Python string.replace equivalent (from Javascript)

Question

I am trying to pick up python and as someone coming from Javascript I haven't really been able to understand python's regex package re

What I am trying to do is something I have done in javascript to build a very very simple templating "engine" (I understand AST is the way to go for anything more complex):

In javascript:

var rawString = 
  "{{prefix_HelloWorld}}   testing this. {{_thiswillNotMatch}} \ 
  {{prefix_Okay}}";

rawString.replace(
   /\{\{prefix_(.+?)\}\}/g,
   function(match, innerCapture){
     return "One To Rule All";
});

In Javascript that will result in:

"One To Rule All testing this. {{_thiswillNotMatch}} One To Rule All"

And the function will get called twice with:

   innerCapture === "HelloWorld"
   match ==== "{{prefix_HelloWorld}}"

and:

   innerCapture === "Okay"
   match ==== "{{prefix_Okay}}"

Now, in python I have tried looking up docs on the re package

import re

Have tried doing something along the lines of:

match = re.search(r'pattern', string)
if match:
  print match.group()
  print match.group(1)

But it really doesn't make sense to me and doesn't work. For one, I'm not clear on what this group() concept means? And how am I to know if there is match.group(n)... group(n+11000)?

Thanks!

Blender · Accepted Answer · 2013-07-31 08:03:04Z

6

Python's re.sub function is just like JavaScript's String.prototype.replace:

import re

def replacer(match):
    return match.group(1).upper()

rawString = "{{prefix_HelloWorld}}   testing this. {{_thiswillNotMatch}} {{prefix_Okay}}"
result = re.sub(r'\{\{prefix_(.+?)\}\}', replacer, rawString)

And the result:

'HELLOWORLD   testing this. {{_thiswillNotMatch}} OKAY'

As for the groups, notice how your replacement function accepts a match argument and an innerCapture argument. The first argument is match.group(0). The second one is match.group(1).

edited Jul 31, 2013 at 8:03

answered Jul 31, 2013 at 7:57

Blender

300k55 gold badges463 silver badges512 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

John Tomson Over a year ago

Thank you that makes a lot of sense. What would group 2, 3, etc be? Or are those not defined.

Blender Over a year ago

@JohnTomson: .group(n) is the nth capturing group.

Jblasco · Accepted Answer · 2013-07-31 08:33:57Z

0

I think you want to substitute all occurrences of {{prefix_*}} where * is basically anything. If so, this code works and is simple.

 pattern = "\{\{prefix_.*?\}\}"
 re.sub(pattern, "One To Rule All", rawString)

Cheers!

answered Jul 31, 2013 at 8:33

Jblasco

3,98724 silver badges26 bronze badges

Comments

samwyse · Accepted Answer · 2014-08-13 03:49:30Z

0

If you will be using the same pattern more than once (such as in a loop), then this is better:

pattern = re.compile("\{\{prefix_.*?\}\}")
# ... later ...
pattern.sub("One To Rule All", rawString)

answered Aug 13, 2014 at 3:49

samwyse

2,9961 gold badge32 silver badges40 bronze badges

Collectives™ on Stack Overflow

Python string.replace equivalent (from Javascript)

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related