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python基础习题答案,Python基础答案

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求python题目解答(初学阶段)

列表lst中有4个元素,看有几个元素,就看逗号就好了,即便是嵌套列表,在两个逗号之间,也算一个元素,你可以使用len(lst)得到结果。

lst[3]的数据类型为列表,列表用[]表示。

lst[3][1][2]=10

lst[-1][-1][1]=9;

lst[-1][-1][3]=12;

lst[-1][-1][-3:]=[9, 10, 12];

lst[-1][-1][-3:][::-1]=[12, 10, 9]  #::-1表示列表反转

求Python高手解答基本Python习题

#!/usr/bin/env python

# coding = utf-8

KNOWTREE = dict(

# does it have a backhone?

True = dict(

# does it give birth to live babies

True = "Mammal",

False = dict(

# does it have feathers

True = "Bird",

False = dict(

# does it have gills

True = "Fish",

False = dict(

# does it lay eggs in water

True = "Amphibian",

Flase = "Reptile",

),

),

),

),

False = dict(

# does it have a shell

True = "Mollusc",

False = dict(

# does it have 6 legs

True = "Insect",

False = "Arachind",

),

),

)

def which_animal(ans):

know = KNOWTREE

while isinstance(know, dict):

know = know[repr(ans.pop(0))]

return know

def movie_price(weekday, dayhour):

if weekday == "Tuesday":

return 10.75

elif weekday == "Wednesday":

return 5.75

elif weekday in ("Monday","Thursday","Friday") and dayhour 17:

return 12.75

else:

return 15.75

print which_animal([True,True,True,True,True,True,True,])

print which_animal([False,False,False,False,False,False,False,])

print movie_price("Tuesday", 4)

print movie_price("Saturday", 15)

print movie_price("Friday", 17)

print movie_price("Friday", 16)

Python中基础练习题?

法一:利用set()函数的去重功能,去重后再使用list()函数将集合转换为我们想要的列表

list1 = [11,22,33]

list2 = [22,33,44]

list3 = list(set(list1 + list2))

list3.sort()

print(list3)

-------------

法二:利用if和for,先遍历list1所有元素追加到list3中,然后遍历list2,条件判断list2中当前元素是否在list3中,如果不在则追加到list3中

list1 = [11,22,33]

list2 = [22,33,44]

list3 = []

for ele1 in list1:

list3.append(ele1)

for ele2 in list2:

if ele2 not in list3:

list3.append(ele2)

print(list3)

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