Find all integers in list of strings and output it in form of a list

Question

How can I find and extract integers (without using the re module) from a list of strings and integers.

['My', '2', 'favorite', 'numbers', 'are', '42', 'and', '69']

How do I extract all the integers (2,42,69) in form of a list? Desired output:

[2, 42, 69]

`[int(s) for s in my_list if s.isdigit()]` – Mateen Ulhaq Jan 20 '18 at 06:34 — Mateen Ulhaq, Jan 20 '18 at 06:34

user1767754 · Accepted Answer · 2018-01-20T20:15:22.867

1

One quick solution would be to try to cast the items to an int and keep those that succeed.

List of any type:

result = []
for i in x:
    try:
        result.append(int(i))
    except Exception:
        pass

print(result)
>>[2, 42, 69]

List Of Strings

As Mateen Ulhaq mentioned, if your input is always string type, it would be more appropriate to use:

[int(s) for s in my_list if s.isdigit()]

Note: This solution won't work if your string has negative numbers such as '-12'

edited Jan 20 '18 at 20:15

answered Jan 20 '18 at 06:30

user1767754

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Of course, one should avoid this solution since exceptions are costly. – Mateen Ulhaq Jan 20 '18 at 06:32
Oh that's true! Just saw your comment, that's a good solution if we expect them all to be strings. – user1767754 Jan 20 '18 at 06:38

score 0 · Answer 2 · answered Jan 20 '18 at 06:33

You can do this by iterating over every element of list and try to convert it into an integer. If you get successful in the conversion then append it to the list of integers.

a = ['My', '2', 'favorite', 'numbers', 'are', '42', 'and', '69']
ints = []
for i in a:
    try:
        z = int(i)
        ints.append(z)
    except:
        pass

Or you could go with a oneliner from here:

a = ['My', '2', 'favorite', 'numbers', 'are', '42', 'and', '69']
[int(s) for s in a if s.isdigit()]
#[2, 42, 69]

Find all integers in list of strings and output it in form of a list

2 Answers2

List of any type:

List Of Strings