1.extend和列表相加的区别
a = [1, 2, 3]
b = [4, 5, 6]
print(a + b)
[1, 2, 3, 4, 5, 6]
2.查找列表中某个元素第一次出现的索引，从0开始
a = ['how', 'are', 'you']
print(a.index('you'))
2
3.将一个对象插入到列表中
num = [1, 2, 4, 5]
num.insert(2, 'three')
print(num)
[1, 2, 'three', 4, 5]
使用切片方式
num = [1, 2, 4, 5]
num[2:2] = ['three']
print(num)
[1, 2, 'three', 4, 5]
4.删除列表中的元素
num = [1, 2, 3, 4, 5]
num.pop()
print(num)
num.pop(1)
print(num)
[1, 2, 3, 4]
[1, 3, 4]
5.删除列表中指定的元素
num = [1, 2, 3, 4, 5]
num.remove(4)
print(num)
[1, 2, 3, 5]
//remove只会删除第一次出现的元素
num = [1, 2, 3, 4, 5, 4]
num.remove(4)
print(num)
[1, 2, 3, 5, 4]
6.让列表按相反的顺序排列
num = [1, 10, 20, 30, 40, 50]
num.reverse()
print(num)
[50, 40, 30, 20, 10, 1]
num = [1, 10, 20, 30, 40, 50]
print(num[::-1])
[50, 40, 30, 20, 10, 1]
7.表示只包含一个元素的元组
n = (1)
print(type(n))
<class 'int'>
n = (1,)
print(type(n))
<class 'tuple'>
8.批量替换字符串中的元素
a = 'how are you'
print(a.replace('y', 'we'))
how are weou
9.把字符串按照空格进行拆分
a = 'how are you'
print(a.split())
['how', 'are', 'you']
10.取出字符串首位的空格
a = ' how are you'
print(a.strip())
how are you
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