How to compare type of an object in Python?

Question

isinstance()

In your case, isinstance("this is a string", str) will return True.

You may also want to read this: http://www.canonical.org/~kragen/isinstance/

isinstance works:

if isinstance(obj, MyClass): do_foo(obj)

but, keep in mind: if it looks like a duck, and if it sounds like a duck, it is a duck.

EDIT: For the None type, you can simply do:

if obj is None: obj = MyClass()

First, avoid all type comparisons. They're very, very rarely necessary. Sometimes, they help to check parameter types in a function -- even that's rare. Wrong type data will raise an exception, and that's all you'll ever need.

All of the basic conversion functions will map as equal to the type function.

type(9) is int
type(2.5) is float
type('x') is str
type(u'x') is unicode
type(2+3j) is complex

There are a few other cases.

isinstance( 'x', basestring )
isinstance( u'u', basestring )
isinstance( 9, int )
isinstance( 2.5, float )
isinstance( (2+3j), complex )

None, BTW, never needs any of this kind of type checking. None is the only instance of NoneType. The None object is a Singleton. Just check for None

variable is None

BTW, do not use the above in general. Use ordinary exceptions and Python's own natural polymorphism.

For other types, check out the types module:

>>> import types
>>> x = "mystring"
>>> isinstance(x, types.StringType)
True
>>> x = 5
>>> isinstance(x, types.IntType)
True
>>> x = None
>>> isinstance(x, types.NoneType)
True

P.S. Typechecking is a bad idea.

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.

You're very close! string is a module, not a type. You probably want to compare the type of obj against the type object for strings, namely str:

type(obj) == str  # this works because str is already a type

Alternatively:

type(obj) == type('')

Note, in Python 2, if obj is a unicode type, then neither of the above will work. Nor will isinstance(). See John's comments to this post for how to get around this... I've been trying to remember it for about 10 minutes now, but was having a memory block!

Use str instead of string

type ( obj ) == str

Explanation

>>> a = "Hello"
>>> type(a)==str
True
>>> type(a)
<type 'str'>
>>>