python全栈开发从入门到放弃之列表的内置方法

1、列表切片

l=['a','b','c','d','e','f']
print(l[1:5]) # 根据索引号来切片，但顾头不顾尾
['b', 'c', 'd', 'e']
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
['b', 'd']
print(l[2:5] )
['c', 'd', 'e']
print(l[-1]) #-1代表从后往前取 -1代表最后一个
f


l=['a','b','c','d','e','f']
print(l[-4:-1]) #从后往前取
['c', 'd', 'e'] 
print(l[2:5]) #
['c', 'd', 'e']
print(l[-1]) 
f
print([l[-4:]]) # -4 从后往前取倒数4个
[['c', 'd', 'e', 'f']]

 

2、append #追加

hobbies=['play','eat','sleep','study']
hobbies.append('girls')
print(hobbies)

 

3、pop 删除

根据索引号来删除
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
hobbies=['play','eat','sleep','study']
x=hobbies.pop(1)
print(hobbies) 
['play', 'sleep', 'study']
print(x)
eat

 

4、#队列：先进先出

入队
queue_l.append('first')
queue_l.append('second')
queue_l.append('third')
print(queue_l)
出队
print(queue_l.pop(0))
print(queue_l.pop(0))
print(queue_l.pop(0))
print(queue_l.pop(0))
print(queue_l.pop(0))
print(queue_l.pop(0))

5、del、remove删除

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

6、len 长度

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

 

7、in 包含

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

 

8、insert     #英文意思 插入嵌入

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)

9、count      #英文意思 计算

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'))

10、extend      #英文意思 延伸

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

11、index          #指标，索引

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
hobbies=['play','eat','sleep','study','eat','eat']
print(hobbies.index('eat'))

12、sort        #英文意思 种类

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]