string operation/regex - find and replace

user_12513693 0
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I am trying to replace each word after . in the txt file below:

line1
line2
field: [orders.cancelled,orders.delivered,orders.reached
orders.pickup,orders.time]
some line
some line

I have a dictionary:

   d = {'cancelled':'cancelled_at', 'deliver':'xxx'}

I am running the following code. However, I am getting the results for partial match i.e

I see the new file has the following words

field: [orders.cancelled_at, orders.xxxed ..........

here from the word delivered the program is still replacing the first 7 words(deliver) and adding 'ed' in the end. I am not sure why

with open('list.txt', 'r') as g:
    text = g.read()
    for k in d:
        before = f'.{k}'
        after = f'.{d[k]}
        #print(before)
        #print(after)
        text = text.replace(before, after)
        #print(text)

with open('new_list.txt', 'w') as w:
    w.write(text)

I also tried this one and I get the same results

import re

with open('list.txt', 'r') as f:
    text = f.read()
    for k in d:
        before = f'.{k}(?!=\w)'
        print(before)
        after = f'.{d[k]}'
        print(after)
        text = re.sub(before, after, text)

with open('new_list.txt', 'w') as w:
    w.write(text)
dictionary python replace regex string
1 Answers

You can use

import re

d = {'cancelled':'cancelled_at', 'deliver':'xxx'}
rx = re.compile(fr"(?<=\.)(?:{'|'.join(d)})\b")

with open('list.txt', 'r') as f:
    print( re.sub(rx, lambda x: d[x.group()], f.read()) )

See the Python demo

The regex generated by the code looks like

(?<=\.)(?:cancelled|deliver)\b

See the regex demo. Details:

  • (?<=\.) - a positive lookbehind that matches a location immediately preceded with a literal .
  • (?:cancelled|deliver) - two alternatives: cancelled or deliver
  • \b - as whole words, \b is a word boundary.

The lambda x: d[x.group()] replacement replaces the matched word with the corresponding dictionary key value.

user_38329700
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