Search and replace a line in a file in Python

Question

I want to loop over the contents of a text file and do a search and replace on some lines and write the result back to the file. I could first load the whole file in memory and then write it back, but that probably is not the best way to do it.

What is the best way to do this, within the following code?

f = open(file)
for line in f:
    if line.contains('foo'):
        newline = line.replace('foo', 'bar')
        # how to write this newline back to the file

Answer 1

I guess something like this should do it. It basically writes the content to a new file and replaces the old file with the new file:

from tempfile import mkstemp
from shutil import move
from os import fdopen, remove

def replace(file_path, pattern, subst):
    #Create temp file
    fh, abs_path = mkstemp()
    with fdopen(fh,'w') as new_file:
        with open(file_path) as old_file:
            for line in old_file:
                new_file.write(line.replace(pattern, subst))
    #Remove original file
    remove(file_path)
    #Move new file
    move(abs_path, file_path)

Answer 2

If you're wanting a generic function that replaces any text with some other text, this is likely the best way to go, particularly if you're a fan of regex's:

import re
def replace( filePath, text, subs, flags=0 ):
    with open( filePath, "r+" ) as file:
        fileContents = file.read()
        textPattern = re.compile( re.escape( text ), flags )
        fileContents = textPattern.sub( subs, fileContents )
        file.seek( 0 )
        file.truncate()
        file.write( fileContents )

Answer 3

For Linux users:

import os
os.system('sed -i \'s/foo/bar/\' '+file_path)

Answer 4

A more pythonic way would be to use context managers like the code below:

from tempfile import mkstemp
from shutil import move
from os import remove

def replace(source_file_path, pattern, substring):
    fh, target_file_path = mkstemp()
    with open(target_file_path, 'w') as target_file:
        with open(source_file_path, 'r') as source_file:
            for line in source_file:
                target_file.write(line.replace(pattern, substring))
    remove(source_file_path)
    move(target_file_path, source_file_path)

You can find the full snippet here.

Answer 5

Here's another example that was tested, and will match search & replace patterns:

import fileinput
import sys

def replaceAll(file,searchExp,replaceExp):
    for line in fileinput.input(file, inplace=1):
        if searchExp in line:
            line = line.replace(searchExp,replaceExp)
        sys.stdout.write(line)

Example use:

replaceAll("/fooBar.txt","Hello\sWorld!$","Goodbye\sWorld.")

Answer 6

Based on the answer by Thomas Watnedal. However, this does not answer the line-to-line part of the original question exactly. The function can still replace on a line-to-line basis

This implementation replaces the file contents without using temporary files, as a consequence file permissions remain unchanged.

Also re.sub instead of replace, allows regex replacement instead of plain text replacement only.

Reading the file as a single string instead of line by line allows for multiline match and replacement.

import re

def replace(file, pattern, subst):
    # Read contents from file as a single string
    file_handle = open(file, 'r')
    file_string = file_handle.read()
    file_handle.close()

    # Use RE package to allow for replacement (also allowing for (multiline) REGEX)
    file_string = (re.sub(pattern, subst, file_string))

    # Write contents to file.
    # Using mode 'w' truncates the file.
    file_handle = open(file, 'w')
    file_handle.write(file_string)
    file_handle.close()