Python - isinstance returns false

Question

I'm having an issue with isinstance().

I'm using Python 2.7.8, and running scripts from the shell.

The array element I'm testing for contains a number, but this function returns false; using number.Numbers:

import numbers
...
print array[x][i] 
>> 1
...
print isinstance(array[x][i], numbers.Number)
>>> False

Also tried this, from this post

import types
...
print isinstance(array[x][i], (types.IntType, types.LongType, types.FloatType, types.ComplexType))
>>> False

From the same post, I tried

isinstance(array[x][i], (int, float, long, complex))

I also tried this solution did not work.

All return false.

What does `print repr(array[x][i]), type(array[x][i])` produce? — Martijn Pieters, Apr 12 '15 at 01:53
You could have a *string* for example; if `array[x][i]` stores `'1'`, then `print` will produce that exact same output. — Martijn Pieters, Apr 12 '15 at 01:54

score 8 · Accepted Answer · answered Apr 12 '15 at 01:57

You don't have a number; you most probably have a string instead, containing the digit '1':

>>> value = '1'
>>> print value
1
>>> print 1
1

This is not a number; it is a string instead. Note that printing that string is indistinguishable from printing an integer.

Use repr() to print out Python representations instead, and / or use the type() function to produce the type object for a given value:

>>> print repr(value)
'1'
>>> print type(value)
<type 'str'>

Now it is clear that the value is a string, not an integer, even though it looks the same when printed.

For actual numeric values, isinstance() together with numbers.Number works as expected:

>>> from numbers import Number
>>> isinstance(value, Number)
False
>>> isinstance(1, Number)
True

Thanks for putting that in context. It only looked like an integer. — JasTonAChair, Apr 12 '15 at 02:09

1 Answers1