列表的增删改查

列表的用法和字符串不一样，不要搞混了！

思维导图：

这里要纠错，sort默认参数是sort（reverse=False），这里是正序排序，如果把reverse=True，那么就得到的是反序。

例：

l = [2,3,6,1,7,8]
l.sort()
print(l)  #得到的打印结果是：[1, 2, 3, 6, 7, 8]

l1 = l.sort(reverse=True)
print（l） #得到的结果是：[8, 7, 6, 3, 2, 1]

列表的增删改查：

# [1,2,'c','dfas',True]
#索引和切片。
# li = [1,2,3,'af','re',4,'45']
# print(l[0])
# print(l[3])
# print(l[-1])
#切片：
# print(li[0:3])
# print(li[2:5])
# print(li[0:5:2])
# print(li[-2::-2])
# print(li[5:0:-2])
#苑昊
# li = ['taibai','alex','wusir']
#1增加 append 在最后增加一个元素
# print(li.append('yuanhao'))
# li.append([1,2,3,'www'])
# print(li)
# while True:
#     username = input('请输入员工姓名：')
#     if username.lower() == 'q':break
#     li.append(username)
#     print(li)
#insert 插入
# li = ['taibai','alex','wusir']
# li.insert(1,'日天')
# print(li)
#extend 迭代的添加
# li.extend('q')
# li.extend('asdt')
# li.extend([1,2,3,'www'])
# print(li)
li = ['taibai','alex','wusir','egon','hulu','jingnvshen']
#删除 pop 按照索引去删除，有返回值
name = li.pop(1)
# print(li)
# print(name)
#remove 按照元素去删除
# li.remove('alex')
# print(li)
#clear 清空列表
# li.clear()
# print(li)
#del
# li = ['taibai','alex','wusir','egon','hulu','jingnvshen']
# del li[2:4]
# print(li)
# del li[0:4:2]
# print(li)
li = ['taibai','alex','wusir','egon','hulu','jingnvshen']
#改 切片先删除，迭代着添加
# li[0] = '男神'
# print(li)
# li[0:3] = '都是男人'
# print(li)
li[:] = '都是男人'
print(li)
# li[0:2] = ['asd']
# print(li)
# li[0:2] = ['asd','fdsa',123]
# print(li)
#查
# print(li[1:4])
# for i in li:
#     print(i)
# li = ['taibai','taibai','wusir','egon','taibai','jingnvshen']
# del li
# print(li)

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列表的其他方法：

一行代码实现删除列表中重复的值:

from collections import OrderedDict 用我们collection模块中的有序字典来实现,
x=OrderedDict.fromkeys(['a','b','c','a','b','c','c'])
"""
这里是fromkeys的源码

def fromkeys(cls, iterable, value=None):
    '''OD.fromkeys(S[, v]) -> New ordered dictionary with keys from S.
    If not specified, the value defaults to None.

    '''
    self = cls()
    for key in iterable:
        self[key] = value
    return self


"""
for i in x:
    print(i)

 1 列表的嵌套：
 2 li = [1,2,5,'taibai','yuanhao',[1,'alex',3,],True]
 3 '''
 4 # print(li[3])
 5 # print(li[3][3])
 6 # s = li[4].capitalize()
 7 # li[4] = s
 8 #li[4] = li[4].capitalize()
 9 # li[4] = 'Yuanhao'
10 # print(li)
11 # s2 = li[4].replace('hao','日天')
12 # li[4] = s2
13 # li[4] = li[4][0:4] + 'ritian'
14 # print(li)
15 # l2 = li[5]
16 # li[5][0] = '文杰'
17 # print(li)
18 '''
19 li[5][1] = li[5][1].upper()
20 print(li)
21 
22 
23 
24 # li = [1,2,3,1,'a',1,'4']
25 #count 计数
26 # print(li.count('fasd'))
27 li = [2,3,1,5,6,8,9,7,4,10]
28 #sort 正序排序
29 # li.sort()
30 # print(li)
31 #li.sort(reverse=True) 倒叙排序
32 # reverse 反转
33 # li.reverse()
34 # print(li)

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元祖：

元祖是可哈西不可变的数据类型

 1 #1,元祖tupe()
 2 # t = (1,2,3,'adsf',True,[12,3,'苑昊','taibai'],('fdsa',2,3))
 3 # print(t[4])
 4 # print(t[1:4])
 5 # for i in t:
 6 #     print(i)
 7 # t[4] = False 儿子不能更改
 8 # print(t)
 9 # t[5][2] = '苑日天'
10 # print(t)

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range：

 1 #range范围，相当于可迭代对象
 2 # for i in range(1,10):
 3 #     print(i)
 4 # for i in range(1,10,2):  # 步长
 5 #     print(i)
 6 # for i in range(10,1,-2):  # 步长
 7 #     print(i)
 8 li = [1,2,'a',4,[1,2,'太白','alex'],2]
 9 #range,len
10 # for i in range(0,len(li)):
11 #     if i == 4:  # type(li[i]) == list
12 #         for j in li[i]:  # [1,2,'太白','alex']
13 #             print(j)
14 #     else:print(li[i])
15 
16 print(li.index('a',3,6))
17 # for i in li:
18 #     print(i)

View Code