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