How to compare type of an object in Python?

Question

Basically I want to do this:

obj = \'str\'
type ( obj ) == string

I tried:

type ( obj ) == type ( string )

<

Answer 1

You can always use the type(x) == type(y) trick, where y is something with known type.

# check if x is a regular string
type(x) == type('')
# check if x is an integer
type(x) == type(1)
# check if x is a NoneType
type(x) == type(None)

Often there are better ways of doing that, particularly with any recent python. But if you only want to remember one thing, you can remember that.

In this case, the better ways would be:

# check if x is a regular string
type(x) == str
# check if x is either a regular string or a unicode string
type(x) in [str, unicode]
# alternatively:
isinstance(x, basestring)
# check if x is an integer
type(x) == int
# check if x is a NoneType
x is None

Note the last case: there is only one instance of NoneType in python, and that is None. You'll see NoneType a lot in exceptions (TypeError: 'NoneType' object is unsubscriptable -- happens to me all the time..) but you'll hardly ever need to refer to it in code.

Finally, as fengshaun points out, type checking in python is not always a good idea. It's more pythonic to just use the value as though it is the type you expect, and catch (or allow to propagate) exceptions that result from it.