python基础三

join

s = 'abcdefghigk'
s1 = '-'.join(s)
print(s1) 
#结果：a-b-c-d-e-f-g-h-i-g-k

list操作

列表里可以存储所有数据类型  int.str.boor.[].().{}

切片，索引与字符串一样（不改变原数据类型）

增：①append 在最后加      

list = ["a","b","c","d","e","f","g",1,2,3]
list.append("qqq")
print(list)
#在后面加，结果为：['a', 'b', 'c', 'd', 'e', 'f', 'g', 1, 2, 3, 'qqq']

     ②insert 插入

list = ["a","b","c","d","e","f","g",1,2,3]
list.insert(3,'X')
print(list)
#3：索引 
#结果：['a', 'b', 'c', 'X', 'd', 'e', 'f', 'g', 1, 2, 3]

    ③extend 迭代插入

list = ["a","b","c","d","e","f","g",1,2,3]
list.extend('我是你爸爸')
list.extend([1,2,33,4])
print(list)
#结果：['a', 'b', 'c', 'd', 'e', 'f', 'g', 1, 2, 3, '我', '是', '你', '爸', '爸', 1, 2, 33, 4]
#把所有内容拆开，挨个插入

删除

①pop 按索引删除，有返回值

list = ["a","b","c","d","e","f","g",1,2,3]
list.pop(2)
ret = list.pop(2)
print(list)
print(ret)
#结果：['a', 'b', 'd', 'e', 'f', 'g', 1, 2, 3]
#   c

②remove 按元素删除

list = ["a","b","c","d","e","f","g",1,2,3]
list.remove('b')
print(list)
#结果：['a', 'c', 'd', 'e', 'f', 'g', 1, 2, 3]

③clear清空列表

list = ["a","b","c","d","e","f","g",1,2,3]
list.clear()
print(list)
#清空所有数据，但保留[]
#输出：[]

④del 删除列表

#功能一：删除列表
list = ["a","b","c","d","e","f","g",1,2,3]
del list
print(list)

功能二：按照索引，切片去删除：
list = ["a","b","c","d","e","f","g",1,2,3]
del list[0]       #删除第一个字母
print(list)
del list[0:3]     #删除前四个
print(list)
del list[1:6:2]    #2为步长
print(list)
#结果：['a', 'c', 'e', 'g', 1, 2, 3]

改

#按照索引改
list = ["a","b","c","d","e","f","g",1,2,3]
list[0]="aaa"
print(list)
#结果：['aaa', 'b', 'c', 'd', 'e', 'f', 'g', 1, 2, 3]

#按照切片改  （按照迭代）
li = ["a","b","c","d"]
li[2:3] = "python"
print(li)
#结果：['a', 'b', 'p', 'y', 't', 'h', 'o', 'n', 'd']

查

l1 = ["a","b","c","d"]
print(l1[1])
print(l1[1:3])
#结果：b
           ['b', 'c']

#循环查
for i in l1:
    print(i)

公共方法

len:长度
l1 = ["a","b","c","d"]
print(len(l1))
#输出：4
count：元素出现次数
l1 = ["a","a","a","b","c","d"]
print(l1.count("a"))
#输出：3
index：通过元素找索引，可以切片
l1 = ["a","b","c","d"]
print(l1.index('d'))
#输出3

sort排序，从小到大:

l1 = [1,3,6,9,8,4,7,2]
l1.sort()
print(l1)
#输出：1，2，3，4，5，6，7，8，9

从大到小：

l1 = [1,3,6,9,8,4,7,2]
l1.sort(reverse = True)
print(l1)
#输出：9，8，7，6，5，4，3，2，1

倒叙：从后往前打印

l1 = [1,3,6,9,8,4,7,2]
l1.reverse()
print(l1)
#结果：[2, 7, 4, 8, 9, 6, 3, 1]

列表嵌套：

l1 = ['a','b','c',['d','e'],1,2,3]
l1[3][0] ="X"  #选中的是‘d'
print(l1)
#输出：['a', 'b', 'c', ['X', 'e'], 1, 2, 3]

元组：只能读不能改

如果元组里面有列表，列表可以改

l1 = ('a','b','c',['d','e'],1,2,3)
l1[3][0] = "X"
print(l1)
#输出：('a', 'b', 'c', ['X', 'e'], 1, 2, 3)

range 范围：

里面的元素是数字，且可控，一般与for配合使用，有步长

for i in range(1,10):
    print(i)
    #输出 1，2.....8,9
for i in range(11):
    print(i)
    #输出 0，1，2....10
for i in range(1,10,2):  #2是步长
    print(i)
    #输出： [1,3,5,7,9]
for i in range(10, 0, -2):
    print(i)
    #输出： 10，8，6，4，2
    #从后往前 步长为2

把列表里的列表拆开打印：

l1 = [1,2,3,'abc',[2,3,'qwe'],7]
for i in l1:
    if type(i) == list:
        for j in i:
            print(j)
    else:
        print(i)
#输出：     1
           2
           3
           abc
           2
           3
           qwe
           7