Question

我很难弄清楚如何从列表中找到min 例如

somelist = [1,12,2,53,23,6,17]

如何通过定义（def）函数

来查找此列表的最小值和最大值

我不想使用内置函数min

Answer 1

from __future__ import division

somelist =  [1,12,2,53,23,6,17] 
max_value = max(somelist)
min_value = min(somelist)
avg_value = sum(somelist)/len(somelist)

如果您想手动将最小值作为函数找到：

somelist =  [1,12,2,53,23,6,17] 

def my_min_function(somelist):
    min_value = None
    for value in somelist:
        if not min_value:
            min_value = value
        elif value < min_value:
            min_value = value
    return min_value

Python 3.4引入了statistics包，它提供了mean和其他统计数据：

from statistics import mean, median

somelist =  [1,12,2,53,23,6,17]
avg_value = mean(somelist)
median_value = median(somelist)

Answer 2

返回元组中的最小值和最大值：

def side_values(num_list):
    results_list = sorted(num_list)
    return results_list[0], results_list[-1]


somelist = side_values([1,12,2,53,23,6,17])
print(somelist)

Answer 3

只有一位老师会要求您做这样的愚蠢的事情。您可以提供预期的答案。或一个独特的解决方案，而课程的其余部分将（打哈欠）相同...

from operator import lt, gt
def ultimate (l,op,c=1,u=0):
    try:
        if op(l[c],l[u]): 
            u = c
        c += 1
        return ultimate(l,op,c,u)
    except IndexError:
        return l[u]
def minimum (l):
    return ultimate(l,lt)
def maximum (l):
    return ultimate(l,gt)

解决方案很简单。用它来使自己远离明显的选择。

Answer 4

list=[]
n = int(input("Enter the length of your list: "))
for i in range (1,n+1):
    a=int(input("Enter the %d number: " %i ))
    list.append(a)
min=list[0]
max=list[0]
for i in range(1,n):
    if max<list[i]:
        max=list[i]
    if min>list[i]:
        min=list[i]
print(" %d is the biggest number " %max)
print(" %d is the smallest number " %min)

Python查找列表的最小最大值和平均值（数组）

4 个答案: