Python中是否有算子计算百分比？

Question

I've recently learned that the " % " sign is used to calculate the remainder of an integer in Python. However I was unable to determine if there's another operator or method to calculate percent in Python.

Like with " / " which will give you the quotient, if you just use a float for one of the integers, it will actually give you the answer like traditional division. So is there a method to work out percentage?

Answer 1

You could just divide your two numbers and multiply by 100. Note that this will throw an error if "whole" is 0, as asking what percentage of 0 a number is does not make sense:

def percentage(part, whole):
  return 100 * float(part)/float(whole)

Or with a % at the end:

 def percentage(part, whole):
  Percentage = 100 * float(part)/float(whole)
  return str(Percentage) + “%”

Or if the question you wanted it to answer was "what is 5% of 20", rather than "what percentage is 5 of 20" (a different interpretation of the question inspired by Carl Smith's answer ), you would write:

def percentage(percent, whole):
  return (percent * whole) / 100.0

Answer 2

There is no such operator in Python, but it is trivial to implement on your own. In practice in computing, percentages are not nearly as useful as a modulo, so no language that I can think of implements one.

Answer 3

use of %

def percent(expression):
    if "%" in expression:
        expression = expression.replace("%","/100")
    return eval(expression)

>>> percent("1500*20%")
300.0

Somthing simple

>>> p = lambda x: x/100
>>> p(20)
0.2
>>> 100*p(20)
20.0
>>>

Answer 4

Brian's answer (a custom function) is the correct and simplest thing to do in general.

But if you really wanted to define a numeric type with a (non-standard) '%' operator, like desk calculators do, so that 'X % Y' means X * Y / 100.0, then from Python 2.6 onwards you can redefine the mod () operator :

import numbers

class MyNumberClasswithPct(numbers.Real):
    def __mod__(self,other):
        """Override the builtin % to give X * Y / 100.0 """
        return (self * other)/ 100.0
    # Gotta define the other 21 numeric methods...
    def __mul__(self,other):
        return self * other # ... which should invoke other.__rmul__(self)
    #...

This could be dangerous if you ever use the '%' operator across a mixture of MyNumberClasswithPct with ordinary integers or floats.

What's also tedious about this code is you also have to define all the 21 other methods of an Integral or Real, to avoid the following annoying and obscure TypeError when you instantiate it

("Can't instantiate abstract class MyNumberClasswithPct with abstract methods __abs__,  __add__, __div__, __eq__, __float__, __floordiv__, __le__, __lt__, __mul__,  __neg__, __pos__, __pow__, __radd__, __rdiv__, __rfloordiv__, __rmod__, __rmul__,  __rpow__, __rtruediv__, __truediv__, __trunc__")

Answer 5

Very quickly and sortly-code implementation by using the lambda operator.

In [17]: percent = lambda part, whole:float(whole) / 100 * float(part)
In [18]: percent(5,400)
Out[18]: 20.0
In [19]: percent(5,435)
Out[19]: 21.75

Answer 6

def percent(part, whole):
    try:
        return 100 * float(part) / float(whole)
    except ZeroDivisionError:
        return 0

Answer 7

I have found that the most useful case for me is to compute the calculations as ratios and then printing the results formatting them as percentages thanks to Python formatting options :

result = 1.0/2.0            # result is 0.5
print(f'{result:.0%}')      # prints "50%"

Answer 8

Just use this

score_in_percentage = round( score *100 , 2 )
print(f' {score_in_percentage}% ') #50%