Check if all elements of a list are of the same type

Try using all in conjunction with isinstance:

all(isinstance(x, int) for x in lst)

You can even check for multiple types with isinstance if that is desireable:

all(isinstance(x, (int, long)) for x in lst)

Not that this will pick up inherited classes as well. e.g.:

class MyInt(int):
     pass

print(isinstance(MyInt('3'),int)) #True

If you need to restrict yourself to just integers, you could use all(type(x) is int for x in lst). But that is a VERY rare scenario.

A fun function you could write with this is one which would return the type of the first element in a sequence if all the other elements are the same type:

def homogeneous_type(seq):
    iseq = iter(seq)
    first_type = type(next(iseq))
    return first_type if all( (type(x) is first_type) for x in iseq ) else False

This will work for any arbitrary iterable, but it will consume "iterators" in the process.

Another fun function in the same vein which returns the set of common bases:

import inspect
def common_bases(seq):
    iseq = iter(seq)
    bases = set(inspect.getmro(type(next(iseq))))
    for item in iseq:
        bases = bases.intersection(inspect.getmro(type(item)))
        if not bases:
           break
    return bases

Using any(), no need to traverse whole list. Just break as soon as object which is not int or long is found:

>>> not any(not isinstance(y,(int,long)) for y in [1,2,3])
True
>>> not any(not isinstance(y,(int,long)) for y in [1,'a',2,3])
False

Combining some of the answers already given, using a combination of map(), type() and set() provides a imho rather readable answer. Assuming the limitation of not checking for type polymorphisms is ok. Also not the most computationally efficient answer, but it allows to easily check whether all elements are of the same type.

# To check whether all elements in a list are integers
set(map(type, [1,2,3])) == {int}
# To check whether all elements are of the same type
len(set(map(type, [1,2,3]))) == 1