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

Answer 1

Using hamishmcn's answer as a template I was able to search for a line in a file that match my regex and replacing it with empty string.

import re 

fin = open("in.txt", 'r') # in file
fout = open("out.txt", 'w') # out file
for line in fin:
    p = re.compile('[-][0-9]*[.][0-9]*[,]|[-][0-9]*[,]') # pattern
    newline = p.sub('',line) # replace matching strings with empty string
    print newline
    fout.write(newline)
fin.close()
fout.close()

Answer 2

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, copymode
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))
    #Copy the file permissions from the old file to the new file
    copymode(file_path, abs_path)
    #Remove original file
    remove(file_path)
    #Move new file
    move(abs_path, file_path)

Answer 3

Expanding on @Kiran's answer, which I agree is more succinct and Pythonic, this adds codecs to support the reading and writing of UTF-8:

import codecs 

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 codecs.open(target_file_path, 'w', 'utf-8') as target_file:
        with codecs.open(source_file_path, 'r', 'utf-8') 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)

Answer 4

fileinput is quite straightforward as mentioned on previous answers:

import fileinput

def replace_in_file(file_path, search_text, new_text):
    with fileinput.input(file_path, inplace=True) as f:
        for line in f:
            new_line = line.replace(search_text, new_text)
            print(new_line, end='')

Explanation:

• fileinput can accept multiple files, but I prefer to close each single file as soon as it is being processed. So placed single file_path in with statement.
• print statement does not print anything when inplace=True, because STDOUT is being forwarded to the original file.
• end='' in print statement is to eliminate intermediate blank new lines.

Can be used as follows:

file_path = '/path/to/my/file'
replace_in_file(file_path, 'old-text', 'new-text')

Answer 5

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 6

This should work: (inplace editing)

import fileinput

# Does a list of files, and
# redirects STDOUT to the file in question
for line in fileinput.input(files, inplace = 1): 
      print line.replace("foo", "bar"),