def function():
n=123.456
x=int(n)
y=n-int(n)
print(x,y)
result:
x= 123
y= 0.45600000000000307
how to get exactly .456
without using library function,
n
can be any floating number
If you know from the outset that the number of decimal places is 3, then:
y = round(n - int(n), 3)
If you don't know the number of decimal places, then you can work it out, as so:
y = round(n - int(n), str(n)[::-1].find('.'))
As furas pointed out, you can also use the decimal
package:
from decimal import Decimal
n = Decimal('123.456')
y = n - int(n)
You can also use the re
module:
import re
def get_decimcal(n: float) -> float:
return float(re.search(r'\.\d+', str(n)).group(0))
def get_decimcal_2(n: float) -> float:
return float(re.findall(r'\.\d+', str(n))[0])
def get_int(n: float) -> int:
return int(n)
print(get_decimcal(123.456))
print(get_decimcal_2(123.456))
print(get_int(123.456))
Output
0.456
0.456
123
Asha Anandan
You can use %f
to round of the floating value to required digits.
def function(n):
x = int(n)
y = n-int(n)
print(x,"%.2f" % y)
function(123.456)
Output:
123
0.456
gañañufla
Try with round(y,n)
, and n=3
its the numbers of decimals.
来源:https://stackoverflow.com/questions/58687240/how-to-get-rid-of-additional-floating-numbers-in-python-subtraction