In Python, how do I determine if an object is iterable?

Question

Is there a method like isiterable? The only solution I have found so far is to call

hasattr(myObj, \'__iter__\')

But I am not

Answer 1

def is_iterable(x):
    try:
        0 in x
    except TypeError:
        return False
    else:
        return True

This will say yes to all manner of iterable objects, but it will say no to strings in Python 2. (That's what I want for example when a recursive function could take a string or a container of strings. In that situation, asking forgiveness may lead to obfuscode, and it's better to ask permission first.)

import numpy

class Yes:
    def __iter__(self):
        yield 1;
        yield 2;
        yield 3;

class No:
    pass

class Nope:
    def __iter__(self):
        return 'nonsense'

assert is_iterable(Yes())
assert is_iterable(range(3))
assert is_iterable((1,2,3))   # tuple
assert is_iterable([1,2,3])   # list
assert is_iterable({1,2,3})   # set
assert is_iterable({1:'one', 2:'two', 3:'three'})   # dictionary
assert is_iterable(numpy.array([1,2,3]))
assert is_iterable(bytearray("not really a string", 'utf-8'))

assert not is_iterable(No())
assert not is_iterable(Nope())
assert not is_iterable("string")
assert not is_iterable(42)
assert not is_iterable(True)
assert not is_iterable(None)

Many other strategies here will say yes to strings. Use them if that's what you want.

import collections
import numpy

assert isinstance("string", collections.Iterable)
assert isinstance("string", collections.Sequence)
assert numpy.iterable("string")
assert iter("string")
assert hasattr("string", '__getitem__')

Note: is_iterable() will say yes to strings of type bytes and bytearray.

• bytes objects in Python 3 are iterable True == is_iterable(b"string") == is_iterable("string".encode('utf-8')) There is no such type in Python 2.
• bytearray objects in Python 2 and 3 are iterable True == is_iterable(bytearray(b"abc"))

The O.P. hasattr(x, '__iter__') approach will say yes to strings in Python 3 and no in Python 2 (no matter whether '' or b'' or u''). Thanks to @LuisMasuelli for noticing it will also let you down on a buggy __iter__.