How do I see the type of a variable? (e.g. unsigned 32 bit)
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
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
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:
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'
__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)
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)
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
print type(str)
returns an error in Python 3.6. Use type(str)
- anyone print(type(str))
- anyone 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
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
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
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
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). int
s 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
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
Speed is not the only difference between these two functions. There is actually an important distinction between how they work:
isinstance
checks if a given object (first parameter) is:
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.
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
"123"
, and you don't know in advance what the type is? - anyone "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.
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 if
s here and there, you can use __class__
property or type
function to do it, but soon you will find yourself updating all these if
s 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