How to determine a Python variable's type?

Asked 2023-09-20 20:56:16 View 750,694

How do I see the type of a variable? (e.g. unsigned 32 bit)

Answers

Use the type() builtin function:

>>> i = 123
>>> type(i)
<type 'int'>
>>> type(i) is int
True
>>> i = 123.456
>>> type(i)
<type 'float'>
>>> type(i) is float
True

To check if a variable is of a given type, use isinstance:

>>> i = 123
>>> isinstance(i, int)
True
>>> isinstance(i, (float, str, set, dict))
False

Note that Python doesn't have the same types as C/C++, which appears to be your question.

Answered   2023-09-20 20:56:16

  • A point to note (since the question asked about numbers) - bool is a subclass of int. So isinstance(True, int) or isinstance(False, int) will return True. - anyone
  • Furthermore, due to bool being a subclass of int, certain surprising things hold true, like 0 == False and {1, True} == {True} - anyone

You may be looking for the type() built-in function.

See the examples below, but there's no "unsigned" type in Python just like Java.

Positive integer:

>>> v = 10
>>> type(v)
<type 'int'>

Large positive integer:

>>> v = 100000000000000
>>> type(v)
<type 'long'>

Negative integer:

>>> v = -10
>>> type(v)
<type 'int'>

Literal sequence of characters:

>>> v = 'hi'
>>> type(v)
<type 'str'>

Floating point number:

>>> v = 3.14159
>>> type(v)
<type 'float'>

Answered   2023-09-20 20:56:16

  • I had to double take when I saw this. Java SE8 now contains unsigned integers, and I've developed so much with it that it seems sinful that Java never had unsigned integers before SE8. - anyone

It is so simple. You do it like this.

print(type(variable_name))

Answered   2023-09-20 20:56:16

How to determine the variable type in Python?

So if you have a variable, for example:

one = 1

You want to know its type?

There are right ways and wrong ways to do just about everything in Python. Here's the right way:

Use type

>>> type(one)
<type 'int'>

You can use the __name__ attribute to get the name of the object. (This is one of the few special attributes that you need to use the __dunder__ name to get to - there's not even a method for it in the inspect module.)

>>> type(one).__name__
'int'

Don't use __class__

In Python, names that start with underscores are semantically not a part of the public API, and it's a best practice for users to avoid using them. (Except when absolutely necessary.)

Since type gives us the class of the object, we should avoid getting this directly. :

>>> one.__class__

This is usually the first idea people have when accessing the type of an object in a method - they're already looking for attributes, so type seems weird. For example:

class Foo(object):
    def foo(self):
        self.__class__

Don't. Instead, do type(self):

class Foo(object):
    def foo(self):
        type(self)

Implementation details of ints and floats

How do I see the type of a variable whether it is unsigned 32 bit, signed 16 bit, etc.?

In Python, these specifics are implementation details. So, in general, we don't usually worry about this in Python. However, to sate your curiosity...

In Python 2, int is usually a signed integer equal to the implementation's word width (limited by the system). It's usually implemented as a long in C. When integers get bigger than this, we usually convert them to Python longs (with unlimited precision, not to be confused with C longs).

For example, in a 32 bit Python 2, we can deduce that int is a signed 32 bit integer:

>>> import sys

>>> format(sys.maxint, '032b')
'01111111111111111111111111111111'
>>> format(-sys.maxint - 1, '032b') # minimum value, see docs.
'-10000000000000000000000000000000'

In Python 3, the old int goes away, and we just use (Python's) long as int, which has unlimited precision.

We can also get some information about Python's floats, which are usually implemented as a double in C:

>>> sys.float_info
sys.floatinfo(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, 
min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, 
mant_dig=53, epsilon=2.2204460492503131e-16, radix=2, rounds=1)

Conclusion

Don't use __class__, a semantically nonpublic API, to get the type of a variable. Use type instead.

And don't worry too much about the implementation details of Python. I've not had to deal with issues around this myself. You probably won't either, and if you really do, you should know enough not to be looking to this answer for what to do.

Answered   2023-09-20 20:56:16

print type(variable_name)

I also highly recommend the IPython interactive interpreter when dealing with questions like this. It lets you type variable_name? and will return a whole list of information about the object including the type and the doc string for the type.

e.g.

In [9]: var = 123

In [10]: var?
Type:       int
Base Class: <type 'int'>
String Form:    123
Namespace:  Interactive
Docstring:
    int(x[, base]) -> integer

Convert a string or number to an integer, if possible. A floating point argument will be truncated towards zero (this does not include a string representation of a floating point number!) When converting a string, use the optional base. It is an error to supply a base when converting a non-string. If the argument is outside the integer range a long object will be returned instead.

Answered   2023-09-20 20:56:16

  • I asked for unsigned int,signed int etc - anyone
  • print type(str) returns an error in Python 3.6. Use type(str) - anyone
  • @KolobCanyon that's because in 3.x, print needs parenthesis: print(type(str)) - anyone
  • You should really modify the print type(var) erroneous code. - anyone
a = "cool"
type(a)

//result 'str'
<class 'str'>
or 
do 
`dir(a)` 
to see the list of inbuilt methods you can have on the variable.

Answered   2023-09-20 20:56:16

Examples of simple type checking in Python:

assert type(variable_name) == int

assert type(variable_name) == bool

assert type(variable_name) == list

Answered   2023-09-20 20:56:16

  • This should be the accepted answer since it shows how to actually use the returned value. Nobody cares about seeing the result of type() in the terminal, we want it to do type-checking, and it's not obvious how (I assumed you had to compare to the type as a string: if (type(var)=='str'):) - anyone

One more way using __class__:

>>> a = [1, 2, 3, 4]
>>> a.__class__
<type 'list'>
>>> b = {'key1': 'val1'}
>>> b.__class__
<type 'dict'>
>>> c = 12
>>> c.__class__
<type 'int'>

Answered   2023-09-20 20:56:16

  • re: comment "names that start with underscores..." (single) underscores are very different than double underscores ("dunders") and they're definitely part of the public API and it's certainly fine to use them. This may be of help... youtu.be/wf-BqAjZb8M - anyone

It may be little irrelevant. but you can check types of an object with isinstance(object, type) as mentioned here.

Answered   2023-09-20 20:56:16

  • I think this one is way better than type function to use in checking process. If you want to check and act with the result you have to use try catch with type but isinstance returns true or false - anyone

The question is somewhat ambiguous -- I'm not sure what you mean by "view". If you are trying to query the type of a native Python object, @atzz's answer will steer you in the right direction.

However, if you are trying to generate Python objects that have the semantics of primitive C-types, (such as uint32_t, int16_t), use the struct module. You can determine the number of bits in a given C-type primitive thusly:

>>> struct.calcsize('c') # char
1
>>> struct.calcsize('h') # short
2
>>> struct.calcsize('i') # int
4
>>> struct.calcsize('l') # long
4

This is also reflected in the array module, which can make arrays of these lower-level types:

>>> array.array('c').itemsize # char
1

The maximum integer supported (Python 2's int) is given by sys.maxint.

>>> import sys, math
>>> math.ceil(math.log(sys.maxint, 2)) + 1 # Signedness
32.0

There is also sys.getsizeof, which returns the actual size of the Python object in residual memory:

>>> a = 5
>>> sys.getsizeof(a) # Residual memory.
12

For float data and precision data, use sys.float_info:

>>> sys.float_info
sys.floatinfo(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53, epsilon=2.2204460492503131e-16, radix=2, rounds=1)

Answered   2023-09-20 20:56:16

  • The question, as I understand it, asks about querying the type of a "variable" in Python. Your answer is generally correct but off-topic. - anyone

Do you mean in Python or using ctypes?

In the first case, you simply cannot - because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more reference on ctypes, an its type, see the official documentation.

Answered   2023-09-20 20:56:17

Python doesn't have such types as you describe. There are two types used to represent integral values: int, which corresponds to platform's int type in C, and long, which is an arbitrary precision integer (i.e. it grows as needed and doesn't have an upper limit). ints are silently converted to long if an expression produces result which cannot be stored in int.

Answered   2023-09-20 20:56:17

Simple, for python 3.4 and above

print (type(variable_name))

Python 2.7 and above

print type(variable_name)

Answered   2023-09-20 20:56:17

It really depends on what level you mean. In Python 2.x, there are two integer types, int (constrained to sys.maxint) and long (unlimited precision), for historical reasons. In Python code, this shouldn't make a bit of difference because the interpreter automatically converts to long when a number is too large. If you want to know about the actual data types used in the underlying interpreter, that's implementation dependent. (CPython's are located in Objects/intobject.c and Objects/longobject.c.) To find out about the systems types look at cdleary answer for using the struct module.

Answered   2023-09-20 20:56:17

For python2.x, use

print type(variable_name)

For python3.x, use

print(type(variable_name))

Answered   2023-09-20 20:56:17

You should use the type() function. Like so:

my_variable = 5

print(type(my_variable)) # Would print out <class 'int'>

This function will view the type of any variable, whether it's a list or a class. Check this website for more information: https://www.w3schools.com/python/ref_func_type.asp

Answered   2023-09-20 20:56:17

  • Please add further details to expand on your answer, such as working code or documentation citations. - anyone

Python is a dynamically typed language. A variable, initially created as a string, can be later reassigned to an integer or a float. And the interpreter won’t complain:

name = "AnyValue"
# Dynamically typed language lets you do this:
name = 21
name = None
name = Exception()

To check the type of a variable, you can use either type() or isinstance() built-in function. Let’s see them in action:

Python3 example:

variable = "hello_world"
print(type(variable) is str) # True
print(isinstance(variable, str)) # True

Let's compare both methods performances in python3

python3 -m timeit -s "variable = 'hello_world'" "type(variable) is int"
5000000 loops, best of 5: 54.5 nsec per loop

python3 -m timeit -s "variable = 'hello_world'" "isinstance(variable, str)"
10000000 loops, best of 5: 39.2 nsec per loop

type is 40% slower approximately (54.5/39.2 = 1.390).

We could use type(variable) == str instead. It would work, but it’s a bad idea:

  • == should be used when you want to check the value of a variable. We would use it to see if the value of the variable is equal to "hello_world". But when we want to check if the variable is a string, is the operator is more appropriate. For a more detailed explanation of when to use one or the other, check this article.
  • == is slower: python3 -m timeit -s "variable = 'hello_world'" "type(variable) == str" 5000000 loops, best of 5: 64.4 nsec per loop

Difference between isinstance and type

Speed is not the only difference between these two functions. There is actually an important distinction between how they work:

  • type only returns the type of an object (it's class). We can use it to check if the variable is of type str.
  • isinstance checks if a given object (first parameter) is:
    • an instance of a class specified as a second parameter. For example, is variable an instance of the str class?
    • or an instance of a subclass of a class specified as a second parameter. In other words - is variable an instance of a subclass of str?

What does it mean in practice? Let’s say we want to have a custom class that acts as a list but has some additional methods. So we might subclass the list type and add custom functions inside:

class MyAwesomeList(list):
    # Add additional functions here
    pass

But now the type and isinstance return different results if we compare this new class to a list!

my_list = MyAwesomeList()
print(type(my_list) is list) # False
print(isinstance(my_list, list)) # True

We get different results because isinstance checks if my_list is an instance of the list (it’s not) or a subclass of the list (it is because MyAwesomeList is a subclass of the list). If you forget about this difference, it can lead to some subtle bugs in your code.

Conclusions

isinstance is usually the preferred way to compare types. It’s not only faster but also considers inheritance, which is often the desired behavior. In Python, you usually want to check if a given object behaves like a string or a list, not necessarily if it’s exactly a string. So instead of checking for string and all its custom subclasses, you can just use isinstance.

On the other hand, when you want to explicitly check that a given variable is of a specific type (and not its subclass) - use type. And when you use it, use it like this: type(var) is some_type not like this: type(var) == some_type.

Answered   2023-09-20 20:56:17

  • This is great, but what would you do if (for example) an integer is given as a string, like "123", and you don't know in advance what the type is? - anyone
  • Let's take the value as "123". It can be checked during type conversion like int(value). It will throw the ValueError: invalid literal for int() with base 10: if it isn't an integer value. This exception can also be caught gracefully by using the try-except block like try: value = int(value) except ValueError: print("Invalid integer value") - anyone

I saw this one when I was new to Python (I still am):

x = ...
print(type(x))

Answered   2023-09-20 20:56:17

There's no 32bit and 64bit and 16bit, python is simple, you don't have to worry about it. See how to check the type:

integer = 1
print(type(integer))  # Result: <class 'int'>, and if it's a string then class will be str and so on.

# Checking the type
float_class = 1.3
print(isinstance(float_class, float))  # True

But if you really have to, you can use Ctypes library which has types like unsigned integer.

Ctypes types documentation

You can use it like this:

from ctypes import *
uint = c_uint(1)  # Unsigned integer
print(uint)  # Output: c_uint(1)

# To actually get the value, you have to call .value
print(uint.value)

# Change value
uint.value = 2
print(uint.value)  # 2

Answered   2023-09-20 20:56:17

There are many data types in python like:

Text Type:  str
Numeric Types:  int, float, complex
Sequence Types: list, tuple, range
Mapping Type:   dict
Set Types:  set, frozenset
Boolean Type:   bool
Binary Types:   bytes, bytearray, memoryview
None Type:  NoneType

Here I have written a code having a list containing all type of data types example and printing their type

L = [
    "Hello World",
    20,
    20.5,
    1j,
    ["apple", "banana", "cherry"],
    ("apple", "banana", "cherry"),
    range(6),
    {"name" : "John", "age" : 36},
    {"apple", "banana", "cherry"},
    frozenset({"apple", "banana", "cherry"}),
    True,
    b"Hello",
    bytearray(5),
    memoryview(bytes(5)),
    None
]
for _ in range(len(L)):
    print(type(L[_]))

OUTPUT:

<class 'str'>
<class 'int'>
<class 'float'>
<class 'complex'>
<class 'list'>
<class 'tuple'>
<class 'range'>
<class 'dict'>
<class 'set'>
<class 'frozenset'>
<class 'bool'>
<class 'bytes'>
<class 'bytearray'>
<class 'memoryview'>
<class 'NoneType'>

Answered   2023-09-20 20:56:17

Python got a type() function for this. You can simply use it like:

variable = "I love apples."
print(type(variable))  # Output will be <class 'str'>

That means your variable is a string.

You can try it yourself with integers, floats, lists etc.

Also with the usage of libraries like numpy, you can see other data types like:

import numpy as np

matrix = np.ones(shape=(3,3), dtype="uint8")
print(type(matrix))  # Output will be <class 'numpy.ndarray'>
print(type(matrix[0][0]))  # Outpu will be <class 'numpy.uint8'>

You can learn more about Data Types in Python from here.

Answered   2023-09-20 20:56:17

Just do not do it. Asking for something's type is wrong in itself. Instead use polymorphism. Find or if necessary define by yourself the method that does what you want for any possible type of input and just call it without asking about anything. If you need to work with built-in types or types defined by a third-party library, you can always inherit from them and use your own derivatives instead. Or you can wrap them inside your own class. This is the object-oriented way to resolve such problems.

If you insist on checking exact type and placing some dirty ifs here and there, you can use __class__ property or type function to do it, but soon you will find yourself updating all these ifs with additional cases every two or three commits. Doing it the OO way prevents that and lets you only define a new class for a new type of input instead.

Answered   2023-09-20 20:56:17