\.\d+))|\W+"
print(re.sub(pattern, lambda x: x.group(1) if x.group(1) else " ", s))

Output

xyz aabc 123.56 cancer s15.2 date 12 03 2021 dd hospital www someurl ocr 5rror 123 sometext

replace non-alphamumeric and multiple spaces

To achieve this in a single regular expression, you can alternate with capturing the space in a group, and then replace with that group (which will be the empty string if the other alternation was used):

const replace = str => str.replace(/( )+|[^\da-z ]+/gi, '$1');console.log(replace('foobar'));console.log(replace('foo       bar'));console.log(replace('foo###bar'));

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