day3之笔记

Posted 2020-07-09

list方法：

append添加，

以下分别插入了不同的元素，从屁股的位置开始插入：

name=[‘alex‘,‘eric‘,‘tony‘]
name.append(‘Hanmeimie‘)
nam_2=[‘Jim‘,‘Tom‘]
name.append(nam_2)
name.append(1)
name_dic={‘k1‘:‘Lilei‘}
name.append(name_dic)
for i in range(10):
    name.append(i)
    name

 

index 获取某个值的索引：

前面三个都会返回正确的index，后面三个都会pop up error，注意最小单位，顺序及大小写，都是严格匹配的容不得半点儿马虎；

另外如果有超过两个相同的最小单位，index只会返回最左边的第一个

 

name=[‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
name.index(‘alex‘)
name.index(1)
name.index([‘Jim‘,‘Tom‘])


name.index(‘Tom‘,‘JIm‘)
name.index(‘Tom‘,‘Jim‘)

name.index(‘Tom‘)

>>> name
[‘Alex‘, ‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, ‘T_bag‘]
>>> name.append(‘alex‘)
>>> name.index(‘alex‘)
1

 

 

 

insert插入索引值+veriable：

都是在索引值的前面插入元素

 

name=[‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
name.insert(0,‘Alex‘)
#result
[‘Alex‘, ‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

name.insert((len(name)-1), ‘T_bag‘)
# result    
‘Alex‘, ‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, ‘T_bag‘, 9]

 

 

 

pop，切掉尾部的那个，并赋值给另外一个，remove+ assign；

[‘Alex‘, ‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, ‘T_bag‘, 9]
>>> adbd=name.pop()
>>> adbd
9
>>> name
[‘Alex‘, ‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, ‘T_bag‘]
>>> 

 

remove：name_list.remove("seven"),移除从左边找到的第一个，重复的

 

>>> name
[‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> name.insert(0,‘alex‘)
>>> name
[‘alex‘, ‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> name.remove(‘alex‘)
>>> name
[‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

 

reverse（）：反向

 

>>> name
[‘alex‘, ‘eric‘, ‘tony‘, [‘Jim‘, ‘Tom‘], 1, ‘Hanmeimie‘, {‘k1‘: ‘Lilei‘}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> name.reverse()
>>> name
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0, {‘k1‘: ‘Lilei‘}, ‘Hanmeimie‘, 1, [‘Jim‘, ‘Tom‘], ‘tony‘, ‘eric‘, ‘alex‘]
>>> 

 

 

 

sort:排序 str + int + 汉字，会pop error，后面需要用算法来cover；

 

>>> name
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0, {‘k1‘: ‘Lilei‘}, ‘Hanmeimie‘, 1, [‘Jim‘, ‘Tom‘], ‘tony‘, ‘eric‘, ‘alex‘]
>>> name.sort()
>>> name
[0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, {‘k1‘: ‘Lilei‘}, [‘Jim‘, ‘Tom‘], ‘Hanmeimie‘, ‘alex‘, ‘eric‘, ‘tony‘]

 

acount 出现的次数

 

[0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, {‘k1‘: ‘Lilei‘}, [‘Jim‘, ‘Tom‘], ‘Hanmeimie‘, ‘alex‘, ‘eric‘, ‘tony‘]
>>> name.count(‘alex‘)
1
>>> name.index(‘alex‘)
14
>>> name.insert(14,‘alex‘)
>>> name.count(‘alex‘)
2

 

 

 

del name_list[1]，删除第二个，删除指定所以位置，或切片位置所对应的值

 

>>> name
[0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, {‘k1‘: ‘Lilei‘}, [‘Jim‘, ‘Tom‘], ‘Hanmeimie‘, ‘alex‘, ‘alex‘, ‘eric‘, ‘tony‘]
>>> name.index(‘Hanmeimie‘)
13
>>> del name[13]
>>> name.index(‘Hanmeimie‘)

Traceback (most recent call last):
  File "<pyshell#107>", line 1, in <module>
    name.index(‘Hanmeimie‘)
ValueError: ‘Hanmeimie‘ is not in list
>>> name
[0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, {‘k1‘: ‘Lilei‘}, [‘Jim‘, ‘Tom‘], ‘alex‘, ‘alex‘, ‘eric‘, ‘tony‘]

 

元组和列表几乎是一样的，列表是可以修改的元祖是不能修改的；

name_tumple=(‘alex‘,‘eric‘,‘tony‘)