Question

我只是python的新手，我需要打印列表中包含最小和最大项目的列表。

例如，如果我有：

 total_list = [[1, 2, 3], [1, 2, 3, 4], [1,2,3,4,5]]

我需要返回一个列表，列表的长度最小。我怎么能简单地以python的方式来做呢？

我尝试对其进行迭代，但是我所能收到的只是列表中每个项目的唯一len

输出必须为：

total_list[0] and total_list[2]

预先感谢

Answer 1

python中的

max和min函数接受一个key参数，该参数将根据定义为键的值找到可迭代的最大值。所以，试试这个：

max(total_list, key=len)

然后您可以使用total_list.index查找这些索引

Answer 2

total_list = [[1, 2, 3], [1, 2, 3, 4], [1,2,3,4,5]]

print('total_list[{0}] and total_list[{1}]'.format(total_list.index(min(total_list, key=len)), total_list.index(max(total_list, key=len))))

输出：

C:\Users\Desktop>py x.py
total_list[0] and total_list[2]

Answer 3

使用min( iterable, key=len)（min / max）最简单。如果您想自己浏览列表，可以这样做：

tl  = [[1, 2, 3], [1, 2, 3, 4], [1,2,3,4,5]]

# init shortest and longest to 1st element of list
# use a tuple to store:  ( listdata, length, position in original list )
shortest = longest = (tl[0], len(tl[0]), 0)

# iterate over remaining elements
for pos_act,act in enumerate(tl[1:],1): # start enumerate on 1 to fix position due slicing
    # get actual length
    len_act = len(act)

    # is it larger then current largest?
    if len_act > longest[1]:
        longest = (act,len_act, pos_act)

    # is it shorter then current shortest?
    if len_act < shortest[1]:
        shortest = (act, len_act, pos_act)

print ("Shortest: {} with len {} on position {}".format(*shortest))
print ("Longest: {} with len {} on position {}".format(*longest))

输出：

Shortest: [1, 2, 3] with len 3 on position 0
Longest: [1, 2, 3, 4, 5] with len 5 on position 2

Doku：

enumerate(iterable[,start_value])

在python列表中找到最长的列表

3 个答案: