day05_09 列表内置方法

1.0 count(计算元素出现的次数)

t = ['to','be','or','not','to','be'].count('to')
print(t)
#>>>2

　　

2.0 extend

a = [1,2,3]
b = [4,5,6]
a.extend(b)
print(a)
print(b)
#>>>[1, 2, 3, 4, 5, 6]
#>>>[4, 5, 6]

 

3.0 index(查找位置,只能索引一个)

a = ['wuchao','jinxin','xiaohu','sanpang','ligang']
print(a.index('jinxin'))
#>>>1

 

 

4.0 reverse(倒序排列)

a = ['wuchao','jinxin','ligang','xiaohu','sanpang','liqang']
a.reverse()
print(a)
#>>>['liqang', 'sanpang', 'xiaohu', 'ligang', 'jinxin', 'wuchao']

　　

5.0 sorted(正序排序)

x = [4,5,2,1,9,8]
x.sort()
print(x)
#>>>[1, 2, 4, 5, 8, 9]

5.1 sorted(倒序排列)

x = [4,9,1,3,7,9,12,78]
x.sort(reverse=True)
print(x)
#>>>[78, 12, 9, 9, 7, 4, 3, 1]

5.1 字符串排序(按照assic码排序)　　

a = ['wuchao','jinxin','ligang','xiaohu','sanpang']
a.sort()
print(a)
#>>>['jinxin', 'ligang', 'sanpang', 'wuchao', 'xiaohu']

　　

6.0 补充:判断haidilaoge是否在列表中　

a = ['wuchao','jinxin','ligang','xiaohu','sanpang','liqang']
print(a.count("haidilaoge"))
#>>>0

6.1 方法二

a = ['wuchao','jinxin','ligang','xiaohu','sanpang','liqang']
print("haidilao ge" in a)
#>>>False

 

7.0 判断身份

a = ['wuchao','jinxin','ligang','xiaohu','sanpang','liqang']
print(type(a) is list)
#>>>True

 

转载于:https://www.cnblogs.com/darkalex001/p/7495144.html