Replacing words in text file using a dictionary

Question

I\'m trying to open a text file and then read through it replacing certain strings with strings stored in a dictionary.

Based on answers to How do I edit a text file

Answer 1

import fileinput

text = "sample file.txt"
fields = {"pattern 1": "replacement text 1", "pattern 2": "replacement text 2"}

for line in fileinput.input(text, inplace=True):
    line = line.rstrip()
    for field in fields:
        if field in line:
            line = line.replace(field, fields[field])

    print line

Answer 2

If You are more familiar with Python, You can use tips from Official documentation:

7.1. string — Common string operations

And subclass, the Template class, in which you define somehow that every single world will be a new placeholder, and then with safe_substitute() You could get a nice and reliable solution.

Answer 3

This is how I would do it:

fields = {"pattern 1": "replacement text 1", "pattern 2": "replacement text 2"}

with open('yourfile.txt', 'w+') as f:
    s = f.read()
    for key in fields:
        s = s.replace(key, fields[key])
    f.write(s)

Answer 4

I used items() to iterate over key and values of your fields dict.

I skip the blank lines with continue and clean the others with rstrip()

I replace every keys found in the line by the values in your fields dict, and I write every lines with print.

import fileinput

text = "sample file.txt"
fields = {"pattern 1": "replacement text 1", "pattern 2": "replacement text 2"}


for line in fileinput.input(text, inplace=True):
    line = line.rstrip()
    if not line:
        continue
    for f_key, f_value in fields.items():
        if f_key in line:
            line = line.replace(f_key, f_value)
    print line

Answer 5

If you can find a regex pattern covering all your keys, you could use re.sub for a very efficient solution : you only need one pass instead of parsing your whole text for each search term.

In your title, you mention "replacing words". In that case, '\w+' would work just fine.

import re

fields = {"pattern 1": "replacement text 1", "pattern 2": "replacement text 2"}

words_to_replace = r'\bpattern \d+\b'

text = """Based on answers to How do I edit a text file in Python? pattern 1 I could pull out
the dictionary values before doing the replacing, but looping through the dictionary seems more efficient.
Test pattern 2
The code doesn't produce any errors, but also doesn't do any replacing. pattern 3"""

def replace_words_using_dict(matchobj):
    key = matchobj.group(0)
    return fields.get(key, key)

print(re.sub(words_to_replace, replace_words_using_dict, text))

It outputs :

Based on answers to How do I edit a text file in Python? replacement text 1 I could pull out
the dictionary values before doing the replacing, but looping through the dictionary seems more efficient.
Test replacement text 2
The code doesn't produce any errors, but also doesn't do any replacing. pattern 3

Also, be very careful when modifying a file in place. I'd advice you to write a second file with the replacements. Once you are 100% sure that it works perfectly, you could switch to inplace=True.

Answer 6

Just figured out how to replace lots of different words in a txt file at one go, by iterating through a dictionary (whole word matches only). It would be really annoying if I want to replace "1" with "John", but ends up turning "12" into "John2." The following code is what works for me.

import re

match = {}  # create a dictionary of words-to-replace and words-to-replace-with

f = open("filename","r")
data = f.read() # string of all file content

def replace_all(text, dic):
    for i, j in dic.items():
        text = re.sub(r"\b%s\b"%i, j, text) 
        # r"\b%s\b"% enables replacing by whole word matches only
    return text

data = replace_all(data,match)
print(data) # you can copy and paste the result to whatever file you like