python全栈开发从入门到放弃之列表的内置方法

 1 l=[‘a‘,‘b‘,‘c‘,‘d‘,‘e‘,‘f‘]
 2 print(l[1:5]) # 根据索引号来切片，但顾头不顾尾
 3 [‘b‘, ‘c‘, ‘d‘, ‘e‘]
 4 print(l[1:5:2]) # 根据索引号来切片，顾头不顾尾，后面加个haxi值每两个空一个
 5 [‘b‘, ‘d‘]
 6 print(l[2:5] )
 7 [‘c‘, ‘d‘, ‘e‘]
 8 print(l[-1]) #-1代表从后往前取 -1代表最后一个
 9 f
10 
11 
12 l=[‘a‘,‘b‘,‘c‘,‘d‘,‘e‘,‘f‘]
13 print(l[-4:-1]) #从后往前取
14 [‘c‘, ‘d‘, ‘e‘] 
15 print(l[2:5]) #
16 [‘c‘, ‘d‘, ‘e‘]
17 print(l[-1]) 
18 f
19 print([l[-4:]]) # -4 从后往前取倒数4个
20 [[‘c‘, ‘d‘, ‘e‘, ‘f‘]]

1 hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘]
2 hobbies.append(‘girls‘)
3 print(hobbies)

1 根据索引号来删除
2 不是单纯的删除，是删除并且把删除的元素返回，我们可以用一个变量名去接收该返回值
3 hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘]
4 x=hobbies.pop(1)
5 print(hobbies) 
6 [‘play‘, ‘sleep‘, ‘study‘]
7 print(x)
8 eat

 1 入队
 2 queue_l.append(‘first‘)
 3 queue_l.append(‘second‘)
 4 queue_l.append(‘third‘)
 5 print(queue_l)
 6 出队
 7 print(queue_l.pop(0))
 8 print(queue_l.pop(0))
 9 print(queue_l.pop(0))
10 print(queue_l.pop(0))
11 print(queue_l.pop(0))
12 print(queue_l.pop(0))

1 del hobbies[1] #单纯的删除
2 hobbies.remove(‘eat‘) #单纯的删除，并且是指定元素去删除

判断列表里默认第一个的长度
hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘]
print(len(hobbies))

1 判断sleep元素是否存在与hobbies在则返回True,不在则返回False
2 hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘]
3 print(‘sleep‘in hobbies)
4 True

指定序列，来添加元素
hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘,‘eat‘,‘eat‘]
hobbies.insert(1,‘walk‘)
hobbies.insert(1,[‘walkl‘,‘walk2‘,‘walk3‘])
print(hobbies)

查找元素是否在列表中有多少个
hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘,‘eat‘,‘eat‘]
print(hobbies.count(‘eat‘))

1 在列表元素的后面添加元素到列表
2 hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘,‘eat‘,‘eat‘]
3 hobbies.extend([‘walk1‘,‘walk2‘,‘walk3‘])
4 print(hobbies)
5 [‘play‘, ‘eat‘, ‘sleep‘, ‘study‘, ‘eat‘, ‘eat‘, ‘walk1‘, ‘walk2‘, ‘walk3‘]

1 查看元素在列表中的位置返回序列，有多个相同元素则只返回最前面一个
2 hobbies=[‘play‘,‘eat‘,‘sleep‘,‘study‘,‘eat‘,‘eat‘]
3 print(hobbies.index(‘eat‘))

按从小到大顺序排列
l=[100,9,-2,11,32]
l.sort()
print(l)

按从大到小顺序排列
l=[100,9,-2,11,32]
l.sort(reverse=True)
print(l)
[100, 32, 11, 9, -2]