最优雅的进行成绩等级判断的方式
使用Python如何优雅地进行成绩等级判断?
一般想到的方法为:
# -*- coding:utf-8 -*-
# author: Man Yacan
# Email: myxc@live.cn
# Website: https://www.manyacan.com
# datetime: 2020/9/2 20:39
# software: PyCharm
score = int(input("请输入成绩:")) # 输入成绩
# 进行成绩判断
if score >= 90:
grade = 'A'
elif score >= 80:
grade = 'B'
elif score >= 70:
grade = 'C'
elif score >= 60:
grade = 'D'
else:
grade = 'E'
print(f"成绩等级为{grade}") # 打印成绩等级
但是这种方法重复代码太多,十分不优雅。
Python没有像JavaScript或者PHP那样的switch
语句,那么该如何实现呢?
今天在网上看到一段代码,十分优雅的实现了成绩等级判断。
# -*- coding:utf-8 -*-
# author: Man Yacan
# Email: myxc@live.cn
# Website: https://www.manyacan.com
# datetime: 2020/9/2 20:39
# software: PyCharm
from bisect import bisect
def score_grade(score):
breakpoints = [60, 70, 80, 90]
grades = "EDCBA"
i = bisect(breakpoints, score)
return grades[i]
print(score_grade(int(input("请输入成绩:"))))
笔记
了解bisect
类,这个类可以使用二分法在已经过排序的数组中查找元素的位置,返回该值在list中的位置,说不清,直接上代码:
from bisect import bisect, insort
breakpoints = [60, 70, 80]
insort(breakpoints, 90)
print(str(breakpoints)) # [60, 70, 80, 90]
print(bisect(breakpoints, 60)) # 1
print(bisect(breakpoints, 90)) # 4
print(bisect(breakpoints, 61)) # 1
print(bisect(breakpoints, 82)) # 3
数组在使用该方法前必须先完成排序,可以使用.sort()
In [2]: a = [1, 0, 3, 4, 7, 5]
In [3]: a.sort()
In [4]: a
Out[4]: [0, 1, 3, 4, 5, 7]