What is the most pythonic way to check if an object is a number?

Question:

Given an arbitrary python object, what’s the best way to determine whether it is a number? Here is is defined as acts like a number in certain circumstances.

For example, say you are writing a vector class. If given another vector, you want to find the dot product. If given a scalar, you want to scale the whole vector.

Checking if something is int, float, long, bool is annoying and doesn’t cover user-defined objects that might act like numbers. But, checking for __mul__, for example, isn’t good enough because the vector class I just described would define __mul__, but it wouldn’t be the kind of number I want.

Asked By: Claudiu

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Answers:

This is a good example where exceptions really shine. Just do what you would do with the numeric types and catch the TypeError from everything else.

But obviously, this only checks if a operation works, not whether it makes sense! The only real solution for that is to never mix types and always know exactly what typeclass your values belong to.

Answered By: Jochen Ritzel

Probably it’s better to just do it the other way around: You check if it’s a vector. If it is, you do a dot product and in all other cases you attempt scalar multiplication.

Checking for the vector is easy, since it should of your vector class type (or inherited from it). You could also just try first to do a dot-product, and if that fails (= it wasn’t really a vector), then fall back to scalar multiplication.

Answered By: sth

For the hypothetical vector class:

Suppose v is a vector, and we are multiplying it by x. If it makes sense to multiply each component of v by x, we probably meant that, so try that first. If not, maybe we can dot? Otherwise it’s a type error.

EDIT — the below code doesn’t work, because 2*[0]==[0,0] instead of raising a TypeError. I leave it because it was commented-upon.

def __mul__( self, x ):
    try:
        return [ comp * x for comp in self ]
    except TypeError:
        return [ x * y for x, y in itertools.zip_longest( self, x, fillvalue = 0 )
Answered By: Katriel

You want to check if some object

acts like a number in certain
circumstances

If you’re using Python 2.5 or older, the only real way is to check some of those “certain circumstances” and see.

In 2.6 or better, you can use isinstance with numbers.Number — an abstract base class (ABC) that exists exactly for this purpose (lots more ABCs exist in the collections module for various forms of collections/containers, again starting with 2.6; and, also only in those releases, you can easily add your own abstract base classes if you need to).

Bach to 2.5 and earlier,
“can be added to 0 and is not iterable” could be a good definition in some cases. But,
you really need to ask yourself, what it is that you’re asking that what you want to consider “a number” must definitely be able to do, and what it must absolutely be unable to do — and check.

This may also be needed in 2.6 or later, perhaps for the purpose of making your own registrations to add types you care about that haven’t already be registered onto numbers.Numbers — if you want to exclude some types that claim they’re numbers but you just can’t handle, that takes even more care, as ABCs have no unregister method [[for example you could make your own ABC WeirdNum and register there all such weird-for-you types, then first check for isinstance thereof to bail out before you proceed to checking for isinstance of the normal numbers.Number to continue successfully.

BTW, if and when you need to check if x can or cannot do something, you generally have to try something like:

try: 0 + x
except TypeError: canadd=False
else: canadd=True

The presence of __add__ per se tells you nothing useful, since e.g all sequences have it for the purpose of concatenation with other sequences. This check is equivalent to the definition “a number is something such that a sequence of such things is a valid single argument to the builtin function sum“, for example. Totally weird types (e.g. ones that raise the “wrong” exception when summed to 0, such as, say, a ZeroDivisionError or ValueError &c) will propagate exception, but that’s OK, let the user know ASAP that such crazy types are just not acceptable in good company;-); but, a “vector” that’s summable to a scalar (Python’s standard library doesn’t have one, but of course they’re popular as third party extensions) would also give the wrong result here, so (e.g.) this check should come after the “not allowed to be iterable” one (e.g., check that iter(x) raises TypeError, or for the presence of special method __iter__ — if you’re in 2.5 or earlier and thus need your own checks).

A brief glimpse at such complications may be sufficient to motivate you to rely instead on abstract base classes whenever feasible…;-).

Answered By: Alex Martelli

Use Number from the numbers module to test isinstance(n, Number) (available since 2.6).

>>> from numbers import Number
... from decimal import Decimal
... from fractions import Fraction
... for n in [2, 2.0, Decimal('2.0'), complex(2, 0), Fraction(2, 1), '2']:
...     print(f'{n!r:>14} {isinstance(n, Number)}')
              2 True
            2.0 True
 Decimal('2.0') True
         (2+0j) True
 Fraction(2, 1) True
            '2' False

This is, of course, contrary to duck typing. If you are more concerned about how an object acts rather than what it is, perform your operations as if you have a number and use exceptions to tell you otherwise.

Answered By: Steven Rumbalski

Just to add upon.
Perhaps we can use a combination of isinstance and isdigit as follows to find whether a value is a number (int, float, etc)

if isinstance(num1, int) or isinstance(num1 , float) or num1.isdigit():

Answered By: shadab.tughlaq

I had a similar issue, when implementing a sort of vector class. One way to check for a number is to just convert to one, i.e. by using

float(x)

This should reject cases where x cannot be converted to a number; but may also reject other kinds of number-like structures that could be valid, for example complex numbers.

Answered By: Ant6n

To rephrase your question, you are trying to determine whether something is a collection or a single value. Trying to compare whether something is a vector or a number is comparing apples to oranges – I can have a vector of strings or numbers, and I can have a single string or single number. You are interested in how many you have (1 or more), not what type you actually have.

my solution for this problem is to check whether the input is a single value or a collection by checking the presence of __len__. For example:

def do_mult(foo, a_vector):
    if hasattr(foo, '__len__'):
        return sum([a*b for a,b in zip(foo, a_vector)])
    else:
        return [foo*b for b in a_vector]

Or, for the duck-typing approach, you can try iterating on foo first:

def do_mult(foo, a_vector):
    try:
        return sum([a*b for a,b in zip(foo, a_vector)])
    except TypeError:
        return [foo*b for b in a_vector]

Ultimately, it is easier to test whether something is vector-like than to test whether something is scalar-like. If you have values of different type (i.e. string, numeric, etc.) coming through, then the logic of your program may need some work – how did you end up trying to multiply a string by a numeric vector in the first place?

Answered By: Gordon Bean

Multiply the object by zero. Any number times zero is zero. Any other result means that the object is not a number (including exceptions)

def isNumber(x):
    try:
        return bool(0 == x*0)
    except:
        return False

Using isNumber thusly will give the following output:

class A: pass 

def foo(): return 1

for x in [1,1.4, A(), range(10), foo, foo()]:
    answer = isNumber(x)
    print('{answer} == isNumber({x})'.format(**locals()))

Output:

True == isNumber(1)
True == isNumber(1.4)
False == isNumber(<__main__.A instance at 0x7ff52c15d878>)
False == isNumber([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
False == isNumber(<function foo at 0x7ff52c121488>)
True == isNumber(1)

There probably are some non-number objects in the world that define __mul__ to return zero when multiplied by zero but that is an extreme exception. This solution should cover all normal and sane code that you generate/encouter.

numpy.array example:

import numpy as np

def isNumber(x):
    try:
        return bool(x*0 == 0)
    except:
        return False

x = np.array([0,1])

answer = isNumber(x)
print('{answer} == isNumber({x})'.format(**locals()))

output:

False == isNumber([0 1])
Answered By: shrewmouse

You could use the isdigit() function.

>>> x = "01234"
>>> a.isdigit()
True
>>> y = "1234abcd"
>>> y.isdigit()
False
Answered By: rsy

To summarize / evaluate existing methods:

Candidate    | type                      | delnan | mat | shrewmouse | ant6n
-------------------------------------------------------------------------
0            | <type 'int'>              |      1 |   1 |          1 |     1
0.0          | <type 'float'>            |      1 |   1 |          1 |     1
0j           | <type 'complex'>          |      1 |   1 |          1 |     0
Decimal('0') | <class 'decimal.Decimal'> |      1 |   0 |          1 |     1
True         | <type 'bool'>             |      1 |   1 |          1 |     1
False        | <type 'bool'>             |      1 |   1 |          1 |     1
''           | <type 'str'>              |      0 |   0 |          0 |     0
None         | <type 'NoneType'>         |      0 |   0 |          0 |     0
'0'          | <type 'str'>              |      0 |   0 |          0 |     1
'1'          | <type 'str'>              |      0 |   0 |          0 |     1
[]           | <type 'list'>             |      0 |   0 |          0 |     0
[1]          | <type 'list'>             |      0 |   0 |          0 |     0
[1, 2]       | <type 'list'>             |      0 |   0 |          0 |     0
(1,)         | <type 'tuple'>            |      0 |   0 |          0 |     0
(1, 2)       | <type 'tuple'>            |      0 |   0 |          0 |     0

(I came here by this question)

Code

#!/usr/bin/env python

"""Check if a variable is a number."""

import decimal


def delnan_is_number(candidate):
    import numbers
    return isinstance(candidate, numbers.Number)


def mat_is_number(candidate):
    return isinstance(candidate, (int, long, float, complex))


def shrewmouse_is_number(candidate):
    try:
        return 0 == candidate * 0
    except:
        return False


def ant6n_is_number(candidate):
    try:
        float(candidate)
        return True
    except:
        return False

# Test
candidates = (0, 0.0, 0j, decimal.Decimal(0),
              True, False, '', None, '0', '1', [], [1], [1, 2], (1, ), (1, 2))

methods = [delnan_is_number, mat_is_number, shrewmouse_is_number, ant6n_is_number]

print("Candidate    | type                      | delnan | mat | shrewmouse | ant6n")
print("-------------------------------------------------------------------------")
for candidate in candidates:
    results = [m(candidate) for m in methods]
    print("{:<12} | {:<25} | {:>6} | {:>3} | {:>10} | {:>5}"
          .format(repr(candidate), type(candidate), *results))
Answered By: Martin Thoma

If you want to call different methods depending on the argument type(s), look into multipledispatch.

For example, say you are writing a vector class. If given another vector, you want to find the dot product. If given a scalar, you want to scale the whole vector.

from multipledispatch import dispatch

class Vector(list):

    @dispatch(object)
    def __mul__(self, scalar):
        return Vector( x*scalar for x in self)

    @dispatch(list)
    def __mul__(self, other):
        return sum(x*y for x,y in zip(self, other))


>>> Vector([1,2,3]) * Vector([2,4,5])   # Vector time Vector is dot product
25
>>> Vector([1,2,3]) * 2                 # Vector times scalar is scaling
[2, 4, 6]

Unfortunately, (to my knowledge) we can’t write @dispatch(Vector) since we are still defining the type Vector, so that type name is not yet defined. Instead, I’m using the base type list, which allows you to even find the dot product of a Vector and a list.

Answered By: AJNeufeld

Short and simple way :

obj = 12345
print(isinstance(obj,int))

Output :

True

If the object is a string, ‘False’ will be returned :

obj = 'some string'
print(isinstance(obj,int))

Output :

False
Answered By: Shekhar

You have a data item, say rec_day that when written to a file will be a float. But during program processing it can be either float, int or str type (the str is used when initializing a new record and contains a dummy flag value).

You can then check to see if you have a number with this

                type(rec_day) != str 

I’ve structured a python program this way and just put in ‘maintenance patch’ using this as a numeric check. Is it the Pythonic way? Most likely not since I used to program in COBOL.

Answered By: CopyPasteIt

can be implemented in a simple try exception block

def check_if_number(str1):
    try:
        int(float(str1))
        return 'number'
    except:
        return 'not a number'

a = check_if_number('32322')
print (a)
# number
Answered By: SuperNova

You can use numbers.Number to check if an object is a number.

For numbers, Python 3 supports 3 types int, float and complex types so if checking the 3 types of values with numbers.Number as shown below:

import numbers

print(type(100), isinstance(100, numbers.Number))
print(type(100.23), isinstance(100.23, numbers.Number))
print(type(100 + 2j), isinstance(100 + 2j, numbers.Number))

All return True as shown below:

<class 'int'> True
<class 'float'> True
<class 'complex'> True

And, for numbers, Python 2 supperts 4 types int, long, float and complex types so if checking the 4 types of values with numbers.Number as shown below::

import numbers

print(type(100), isinstance(100, numbers.Number))
print(type(10000000000000000000), isinstance(10000000000000000000, numbers.Number))
print(type(100.23), isinstance(100.23, numbers.Number))
print(type(100 + 2j), isinstance(100 + 2j, numbers.Number))

All return True as shown below:

(<type 'int'>, True)
(<type 'long'>, True)
(<type 'float'>, True)
(<type 'complex'>, True)
Answered By: Kai – Kazuya Ito
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