숫자가 int인지 float인지 확인하십시오.

숫자가 int인지 float인지 확인하십시오.

---

내가 한 방법은 다음과 같습니다.

inNumber = somenumber
inNumberint = int(inNumber)
if inNumber == inNumberint:
    print "this number is an int"
else:
    print "this number is a float"

그런 것.
더 좋은 방법이 있습니까?

---

isinstance를 사용하십시오 .

>>> x = 12
>>> isinstance(x, int)
True
>>> y = 12.0
>>> isinstance(y, float)
True

그래서:

>>> if isinstance(x, int):
        print 'x is a int!'

x is a int!

_편집하다:_

지적한 바와 같이, 긴 정수의 경우 위의 작동하지 않습니다. 따라서 다음을 수행해야합니다.

>>> x = 12L
>>> import numbers
>>> isinstance(x, numbers.Integral)
True
>>> isinstance(x, int)
False

---

@ninjagecko의 답변이 가장 좋습니다.

이것은 또한 작동합니다 :

Python 2.x 용

isinstance(n, (int, long, float)) 

파이썬 3.x는 오래 가지 않습니다

isinstance(n, (int, float))

복소수에 대한 복소수 유형도 있습니다

---

짧막 한 농담:

isinstance(yourNumber, numbers.Real)

이것은 몇 가지 문제를 피합니다.

>>> isinstance(99**10,int)
False

데모:

>>> import numbers

>>> someInt = 10
>>> someLongInt = 100000L
>>> someFloat = 0.5

>>> isinstance(someInt, numbers.Real)
True
>>> isinstance(someLongInt, numbers.Real)
True
>>> isinstance(someFloat, numbers.Real)
True

---

허락을 구하는 것보다 용서를 구하는 것이 더 쉽습니다. 간단히 작업을 수행하십시오. 그것이 작동한다면, 물체는 수용 가능하고 적합하며 적절한 유형이었습니다. 작업이 작동하지 않으면 개체 유형이 적절하지 않은 것입니다. 유형을 아는 것은 거의 도움이되지 않습니다.

작업을 시도하고 작동하는지 확인하십시오.

inNumber = somenumber
try:
    inNumberint = int(inNumber)
    print "this number is an int"
except ValueError:
    pass
try:
    inNumberfloat = float(inNumber)
    print "this number is a float"
except ValueError:
    pass

---

What you can do too is usingtype() Example:

if type(inNumber) == int : print "This number is an int"
elif type(inNumber) == float : print "This number is a float"

---

Here's a piece of code that checks whether a number is an integer or not, it works for both Python 2 and Python 3.

import sys

if sys.version < '3':
    integer_types = (int, long,)
else:
    integer_types = (int,)

isinstance(yourNumber, integer_types)  # returns True if it's an integer
isinstance(yourNumber, float)  # returns True if it's a float

Notice that Python 2 has both types int and long, while Python 3 has only type int. Source.

If you want to check whether your number is a float that represents an int, do this

(isinstance(yourNumber, float) and (yourNumber).is_integer())  # True for 3.0

If you don't need to distinguish between int and float, and are ok with either, then ninjagecko's answer is the way to go

import numbers

isinstance(yourNumber, numbers.Real)

---

You can use modulo to determine if x is an integer numerically. The isinstance(x, int) method only determines if x is an integer by type:

def isInt(x):
    if x%1 == 0:
        print "X is an integer"
    else:
        print "X is not an integer"

---

I know it's an old thread but this is something that I'm using and I thought it might help.

It works in python 2.7 and python 3< .

def is_float(num):
    """
    Checks whether a number is float or integer

    Args:
        num(float or int): The number to check

    Returns:
        True if the number is float
    """
    return not (float(num)).is_integer()


class TestIsFloat(unittest.TestCase):
    def test_float(self):
        self.assertTrue(is_float(2.2))

    def test_int(self):
        self.assertFalse(is_float(2))

---

pls check this: import numbers

import math

a = 1.1 - 0.1
print a 

print isinstance(a, numbers.Integral)
print math.floor( a )
if (math.floor( a ) == a):
    print "It is an integer number"
else:
    print False

Although X is float but the value is integer, so if you want to check the value is integer you cannot use isinstance and you need to compare values not types.

---

how about this solution?

if type(x) in (float, int):
    # do whatever
else:
    # do whatever

---

Tried in Python version 3.6.3 Shell

>>> x = 12
>>> import numbers
>>> isinstance(x, numbers.Integral)
True
>>> isinstance(x,int)
True

Couldn't figure out anything to work for.

---

absolute = abs(x)
rounded = round(absolute)
if absolute - rounded == 0:
  print 'Integer number'
else:
  print 'notInteger number'

---

def is_int(x):
  absolute = abs(x)
  rounded = round(absolute)
  if absolute - rounded == 0:
    print str(x) + " is an integer"
  else:
    print str(x) +" is not an integer"


is_int(7.0) # will print 7.0 is an integer

---

Try this...

def is_int(x):
  absolute = abs(x)
  rounded = round(absolute)
  return absolute - rounded == 0

---

Update: Try this

inNumber = [32, 12.5, 'e', 82, 52, 92, '1224.5', '12,53',
            10000.000, '10,000459', 
           'This is a sentance, with comma number 1 and dot.', '121.124']

try:

    def find_float(num):
        num = num.split('.')
        if num[-1] is not None and num[-1].isdigit():
            return True
        else:
            return False

    for i in inNumber:
        i = str(i).replace(',', '.')
        if '.' in i and find_float(i):
            print('This is float', i)
        elif i.isnumeric():
            print('This is an integer', i)
        else:
            print('This is not a number ?', i)

except Exception as err:
    print(err)

---

variable.isnumeric checks if a value is an integer:

 if myVariable.isnumeric:
    print('this varibale is numeric')
 else:
    print('not numeric')

참고URL : https://stackoverflow.com/questions/4541155/check-if-a-number-is-int-or-float