CODEWITHSUNDEEP
pythonpython-3.xregex-group
prasanna kumar
prasanna kumar

Reputation: 287

How to replace a string with a pattern using reg.expression

I have a address list as :

addr = ['100 NORTH MAIN ROAD',
            '100 BROAD ROAD APT.',
            'SAROJINI DEVI ROAD',
            'BROAD AVENUE ROAD']

I need to do my replacement work in a following function:

def subst(pattern, replace_str, string):

by defining a pattern outside of this function and passing it as an argument to subst.

I need an output like:

addr = ['100 NORTH MAIN RD',
            '100 BROAD RD APT.',
            'SAROJINI DEVI RD ',
            'BROAD AVENUE RD']

where all 'ROAD' strings are replaced with 'RD'

def subst(pattern, replace_str, string):
  #susbstitute pattern and return it
  new=[]
  for x in string:
    new.insert((re.sub(r'^(ROAD)','RD',x)),x)
  return new

def main():
  addr = ['100 NORTH MAIN ROAD',
        '100 BROAD ROAD APT.',
        'SAROJINI DEVI ROAD',
        'BROAD AVENUE ROAD']

  #Create pattern Implementation here 
  pattern=r'^(ROAD)'
  print (pattern)
  #Use subst function to replace 'ROAD' to 'RD.',Store as new_address
  new_address=subst(pattern,'RD',addr)
  return new_address

I have done this and getting below error

Traceback (most recent call last):

File "python", line 23, in File "python", line 20, in main File "python", line 7, in subst

TypeError: 'str' object cannot be interpreted as an integer

Upvotes: 0

Views: 12301

Answers (5)

Rajneesh Bharti
Rajneesh Bharti

Reputation: 1

Try this:

#!/bin/python3

import sys
import os
import io
import re


# Complete the function below.
def subst(pattern, replace_str, string):
    #susbstitute pattern and return it
    new_address=[pattern.sub(replace_str,st) for st in string]
    return new_address


def main():
    addr = ['100 NORTH MAIN ROAD',
            '100 BROAD ROAD APT.',
            'SAROJINI DEVI ROAD',
            'BROAD AVENUE ROAD']
            
    #Create pattern Implementation here 
    pattern=re.compile(r'\bROAD')
    #Use subst function to replace 'ROAD' to 'RD.',Store as new_address
    new_address=subst(pattern,'RD.',addr)
    print(new_address)
    return new_address

'''For testing the code, no input is required'''

if __name__ == "__main__":
  main()
  
  
def bind(func):
    func.data = 9
    return func

@bind
def add(x, y):
    return x + y

print(add(3, 10))
print(add.data)

Upvotes: 0

prasanna kumar
prasanna kumar

Reputation: 287

I got the answer,

pattern=r'\bROAD\b'

passing this a parameter to 

def subst(pattern, replace_str, string):

    #susbstitute pattern and return it
    new=[re.sub(pattern,'RD',x) for x in string]
    return new

we can get the op :)

Upvotes: 0

llllllllll
llllllllll

Reputation: 16404

No need for regex, just use replace:

[x.replace('ROAD', 'RD') for x in addr]

If you only want to replace the ROAD as a word, no in the middle, use:

[re.sub(r'\bROAD\b', 'RD', x) for x in addr]

Upvotes: 2

Leonardo Maffei
Leonardo Maffei

Reputation: 372

Using RegEx.. 

import re
count = 0
for x in addr:
    addr[count] = re.sub('ROAD', 'RD', x)
    count = count + 1

addr    # just to print the result.

Upvotes: 0

Rakesh
Rakesh

Reputation: 82765

Using re 

import re
addr = ['100 NORTH MAIN ROAD',
            '100 BROAD ROAD APT.',
            'SAROJINI DEVI ROAD',
            'BROAD AVENUE ROAD']

for i, v in enumerate(addr):
    addr[i] = re.sub('ROAD', 'RD', v) #v.replace("ROAD", "RD")

print addr

Output:

['100 NORTH MAIN RD', '100 BRD RD APT.', 'SAROJINI DEVI RD', 'BRD AVENUE RD']

Upvotes: 0

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