Replace number into character from string using python

Question

I have a string like this

        convert_text = "tet1+tet2+tet34+tet12+tet3"

I want to replace digits into character from above string.That mapping list available separately.so,When am trying to replace digit 1 with character 'g' using replace like below

       import re
       convert_text = convert_text.replace('1','g')
       print(convert_text)

output is

      "tetg+tet2+tet34+tetg2+tet3"

How to differentiate single digit and two digit values.Is there is any way to do with Regexp or something else?

Is "34" a single replacement or two replacements of "3" and "4"? — Jon Clements
– Jon Clements, Commented Aug 9, 2017 at 10:13

Jon Clements · Accepted Answer · 2017-08-09 10:32:54Z

3

You can use a regular expression with a callable replacement argument to substitute consecutive runs of digits with a value in a lookup table, eg:

import re

# Input text
convert_text = "tet1+tet2+tet34+tet12+tet3"

# to->from of digits to string
replacements = {'1': 'A', '2': 'B', '3': 'C', '12': 'T', '34': 'X'}

# Do actual replacement of digits to string
converted_text = re.sub('(\d+)', lambda m: replacements[m.group()], convert_text)

Which gives you:

'tetA+tetB+tetX+tetT+tetC'

answered Aug 9, 2017 at 10:32

Jon Clements

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Comments

Ando Jurai · Accepted Answer · 2017-08-09 10:22:39Z

0

import re
convert_text = "tet1+tet2+tet34+tet12+tet3"
pattern = re.compile(r'((?<!\d)\d(?!\d))')
convert_text2=pattern.sub('g',convert_text)

convert_text2 Out[2]: 'tetg+tetg+tet34+tet12+tetg'

You have to use negative lookahead and negative lookbehind patterns which are in between parenthesis

(?!pat) and 
(?<!pat),

you have the same with = instead of ! for positive lookahead/lookbehind.

EDIT: if you need replacement of strings of digits, regex is

pattern2 = re.compile(r'\d+')

In any pattern you can replace \d by a specific digit you need.

answered Aug 9, 2017 at 10:22

Ando Jurai

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