How to check if a float value is a whole number
Question:
I am trying to find the largest cube root that is a whole number, that is less than 12,000.
processing = True
n = 12000
while processing:
n -= 1
if n ** (1/3) == #checks to see if this has decimals or not
I am not sure how to check if it is a whole number or not though! I could convert it to a string then use indexing to check the end values and see whether they are zero or not, that seems rather cumbersome though. Is there a simpler way?
Answers:
To check if a float value is a whole number, use the float.is_integer()
method:
>>> (1.0).is_integer()
True
>>> (1.555).is_integer()
False
The method was added to the float
type in Python 2.6.
Take into account that in Python 2, 1/3
is 0
(floor division for integer operands!), and that floating point arithmetic can be imprecise (a float
is an approximation using binary fractions, not a precise real number). But adjusting your loop a little this gives:
>>> for n in range(12000, -1, -1):
... if (n ** (1.0/3)).is_integer():
... print n
...
27
8
1
0
which means that anything over 3 cubed, (including 10648) was missed out due to the aforementioned imprecision:
>>> (4**3) ** (1.0/3)
3.9999999999999996
>>> 10648 ** (1.0/3)
21.999999999999996
You’d have to check for numbers close to the whole number instead, or not use float()
to find your number. Like rounding down the cube root of 12000
:
>>> int(12000 ** (1.0/3))
22
>>> 22 ** 3
10648
If you are using Python 3.5 or newer, you can use the math.isclose()
function to see if a floating point value is within a configurable margin:
>>> from math import isclose
>>> isclose((4**3) ** (1.0/3), 4)
True
>>> isclose(10648 ** (1.0/3), 22)
True
For older versions, the naive implementation of that function (skipping error checking and ignoring infinity and NaN) as mentioned in PEP485:
def isclose(a, b, rel_tol=1e-9, abs_tol=0.0):
return abs(a - b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)
You could use this:
if k == int(k):
print(str(k) + " is a whole number!")
You can use a modulo operation for that.
if (n ** (1.0/3)) % 1 != 0:
print("We have a decimal number here!")
Wouldn’t it be easier to test the cube roots? Start with 20 (20**3 = 8000) and go up to 30 (30**3 = 27000). Then you have to test fewer than 10 integers.
for i in range(20, 30):
print("Trying {0}".format(i))
if i ** 3 > 12000:
print("Maximum integral cube root less than 12000: {0}".format(i - 1))
break
You don’t need to loop or to check anything. Just take a cube root of 12,000 and round it down:
r = int(12000**(1/3.0))
print r*r*r # 10648
How about
if x%1==0:
print "is integer"
The above answers work for many cases but they miss some. Consider the following:
fl = sum([0.1]*10) # this is 0.9999999999999999, but we want to say it IS an int
Using this as a benchmark, some of the other suggestions don’t get the behavior we might want:
fl.is_integer() # False
fl % 1 == 0 # False
Instead try:
def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
return abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)
def is_integer(fl):
return isclose(fl, round(fl))
now we get:
is_integer(fl) # True
isclose
comes with Python 3.5+, and for other Python’s you can use this mostly equivalent definition (as mentioned in the corresponding PEP)
We can use the modulo (%) operator. This tells us how many remainders we have when we divide x by y – expresses as x % y
. Every whole number must divide by 1, so if there is a remainder, it must not be a whole number.
This function will return a boolean, True
or False
, depending on whether n
is a whole number.
def is_whole(n):
return n % 1 == 0
You can use the round
function to compute the value.
Yes in python as many have pointed when we compute the value of a cube root, it will give you an output with a little bit of error. To check if the value is a whole number you can use the following function:
def cube_integer(n):
if round(n**(1.0/3.0))**3 == n:
return True
return False
But remember that int(n)
is equivalent to math.floor
and because of this if you find the int(41063625**(1.0/3.0))
you will get 344 instead of 345.
So please be careful when using int
withe cube roots.
Just a side info, is_integer
is doing internally:
import math
isInteger = (math.floor(x) == x)
Not exactly in python, but the cpython implementation is implemented as mentioned above.
Try using:
int(val) == val
It will give lot more precision than any other methods.
All the answers are good but a sure fire method would be
def whole (n):
return (n*10)%10==0
The function returns True if it’s a whole number else False….I know I’m a bit late but here’s one of the interesting methods which I made…
Edit: as stated by the comment below, a cheaper equivalent test would be:
def whole(n):
return n%1==0
>>> def is_near_integer(n, precision=8, get_integer=False):
... if get_integer:
... return int(round(n, precision))
... else:
... return round(n) == round(n, precision)
...
>>> print(is_near_integer(10648 ** (1.0/3)))
True
>>> print(is_near_integer(10648 ** (1.0/3), get_integer=True))
22
>>> for i in [4.9, 5.1, 4.99, 5.01, 4.999, 5.001, 4.9999, 5.0001, 4.99999, 5.000
01, 4.999999, 5.000001]:
... print(i, is_near_integer(i, 4))
...
4.9 False
5.1 False
4.99 False
5.01 False
4.999 False
5.001 False
4.9999 False
5.0001 False
4.99999 True
5.00001 True
4.999999 True
5.000001 True
>>>
You can use something like:
num = 1.9899
bool(int(num)-num)
#returns True
if it is True, It means it holds some value, hence not a whole number. Else
num = 1.0
bool(int(num)-num)
# returns False
This problem has been solved, but I would like to propose an additional mathematical-based solution for funcies.
The benefit of this approach is that it calculates the whole number part of your number, which may be beneficial depending on your general task.
Algorithm:
- Decompose whole number part of your number its a sum of its decimals (e.g.,
327=3*100+2*10+7*1
)
- take difference between calculated whole number and number itself
- decide whether difference is close enough to be considered an integer.
from math import ceil, log, isclose
def is_whole(x: float) -> bool:
n_digits = ceil(log(x,10)) # number of digits of decimals at or above ones
digits = [(n//(10**i))%10 for i in range(n_digits)] # parse digits of `x` at or above ones decimal
whole = 0 # will equal the whole number part of `x`
for i in range(n_digits):
decimal = 10**i
digit = digits[i]
whole += digit*decimal
diff = whole - x
return isclose(diff, 0.0)
NOTE: the idea of parsing digits of a number was realized from here
I am trying to find the largest cube root that is a whole number, that is less than 12,000.
processing = True
n = 12000
while processing:
n -= 1
if n ** (1/3) == #checks to see if this has decimals or not
I am not sure how to check if it is a whole number or not though! I could convert it to a string then use indexing to check the end values and see whether they are zero or not, that seems rather cumbersome though. Is there a simpler way?
To check if a float value is a whole number, use the float.is_integer()
method:
>>> (1.0).is_integer()
True
>>> (1.555).is_integer()
False
The method was added to the float
type in Python 2.6.
Take into account that in Python 2, 1/3
is 0
(floor division for integer operands!), and that floating point arithmetic can be imprecise (a float
is an approximation using binary fractions, not a precise real number). But adjusting your loop a little this gives:
>>> for n in range(12000, -1, -1):
... if (n ** (1.0/3)).is_integer():
... print n
...
27
8
1
0
which means that anything over 3 cubed, (including 10648) was missed out due to the aforementioned imprecision:
>>> (4**3) ** (1.0/3)
3.9999999999999996
>>> 10648 ** (1.0/3)
21.999999999999996
You’d have to check for numbers close to the whole number instead, or not use float()
to find your number. Like rounding down the cube root of 12000
:
>>> int(12000 ** (1.0/3))
22
>>> 22 ** 3
10648
If you are using Python 3.5 or newer, you can use the math.isclose()
function to see if a floating point value is within a configurable margin:
>>> from math import isclose
>>> isclose((4**3) ** (1.0/3), 4)
True
>>> isclose(10648 ** (1.0/3), 22)
True
For older versions, the naive implementation of that function (skipping error checking and ignoring infinity and NaN) as mentioned in PEP485:
def isclose(a, b, rel_tol=1e-9, abs_tol=0.0):
return abs(a - b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)
You could use this:
if k == int(k):
print(str(k) + " is a whole number!")
You can use a modulo operation for that.
if (n ** (1.0/3)) % 1 != 0:
print("We have a decimal number here!")
Wouldn’t it be easier to test the cube roots? Start with 20 (20**3 = 8000) and go up to 30 (30**3 = 27000). Then you have to test fewer than 10 integers.
for i in range(20, 30):
print("Trying {0}".format(i))
if i ** 3 > 12000:
print("Maximum integral cube root less than 12000: {0}".format(i - 1))
break
You don’t need to loop or to check anything. Just take a cube root of 12,000 and round it down:
r = int(12000**(1/3.0))
print r*r*r # 10648
How about
if x%1==0:
print "is integer"
The above answers work for many cases but they miss some. Consider the following:
fl = sum([0.1]*10) # this is 0.9999999999999999, but we want to say it IS an int
Using this as a benchmark, some of the other suggestions don’t get the behavior we might want:
fl.is_integer() # False
fl % 1 == 0 # False
Instead try:
def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
return abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)
def is_integer(fl):
return isclose(fl, round(fl))
now we get:
is_integer(fl) # True
isclose
comes with Python 3.5+, and for other Python’s you can use this mostly equivalent definition (as mentioned in the corresponding PEP)
We can use the modulo (%) operator. This tells us how many remainders we have when we divide x by y – expresses as x % y
. Every whole number must divide by 1, so if there is a remainder, it must not be a whole number.
This function will return a boolean, True
or False
, depending on whether n
is a whole number.
def is_whole(n):
return n % 1 == 0
You can use the round
function to compute the value.
Yes in python as many have pointed when we compute the value of a cube root, it will give you an output with a little bit of error. To check if the value is a whole number you can use the following function:
def cube_integer(n):
if round(n**(1.0/3.0))**3 == n:
return True
return False
But remember that int(n)
is equivalent to math.floor
and because of this if you find the int(41063625**(1.0/3.0))
you will get 344 instead of 345.
So please be careful when using int
withe cube roots.
Just a side info, is_integer
is doing internally:
import math
isInteger = (math.floor(x) == x)
Not exactly in python, but the cpython implementation is implemented as mentioned above.
Try using:
int(val) == val
It will give lot more precision than any other methods.
All the answers are good but a sure fire method would be
def whole (n):
return (n*10)%10==0
The function returns True if it’s a whole number else False….I know I’m a bit late but here’s one of the interesting methods which I made…
Edit: as stated by the comment below, a cheaper equivalent test would be:
def whole(n):
return n%1==0
>>> def is_near_integer(n, precision=8, get_integer=False):
... if get_integer:
... return int(round(n, precision))
... else:
... return round(n) == round(n, precision)
...
>>> print(is_near_integer(10648 ** (1.0/3)))
True
>>> print(is_near_integer(10648 ** (1.0/3), get_integer=True))
22
>>> for i in [4.9, 5.1, 4.99, 5.01, 4.999, 5.001, 4.9999, 5.0001, 4.99999, 5.000
01, 4.999999, 5.000001]:
... print(i, is_near_integer(i, 4))
...
4.9 False
5.1 False
4.99 False
5.01 False
4.999 False
5.001 False
4.9999 False
5.0001 False
4.99999 True
5.00001 True
4.999999 True
5.000001 True
>>>
You can use something like:
num = 1.9899
bool(int(num)-num)
#returns True
if it is True, It means it holds some value, hence not a whole number. Else
num = 1.0
bool(int(num)-num)
# returns False
This problem has been solved, but I would like to propose an additional mathematical-based solution for funcies.
The benefit of this approach is that it calculates the whole number part of your number, which may be beneficial depending on your general task.
Algorithm:
- Decompose whole number part of your number its a sum of its decimals (e.g.,
327=3*100+2*10+7*1
) - take difference between calculated whole number and number itself
- decide whether difference is close enough to be considered an integer.
from math import ceil, log, isclose
def is_whole(x: float) -> bool:
n_digits = ceil(log(x,10)) # number of digits of decimals at or above ones
digits = [(n//(10**i))%10 for i in range(n_digits)] # parse digits of `x` at or above ones decimal
whole = 0 # will equal the whole number part of `x`
for i in range(n_digits):
decimal = 10**i
digit = digits[i]
whole += digit*decimal
diff = whole - x
return isclose(diff, 0.0)
NOTE: the idea of parsing digits of a number was realized from here