Python: Extract integers from string based on position of specific letters

Question

New to Python

Scenario

A string is a combination of letters and digits.
There is always an integer following a letter. But the number of digits of an integer varies.
For letter "A", there is a pattern to locate its position. For example, if I replace all letters in the string by a delimiter, such as commas, "A" is always on the position of multiple of 3. I have figured a way to extract the integers that following "A".
However, for letter "B", there is no pattern where in the string it might appear.

My Goal

Extract all integers following letter "A" and "B", respectively.
Sum them after extraction, respectively.

Trial and Error

The way I handled "A" pattern is to use modulus operation (%) to achieve my goal

s = 'A4B2d2A14d3B11A20B10d10'
s_replace = s.replace('A', ',').replace('B', ',').replace('d', ',')[1:]
test_n = [int(n) for n in s_replace.split(',') if n.isdigit()]
test_Asum = sum(test_n[i] for i in range(len(test_n)) if i % 3 == 0)

print(test_Asum)
# 38

Please help with "B" issue. I can find the index of "B" in the string. However, how could I extract how ever many digits starting from the index of "B"?

test_B_index = [Bindx for Bindx, char in enumerate(s) if char == "B"]
print(test_B_index)
# [2, 11, 17]

Asked By: Jane Lu

||

Source

Answer 1

My favorite way to extract things is to use regex.
Here an example in your case.
I hope it helps.

import re
s = 'A4B2d2A14d3B11A20B10d10'
m = re.findall(r'B([0-9]+)', s)
print(m)
#['2', '11', '10']
print(sum([int(n) for n in m]))
#23

Answered By: Pauly

Python: Extract integers from string based on position of specific letters

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