python: insert a character 2 positions after every subtring

Question

I have this string: '51.6251.8750.6951.4651.462626200', and want to get a list like this: ['51.62', '51.87', '50.69', '51.46', '51.46', '2626200']

So, basically after find a dot, the string should be split 2 positions later.

For now, I have the following, where I created a list of all dot positions, not sure if that's the way.

import re

string = '51.6251.8750.6951.4651.462626200'

k = 2

# I couldn't find the position of the dot actually, so I had to replace for something first, it may be unecessary

subString = '-'

replacedString = re.sub('[.]', subString, string)

# create a list with all positions of the substring

subStringPosition = [i.start() for i in re.finditer(subString, replacedString)]

# add k to each element of the subStringPosition

subStringAfterK = [x + k for x in subStringPosition]

I'd appreciate if anybody can help me.

Answer 1

You can use the regex methods to achieve this result split and findall with the expiration of \d{2}\.\d{2} that is 2 digits followed by . and another 2 digits.
In findall it will return all the 2 numbers and 2 numbers after the . and in the split[-1] it will return the rest of the string.

string = "51.6251.8750.6951.4651.462626200"
result = re.findall(r"\d{2}\.\d{2}", string)
result.append(re.split(r"\d{2}\.\d{2}", string)[-1])
print(result)

Output

['51.62', '51.87', '50.69', '51.46', '51.46', '2626200']

Answer 2

You can use:

# string = '51.6251.8750.6951.4651.462626200'
matches = re.split(r'(?<=\.\d{2})', string)

Explanation:

Positive Lookbehind (?<=\.\d{2})

• \. matches the character .
• \d{2} matches a digit exactly 2 times

Result:

#print(matches)
['51.62', '51.87', '50.69', '51.46', '51.46', '2626200']

You can test the regular expression here.