如何简化评分系统函数中重复的 if-elif 语句? [英] How can I simplify repetitive if-elif statements in my grading system function?

问题描述

目标是构建一个程序，将分数从0 到 1"系统转换为F 到 A"系统:

The goal is to build a program to convert scores from a '0 to 1' system to an 'F to A' system:

• 如果 score >= 0.9 将打印 'A'
• 如果 score >= 0.8 将打印 'B'
• 0.7，C
• 0.6，D
• 以及低于该点的任何值，打印 F

• If score >= 0.9 would print 'A'
• If score >= 0.8 would print 'B'
• 0.7, C
• 0.6, D
• And any value below that point, print F

这是构建它的方法，它适用于程序，但有点重复:

This is the way to build it and it works on the program, but it's somewhat repetitive:

if scr >= 0.9:
    print('A')
elif scr >= 0.8:
    print('B')
elif scr >= 0.7:
    print('C')
elif scr >= 0.6:
    print('D')
else:
    print('F')

我想知道是否有一种方法可以构建一个函数，这样复合语句就不会重复.

I would like to know if there is a way to build a function so that the compound statements wouldn't be as repetitive.

我是一个完全的初学者，但会在以下几行:

I'm a total beginner, but would something in the lines of :

def convertgrade(scr, numgrd, ltrgrd):
    if scr >= numgrd:
        return ltrgrd
    if scr < numgrd:
        return ltrgrd

有可能吗?

这里的目的是稍后我们可以通过只传递 scr、numbergrade 和 letter grade 作为参数来调用它:

The intention here is that later we can call it by only passing the scr, numbergrade and letter grade as arguments:

convertgrade(scr, 0.9, 'A')
convertgrade(scr, 0.8, 'B')
convertgrade(scr, 0.7, 'C')
convertgrade(scr, 0.6, 'D')
convertgrade(scr, 0.6, 'F')

如果可以传递更少的参数，那就更好了.

If it would be possible to pass fewer arguments, it would be even better.

推荐答案

您可以使用 bisect 模块进行数字表查找:

You can use the bisect module to do a numeric table lookup:

from bisect import bisect 

def grade(score, breakpoints=[60, 70, 80, 90], grades='FDCBA'):
     i = bisect(breakpoints, score)
     return grades[i]

>>> [grade(score) for score in [33, 99, 77, 70, 89, 90, 100]]
['F', 'A', 'C', 'C', 'B', 'A', 'A']

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