How to replace numbers in a string with numbers from another list?

Question

I have a text string that contains numbers. Plus, I have a number list. I want to replace the numbers inside the string with numbers from the list in the order of the string and list.

By using a Regex I extracted from the string the existing numbers and passed them to the list as well and now I have a match between the original number and the alternate number. However, it is still unclear how I can find adjustments and make replacements in order.

with this line, I extract the numbers from the given string:

    list_of_numbers_in_string = [int(x) for x in re.findall('\d+', str)]

And now I wonder how it can be used, or another method to get the desired result and from this input:

    data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
    new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]

get this output:

    corrected_data = 'readingOrder {index:24;} person {offset:0; length:12;} textStyle {offset:0; length:12; underlined:true;} place {offset:58; length:14;} textStyle {offset:58; length:14; underlined:true;}'

Moshe perez · Accepted Answer · 2019-10-20 08:00:19Z

1

The accepted answer is actually incorrect (it is now deleted). data.replace() will replace the first occurrence of the number, which is not always the correct one. For example, when you try to replace 8 with 14, it actually replaces 58 with 514.

Here is my solution:

import re

data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]

offset = 0
for index, match in enumerate(re.finditer('\d+', data)):
    data = data[:match.start() + offset] + str(new_numbers[index]) + data[match.end() + offset:]
    offset += len(str(new_numbers[index])) - match.end() + match.start()

edited Oct 20, 2019 at 8:00

answered Oct 20, 2019 at 7:45

Moshe perez

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1 Comment

Dipen Dadhaniya Over a year ago

This is perfect! Except that the time complexity is O(len(new_numbers) * len(data)). If the time complexity is not an issue, then this is the solution to go with.

Dipen Dadhaniya · Accepted Answer · 2019-10-20 07:54:16Z

If you directly operate on data (as a string) with one new number at a time (i.e. in a for loop you do data = operate on data), then it is likely that the time complexity will be O(len(new_numbers) * len(data)).

One efficient way of doing it in O(len(data)) time is to operate on a list of characters:

def replace_numbers(data, new_numbers):
    new_numbers_idx = 0
    data_as_char_list = []

    skip = False

    for data_ch in data:
        if data_ch.isdigit():
            if not skip:
                # When we encounter any number's first digit in data, we will add the new number, and in the next iterations we will skip rest of the digits in data.
                # e.g. data = 'hi123hi', new_numbers = [444], then when we encounter `1` we will add ['4', '4', '4'] and skip the rest of the digits '2' and '3' from data by setting skip = True.
                new_number_as_char_list = list(str(new_numbers[new_numbers_idx]))
                data_as_char_list.extend(new_number_as_char_list)
                new_numbers_idx += 1
                skip = True
        else:
            data_as_char_list.append(data_ch)
            skip = False

    return ''.join(data_as_char_list)


data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]
data = replace_numbers(data, new_numbers)

corrected_data = 'readingOrder {index:24;} person {offset:0; length:12;} textStyle {offset:0; length:12; underlined:true;} place {offset:58; length:14;} textStyle {offset:58; length:14; underlined:true;}'
assert data == corrected_data

PythonProgrammi · Accepted Answer · 2019-10-21 04:25:06Z

0

alternative

import re

data = 'readingOrder {index:24;} person {offset:0; length:7;} textStyle {offset:0; length:7; underlined:true;} place {offset:52; length:8;} textStyle {offset:52; length:8; underlined:true;}'
new_numbers = [24, 0, 12, 0, 12, 58, 14, 58, 14]
x = re.findall("\d+", data)
data = data.replace("{","{{").replace("}","}}")
for n in x:
    data = data.replace(n,"{}",1)
data = data.format(*new_numbers)
print(data)

[Out]:

readingOrder {index:24;} person {offset:0; length:12;} textStyle {offset:0; length:12; underlined:true;} place {offset:58; length:14;} textStyle {offset:58; length:14; underlined:true;}

edited Oct 21, 2019 at 4:25

answered Oct 21, 2019 at 4:19

PythonProgrammi

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