Python: "Try" inside shortened for...loop

Question

Assume I'd like to convert a list of strings to integer, but it cannot be done for all elements.

I know this works:

a = ['2.0','3.0','4.0','5.0','Cherry']
b = []

for k in a:
    try:
        int(k)
        b.append(int(k))
    except:
        pass

print b
> [2, 3, 4, 5]

But is there also a shorter way of doing this? I thought about something like:

b = [try int(k) for k in a]

This may sound like a silly question since I do have a working solution, but I have often been shown shorter ways of doing the same thing and always appreciated this kind of help. I am using Python 2.7

Thanks!

Edit: sorry, I was also talking about floating point. I just changed my example data

At the very least you can omit the standalone redundant `int(k)`. — deceze, Nov 19 '15 at 10:12
Never, ever, ever do a blank `except`, and especially never ever ever do that only to `pass`. — Daniel Roseman, Nov 19 '15 at 10:19
Edit: my code doesn't work without the except. It raises an error that the following code block was unindented? — offeltoffel, Nov 19 '15 at 10:30

Andrés Pérez-Albela H. · Accepted Answer · 2015-11-19T10:37:42.163

3

There is no way to use try/except clauses inside List Comprehensions but this could help:

a = ['2','3','4','5','Cherry']
print [int(x) for x in a if x.isdigit()]

Output:

['2', '3', '4', '5']

Update (as the question was updated):

This could help but I don't know how good/accurate is to use it:

a = ['2.0','3.0','4.0','5.0', '44545.45', 'Cherry']

[float(x) for x in a if x.split('.')[0].isdigit() and x.split('.')[1].isdigit()]

Output:

[2.0, 3.0, 4.0, 5.0, 44545.45]

edited Nov 19 '15 at 10:37

answered Nov 19 '15 at 10:13

Andrés Pérez-Albela H.

4,003
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2

Shouldn't it be `[int(x) for x in a if x.isdigit()]` since OP wants a list of integers. – JRodDynamite Nov 19 '15 at 10:15
1

Note that this will not work for negative numbers, number with an explicit `+` such as `'+123'`, etc. – Holt Nov 19 '15 at 10:15
This would be a perfect solution, but I should have mentioned that the list also contains floating point numbers. Does a function like "x.isstr()" exist? I could then add "if not x.isstr()" – offeltoffel Nov 19 '15 at 10:17
1

@offeltoffel, yeah you should. Now every answer is based on integers. – Andrés Pérez-Albela H. Nov 19 '15 at 10:18
it seems that for updated answer it won't work. You can't do int('2.0'). – Andrii Rusanov Nov 19 '15 at 10:24
1

@Andrey: yes, I didn't know that either, so it didn't seem important to me - my fault. But still I have learned something new and with these answers I was able to improve my solution – offeltoffel Nov 19 '15 at 10:29
@offeltoffel I've updated my answer. And as I mention, I don't know how accurate it is, but it works for me. – Andrés Pérez-Albela H. Nov 19 '15 at 10:32

Andrii Rusanov · Answer 2 · 2015-11-19T10:23:31.577

0

Try this one.

def is_number(k):
    try:
        float(k)
    except ValueError:
        return False
    return True

[int(float(k)) for k in a if is_number(k)]

edited Nov 19 '15 at 10:23

answered Nov 19 '15 at 10:15

Andrii Rusanov

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`'33'` is not an instance of `int`. – Holt Nov 19 '15 at 10:16

score 0 · Answer 3 · edited May 23 '17 at 11:59

If you want to compress try-except into one line then i think answer is NO and it is answered at here. I would go with mix of regex and isinstance check-It captures all number types e.g. float, long, int and complex-

>>>a=['2', '3', '4', '5', 'Cherry', '23.3', '-3']
>>>filter(bool,[i if isinstance(eval(i),(int, long, float, complex)) else None for i in filter(lambda x: re.findall(r'[-+]?\d+[\.]?\d*',x),a)])
>>>['2', '3', '4', '5', '23.3', '-3']

If you want to capture only floats-

>>>filter (bool,[i if isinstance(eval(i),float) else None for i in filter(lambda x: re.findall(r'[-+]?\d+[\.]?\d*',x),a)])
>>>['23.3']

If you want to capture only int-

>>>filter (bool,[i if isinstance(eval(i),int) else None for i in filter(lambda x: re.findall(r'[-+]?\d+[\.]?\d*',x),a)])
>>>['2', '3', '4', '5', '-3']

Python: "Try" inside shortened for...loop

3 Answers3