将元组列表转换为该列表中使用的所有整数的列表

Question

Python beginner here. I have a list of tuples like so:

[(100, 1), (50, 2), (25, 4), (20, 5), (10, 10)]

and I want to convert it into

[0, 2, 4, 5, 1]

There has to be a faster way to do this than constantly replacing each character:

strfp1 = re.sub('\(','',str(factor_pairs)); strfp2 = re.sub('\)','',strfp1); strfp3 = re.sub('\[','',strfp2); strfp4 = re.sub(']','',strfp3); strfp5 = re.sub(',','',strfp4); strfp6 = re.sub(' ','',strfp5)
factor_numbers = [int(i) for i in set(strfp6)]

Yet, I couldn't even find a way to replace multiple non-adjacent characters at once. Am I missing something obvious?

Answer 1

If you wanted to use your method of replacement, there is indeed an easy way to do this:

import re
factor_pairs = [(100, 1), (50, 2), (25, 4), (20, 5), (10, 10)]
s = re.sub(r'[\[\]\(\), ]', '', str(factor_pairs))
factor_numbers = [int(i) for i in set(s)]

Any character specified in the outer [] 's will be replaced.

Answer 2

You can use a set if order necessarily not matters:

from itertools import chain

lst = [(100, 1), (50, 2), (25, 4), (20, 5), (10, 10)]

flst = map(lambda x: str(x), chain.from_iterable(lst))

s = set()
for x in flst:
    for i in x:
        s.add(i)

print(s)
# {'2', '1', '0', '5', '4'}

If order matters, use a list:

from itertools import chain

lst = [(100, 1), (50, 2), (25, 4), (20, 5), (10, 10)]

flst = map(lambda x: str(x), chain.from_iterable(lst))

s = []
for x in flst:
    for i in x:
        if i not in s:
            s.append(i)

print(s)
# ['1', '0', '5', '2', '4']