CODEWITHSUNDEEP
pythonstring
Lucas
Lucas

Reputation: 14129

How to replace whitespaces with underscore?

I want to replace whitespace with underscore in a string to create nice URLs. So that for example:

"This should be connected"

Should become

"This_should_be_connected"

I am using Python with Django. Can this be solved using regular expressions?

Upvotes: 322

Views: 590276

Answers (13)

Meghaa Yadav
Meghaa Yadav

Reputation: 101

You can try this instead:

mystring.replace(r' ','_')

Upvotes: 10

Kenan Banks
Kenan Banks

Reputation: 211980

Replacing spaces is fine, but I might suggest going a little further to handle other URL-hostile characters like question marks, apostrophes, exclamation points, etc.

Also note that the general consensus among SEO experts is that dashes are preferred to underscores in URLs.

import re

def urlify(s):

    # Remove all non-word characters (everything except numbers and letters)
    s = re.sub(r"[^\w\s]", '', s)

    # Replace all runs of whitespace with a single dash
    s = re.sub(r"\s+", '-', s)

    return s

# Prints: I-cant-get-no-satisfaction"
print(urlify("I can't get no satisfaction!"))

Upvotes: 113

rogeriopvl
rogeriopvl

Reputation: 54056

You don't need regular expressions. Python has a built-in string method that does what you need:

mystring.replace(" ", "_")

Upvotes: 538

Rajesh
Rajesh

Reputation: 41

mystring.replace (" ", "_")

if you assign this value to any variable, it will work

s = mystring.replace (" ", "_")

by default mystring wont have this

Upvotes: 4

Yash
Yash

Reputation: 7064

Surprisingly this library not mentioned yet

python package named python-slugify, which does a pretty good job of slugifying:

pip install python-slugify

Works like this:

from slugify import slugify

txt = "This is a test ---"
r = slugify(txt)
self.assertEquals(r, "this-is-a-test")

txt = "This -- is a ## test ---"
r = slugify(txt)
self.assertEquals(r, "this-is-a-test")

txt = 'C\'est déjà l\'été.'
r = slugify(txt)
self.assertEquals(r, "cest-deja-lete")

txt = 'Nín hǎo. Wǒ shì zhōng guó rén'
r = slugify(txt)
self.assertEquals(r, "nin-hao-wo-shi-zhong-guo-ren")

txt = 'Компьютер'
r = slugify(txt)
self.assertEquals(r, "kompiuter")

txt = 'jaja---lol-méméméoo--a'
r = slugify(txt)
self.assertEquals(r, "jaja-lol-mememeoo-a")

Upvotes: 7

skilleo
skilleo

Reputation: 2481

OP is using python, but in javascript (something to be careful of since the syntaxes are similar.

// only replaces the first instance of ' ' with '_'
"one two three".replace(' ', '_'); 
=> "one_two three"

// replaces all instances of ' ' with '_'
"one two three".replace(/\s/g, '_');
=> "one_two_three"

Upvotes: 2

xOneca
xOneca

Reputation: 931

This takes into account blank characters other than space and I think it's faster than using re module:

url = "_".join( title.split() )

Upvotes: 60

fo0
fo0

Reputation: 

perl -e 'map { $on=$_; s/ /_/; rename($on, $_) or warn $!; } <*>;'

Match et replace space > underscore of all files in current directory

Upvotes: -4

Ionis
Ionis

Reputation:

Python has a built in method on strings called replace which is used as so:

string.replace(old, new)

So you would use:

string.replace(" ", "_")

I had this problem a while ago and I wrote code to replace characters in a string. I have to start remembering to check the python documentation because they've got built in functions for everything.

Upvotes: 7

Armandas
Armandas

Reputation: 2396

I'm using the following piece of code for my friendly urls:

from unicodedata import normalize
from re import sub

def slugify(title):
    name = normalize('NFKD', title).encode('ascii', 'ignore').replace(' ', '-').lower()
    #remove `other` characters
    name = sub('[^a-zA-Z0-9_-]', '', name)
    #nomalize dashes
    name = sub('-+', '-', name)

    return name

It works fine with unicode characters as well.

Upvotes: 5

Daniel Roseman
Daniel Roseman

Reputation: 599610

Django has a 'slugify' function which does this, as well as other URL-friendly optimisations. It's hidden away in the defaultfilters module.

>>> from django.template.defaultfilters import slugify
>>> slugify("This should be connected")

this-should-be-connected

This isn't exactly the output you asked for, but IMO it's better for use in URLs.

Upvotes: 45

mdirolf
mdirolf

Reputation: 7651

use string's replace method:

"this should be connected".replace(" ", "_")

"this_should_be_disconnected".replace("_", " ")

Upvotes: 11

Jarret Hardie
Jarret Hardie

Reputation: 97932

Using the re module:

import re
re.sub('\s+', '_', "This should be connected") # This_should_be_connected
re.sub('\s+', '_', 'And     so\tshould this')  # And_so_should_this

Unless you have multiple spaces or other whitespace possibilities as above, you may just wish to use string.replace as others have suggested.

Upvotes: 30

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