If you need to do this, do
unless you are in Python 2.x in which case you want
isinstance(<var>, (int, long))
Do not use
type. It is almost never the right answer in Python, since it blocks all the flexibility of polymorphism. For instance, if you subclass
int, your new class should register as an
type will not do:
class Spam(int): pass x = Spam(0) type(x) == int # False isinstance(x, int) # True
This adheres to Python's strong polymorphism: you should allow any object that behaves like an
int, instead of mandating that it be one.
The classical Python mentality, though, is that it's easier to ask forgiveness than permission. In other words, don't check whether
x is an integer; assume that it is and catch the exception results if it isn't:
try: x += 1 except TypeError: ...
This mentality is slowly being overtaken by the use of abstract base classes, which let you register exactly what properties your object should have (adding? multiplying? doubling?) by making it inherit from a specially-constructed class. That would be the best solution, since it will permit exactly those objects with the necessary and sufficient attributes, but you will have to read the docs on how to use it.
All proposed answers so far seem to miss the fact that a double (floats in python are actually doubles) can also be an integer (if it has nothing after the decimal point). I use the built-in
is_integer() method on doubles to check this.
Example (to do something every xth time in a for loop):
for index in range(y): # do something if (index/x.).is_integer(): # do something special
You can always convert to a float before calling this method. The three possibilities:
>>> float(5).is_integer() True >>> float(5.1).is_integer() False >>> float(5.0).is_integer() True
Otherwise, you could check if it is an int first like Agostino said:
def is_int(val): if type(val) == int: return True else: if val.is_integer(): return True else: return False