[11,2,3,3,7,9,11,2,3]
list1=[]
for i in list:
if i not in list1:
list1.append(i)
print(list1)
lst=set(list) # 有问题 下面是正确的
ret=lst.sort(key=list.index)
print(ret)
ret = list(set(l1)) # 正确的
print(ret)
ret.sort(key=l1.index) # 按值在l1中的索引进行排序
print(ret)
list2=[{"name":'alex',"age":36},{"name":"god","age":26},{"name":"xiao","age":27}]
# dict={"alex":{"name":'alex',"age":36},
# "god":{"name":"god","age":26},
# "xiao":{"name":"xiao","age":26}}
list2.sort(key=lambda x:x["age"])
print(list2)
c=(250,200)
commd="fire:%s,%s"%c
print(commd)
commd1="fire:{}".format(c)
print(commd1)
name="alex"
age=9000
s="{} is {}".format(name,age)
print(s)
ss=f"{name} is {age}"
print(ss)
def foo(arg,li=[]):
li.append(arg)
return li
list1=foo(21)
list2=foo(21,[1,])
list3=foo(28)
print(list1)
print(list2)
print(list3)
# [21, 28]
# [1, 21]
# [21, 28]
li=[]
def foo(arg):
li.append(arg)
return li
list1=foo(21)
list2=foo(21)
list3=foo(28)
print(list1)
print(list2)
print(list3)
# 结果都是[21,21,28]
# li.append() 没有返回值
def foo(arg,li=[]):
return li.append(arg)
list1=foo(21)
list2=foo(21,[1,])
list3=foo(28)
print(list1)
print(list2)
print(list3)
# None
# None
# None
list5=[11,22,33,44,55]
print(list5[10:])
# []
---------------------------------------------------------------list1 = [11, [22, 3], [4, ], [55, 66], 8, [9, [7, [12, [34, [26]]]]]] ------搞不懂
# 去除多余嵌套的列表,得到[11, 22, 3, 4, 55, 66, 8]
# 小剥皮
# [11, [22, 3]]
# [11, [22, [3, 4]]
def func(x):
return [a for b in x for a in func(b)] if isinstance(x, list) else [x]
def f(x):
ret = []
for b in x:
if isinstance(b, list):
for a in f(b):
ret.append(a)
else:
ret.append(b)
return ret
list2 = [11, 22, [33, 44], [55, [66, 77]], [88, [99, [100, [200, [300]]]]]]
ret = f(list2)
print(ret)
ret2 = func(list2)
print(ret2)
--------------------------------------------------------------
# 随机打乱列表参数顺序
import random
random.shuffle(list5)
print(list5)
相当于一个dom树的一个节点，同一个对象，不停的动态添加（类似于指针）
---------------------------------------------------------------
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# __author__ = "Q1mi"
# Date: 2017/11/24
"""
Python全栈课前练习题
"""
# s = "Alex SB 哈哈 x:1 y:2 z:3 自行车"
# # 问题1：如何取到["Alex SB 哈哈 x:1 y:2 z:3", "自行车"]?
# s1=s.split(" ")
# print(s1)
# # 问题2：如何在上面结果基础上拿到["Alex", "SB", "哈哈"]?
# s1=['Alex SB 哈哈 x:1 y:2 z:3', '自行车']
# s11=s1[0].split(" ")[0].split()
# print(s11)
# # 问题3：如何在上面结果基础上拿到"SB"?
# s2=["Alex", "SB", "哈哈"]
# print(s2[1])
# ------------------------------------------------------------------------------------------
# 有一个列表，他的内部是一些元祖，元祖的第一个元素是姓名，第二个元素是爱好。
# 现在我给你一个姓名，如"Egon",如果有这个姓名，就打印出他的爱好，没有就打印查无此人。
list1 = [
("Alex", "烫头"),
("Egon", "街舞"),
("Yuan", "喝茶")
]
for i in list1:
if "Egon"==i[0]:
print(i[1])
break
else:print("查无此人")
# for i in list1: # 有问题
# dict={}
# dict[i[0]]=i[1]
# print(dict)
# ------------------------------------------------------------------------------------------
# 我有一个HTML文件"login.html"
# 问题1：我如何读取它的内容保存到变量html_s？
# with open("login.html","r",encoding="utf-8") as f:
# html_s=f.read()
# print(html_s)
# 问题2：我如何读取它的二进制内容保存到变量html_b？
# with open("login.html","rb",) as f: # 不加 encoding
# html_s=f.read()
# print(html_s)
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