3.21 DAY4

int会默认去除空格

1.APPEND

list = [1,'a','b',2,3,'老男孩']
list.append('alex')
print(list)

print(li.append('alex') return:none这个是动作,没有返回值

2.INSERT

list = [1,'a','b',2,3,'老男孩']
list.insert(2,'kate')
print(list)

3.EXTEND

li = [1,'a','b',2,3,'老男孩']
li.extend('abc')
print(li)
##HR添加新员工
name_list = ['kate','nero','jake'] while True: name = input('请添加员工姓名:按Q/q退出') if name.upper() == 'Q':break else: name_list.append(name) print('已成功添加新员工%s % name') print(name_list)

#POP唯一一个有返回值的,按索引删,只能删一个元素

li = ['kate','nero','jake',1,2,3]
li.pop()
print(li)
空格默认删除最后一个
li = ['kate','nero','jake',1,2,3]
li.pop(1)
print(li)
['kate', 'jake', 1, 2, 3]
li = ['kate','nero','jake',1,2,3]
print(li.pop(1))
返回值 nero

#remove按元素删,默认就删一个

li = ['kate','nero','jake',1,2,3]
li.remove('kate')
print(li)

# clear 清空内容

li = ['kate','nero','jake',1,2,3]
li.clear()
print(li)

return[]

# del 删除列表

li = ['kate','nero','jake',1,2,3]
del li[0:2]
print(li)

# 跳着删

li = ['kate','nero','jake',1,2,3]
del li[0:5:2]
print(li)

['nero', 1, 3]

1.按照索引改

li = ['kate','nero','jake',1,2,3]
li[1]= 'KATE'
print(li)

['kate', 'KATE', 'jake', 1, 2, 3]
print(li[1]) 只是打印元素 KATE

2.按切片去改

li = ['kate','nero','jake',1,2,3]
li[:3] = '0'
print(li)
li = ['kate','nero','jake',1,2,3]
li[:3] = 'nerosb'
print(li)

['n', 'e', 'r', 'o', 's', 'b', 1, 2, 3]
li = ['kate','nero','jake',1,2,3]
li[:3] = [11,22,33,44]
print(li)

把原来位置清空,再放入新的元素一个一个添加进去
[11, 22, 33, 44, 1, 2, 3]

查:索引,切片步长,查看

li = ['kate','nero','jake',1,2,3]
print(li[:3])
for i in li:
    print(i)

kate
nero
jake
1
2
3

#sort: 正向排序,从小到大

li = [2,7,9,0,2,3,4,5,5]
li.sort()
print(li)

[0, 2, 2, 3, 4, 5, 5, 7, 9]

#反向排序,从大到小

li = [2,7,9,0,2,3,4,5,5]
li.sort(reverse=True)
print(li)

[9, 7, 5, 5, 4, 3, 2, 2, 0]

# 翻转 reverse

li = [2,7,9,0,2,3,4,5,5]
li.reverse()
print(li)

[5, 5, 4, 3, 2, 0, 9, 7, 2]

# len 长度

li = [2,7,9,0,2,3,4,5,5]
print(len(li))

9

# count

li = [2,7,9,0,2,3,4,5,5]

print(li.count(2))

2

# index 通过元素找索引

li = [2,7,9,0,2,3,4,5,5]

print(li.index(2))

0

列表的嵌套

li = [1,2,'alex',['100','wusir',99],22]

li[2]=li[2].capitalize()
print(li)

li[3][1] = li[3][1].upper()
print(li)

li[3][2] = li[3][2] + 1
print(li)

1.将alex变成Alex

2.将wusir变成WUSIR

3.将99变成100

[1, 2, 'Alex', ['100', 'wusir', 99], 22]
[1, 2, 'Alex', ['100', 'WUSIR', 99], 22]
[1, 2, 'Alex', ['100', 'WUSIR', 100], 22]

#元组:不能增删改,只能查,通过切片,索引,for循环来查

tu = (1,2,'alex','oldboy')
print(tu[:2])

print(tu[2])

for i in tu:
    print(i)

#儿子不能改,孙子可能改
tul = (1,2,'alex',[1,'taibai'],'1,2,3','oldboy')
tul[3].append('日天')
print(tul)

元组中含列表,只能改列表

#RANGE:

1.当成数字列表range(1,10,2)范围加步长

for i in range(1,100,1):
    print(i)
打印1-100

2.还可以倒着取值


for i in range(100,0,-1):
print(i)

打印100-1倒叙

3.len

li = input('请输入内容')
for i in range(0,len(li)):
    print(i)

#Join:iterable 对可迭代的对象进行操作,字符串,元组,列表都可用,但是字符串用最方便

s = '*'.join('老男孩')
print(s)

老*男*孩

# split str转list

s = 'wusir alex taibai'
print(s.split())

['wusir', 'alex', 'taibai'] 字符串变列表

# join  list转str

s =['wusir', 'alex', 'taibai']
print(''.join(s))

wusiralextaibai 列表变字符串

思考题:

1.把每一个元素依次打印,两种方法

第一种:
li = [1,2,3,['alex','wusir','老男孩'],4] for i in li: if type(i) == list: for k in li[3]: print(k) else: print(i)

  第二种:

 
原文地址:https://www.cnblogs.com/kateli/p/8618875.html