How to check if a number is a number when it is a 'string'?

Question

This is a part of a larger program. Here is what I am trying to do.

Pass a sentence to the scan method.
Have the sentence contain numbers.
Split the sentence into it's different terms.
append to a list a tuple, the first expression in the tuple being the type of thing that the word or sentence element fits into, the second being the word or number.

Here is what I am trying:

def scan(self, sentence):
    self.term = []

    for word in sentence.split():
        if word in direction:
            self.term.append(('direction', word))
        elif word in verbs:
            self.term.append(('verb', word))
        elif word in stop:
            self.term.append(('stop', word))
        elif word in nouns:
            self.term.append(('noun', word))
        elif type(int(word)) == 'int':
            self.term.append(('number', int(word)))
        else:
            self.term.append(('error', word))

    return self.term



print lexicon.scan('12 1234')

This is a method in a class, the print statement is outside. The part I am concerned with and having trouble with is this:

elif type(int(word)) == int:
    self.term.append(('number', int(word)))

It should work for any natural number [1, infinity)

Edit: I run into a problem when I try to scan('ASDFASDFASDF')

what are you doing exactly? type(int(word)) == int ??? you convert it to int then check if it's int? — zero.zero.seven, May 20 '13 at 17:52
Duplicate of http://stackoverflow.com/questions/5424716/python-how-to-check-if-input-is-a-number-given-that-input-always-returns-stri of http://stackoverflow.com/questions/16488278/how-to-check-if-a-variable-is-an-integer-or-a-string of http://stackoverflow.com/questions/354038/how-do-i-check-if-a-string-is-a-number-in-python of ... — Alexei Sholik, May 20 '13 at 17:58

score 5 · Accepted Answer · answered May 20 '13 at 17:52

5

Since you only need positive integers, then try elif word.isdigit(): (note that this will also accept "0").

answered May 20 '13 at 17:52

PaulMcG

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1

You could do `elif word.isdigit() and word != '0':` to filter out zero. – SethMMorton May 20 '13 at 17:55
2

Possibly: `elif word.ltrim('0').isdigit()` – Jon Clements May 20 '13 at 18:02
`elif word.isdigit() and int(word) > 0:` – Ryan Haining May 20 '13 at 18:10
I'd stick with string-style methods during the 'if' test, such as Jon Clements's suggestion of lstrip, and try to do the int() conversion only once in the elif body once we know we want this value as a positive integer. – PaulMcG May 20 '13 at 18:12

Ryan Haining · Answer 2 · 2013-05-21T20:54:32.200

1

if word.lstrip('0').isdigit(): 
    #append

Using the .lstrip('0') will remove leading 0's and cause strings such as '0' and '000' to not pass the check. Simply doing if word.isdigit() and word !='0' will not exclude '00' or any other string that is just multiple '0's

You could also use a try/except/else to see if it is an int and respond accordingly

try:
    int(s)
except ValueError:
    print s, 'is not an int'
else:
    print s, 'is an int'

edited May 21 '13 at 20:54

answered May 20 '13 at 17:54

Ryan Haining

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`s` is a string, so you mean "and s != '0'". – PaulMcG May 20 '13 at 17:58
@PaulMcGuire I meant `int(s)` actually, but yeah my mistake thanks – Ryan Haining May 20 '13 at 18:02

score 0 · Answer 3 · answered May 20 '13 at 17:52

0

You could apply int to word and catch a ValueError if it's not a number.

answered May 20 '13 at 17:52

George

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Do an `except ValueError:` then what? – Where's-the-pi May 20 '13 at 17:55

How to check if a number is a number when it is a 'string'?

3 Answers3