#创建一个含有元素1，2，4，8，16，32的列表
#方法1
L = [1, 2, 4, 8, 16, 32]
#方法2
L = []
for x in range(0, 6):
  L.append(2**x)
#方法3
L = [2**x for x in range(0, 6)]

l = [1, 2, 3]
l.append('hello')  #得到 [1, 2, 3, 'hello']
l.append(['hello'])  #得到 [1, 2, 3, 'hello', ['hello']]

l1 = [1, 2, 3]
l1.insert(1,9)  #[1, 9, 2, 3]

#第一种
l1 = [1, 2, 3]
l3 = l1 + [4, 5, 6] #这种方式,l1不变，二者返回新的列表，当列表很长时，会消耗大量内存
#第二种
l1.extend([4, 5, 6]) #直接扩增l1
#相当于
l1 += [4,5,6]

l1 = [1, 2, 3, 4, 5, 6]
del l1[0]  #得到[2, 3, 4, 5, 6]
del l1[0:2] #得到[4, 5, 6]

l1 = [1,2,3,1,2]
l1.remove(1) #得到[2,3,1,2]

l1 = [1, 2, 3, 4, 5]
a = l1.pop(1)  #a=2
b = l1.pop()  #a=5

l1 = [1, 2, 3, 4]
l1[0] = 0  #[0,2,3,4]

l1= [1,2,3,4]
l1[0:2] = [7,8,9] #[7,8,9,3,4]

l1[:] = []  #清空了

l1 = [1,2,3,4,5]
l1[0] #1
l1[-1] #5，负数的索引从尾部开始计数,最后一个元素为-1

l1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
l1[0:2]  #[1, 2],  取区间[i,j) ，左闭右开
l1[:2]   #同上，可省略第一位
l1[2:]   #[3, 4, 5, 6, 7, 8, 9, 10, 11]
l1[2:-1]  #[3, 4, 5, 6, 7, 8, 9, 10]
l1[:]    #同l1，相当于复制一份

l1[::2]  #步长2，[1, 3, 5, 7, 9, 11]
l1[0:7:2] #[1, 3, 5, 7]
l1[7:0:-2] #[8, 6, 4, 2]  注意步长为负、理解起来相当于从7到1，倒序步长2

l1 = [5,3,2,1,4,6]
l1.sort()  #得到[1,2,3,4,5,6]  默认升序

> l1 = [(1,99),(3,97),(2,98),(4,96)]
> l1.sort(key=lambda x: x[1])
> l1
[(4, 96), (3, 97), (2, 98), (1, 99)]
> l1.sort(key=lambda x: x[1], reverse=True)
> l1
[(1, 99), (2, 98), (3, 97), (4, 96)]

sorted(l,key=str.lower,reverse=True)

l1.reverse() #l1反序

l1 = [1, 2, 3, 4]
1 in l1 #True
1 not in l1 #False

l1 = [1, 2, 3, 4]
l1.index(1)  #0
> l1.index(5)  #特别注意，当值不存在于列表，用index将抛ValueError

Traceback (most recent call last):
 File "<pyshell#44>", line 1, in <module>
  l1.index(5)
ValueError: 5 is not in list

l1 = [1, 2, 3, 4, 1]
l1.count(1)  #2

l1 = [1, 2, 3, 4, 5]
for i in l1:
   print (i)

l1 = [1, 2, 3, 4, 5]
for index,value in enumerate(l1):
   print (index,value)

len(l)  #列表长度
l*3 重复
l1 = [1, 2]
l1*3  #[1,2,1,2,1,2]

l1 = []
l1[:] = []
del l1[:]

l2 = l1[:]