零基础入门学习Python（12）--列表：一个打了激素的数组(3)

前言

知识点

Python常用操作符

>>> list1 = [123]
>>> list2 = [234]
>>> list1 > list2
False

>>> list1 = [1]
>>> list2 = ['b']
>>> list1 > list2
Traceback (most recent call last):
File "<pyshell#9>", line 1, in <module>
list1 > list2
TypeError: '>' not supported between instances of 'int' and 'str'
>>> list2 = [3,'b']
>>> list1 > list2
False

我们发现列表还会比较大小，那如果有两个元素的列表，或者多个元素的列表，怎么比较呢？
>>> list1 = [123,456]
>>> list2 = [234,123]
>>>
>>> list1 > list2
False
>>>
当有多个元素时，默认是从索引位置0元素开始比较，不用考虑后面的元素

字符串比较就是ASCII码大小
>>> list1 = ['a']
>>> list2 = ['b']
>>> list1 > list2
False
>>> list3 = ['B']
>>> list2 > list3
True
>>> list1 > list2
False
>>> list1 > list3
True
>>> list4 = ['A']
>>> list4 > list3
False

>>> list1 = [123,456]
>>> list2 = [234,123]
>>> list3 = [123,456]
>>> (list1 < list2) and (list1 == list2)
False
>>> (list1 < list2) and (list1 == list3)
True
>>>

>>> list4 = list1 + list2
>>> list4
[123, 456, 234, 123]

>>> list1 + '小甲鱼'
Traceback (most recent call last):
File "<pyshell#9>", line 1, in <module>
list1 + '小甲鱼'
TypeError: can only concatenate list (not "str") to list

+ 号两边 数据类型必须一致。

>>> list3
[123, 456]
>>> list3 * 3
[123, 456, 123, 456, 123, 456]
>>> list3 *= 3
>>> list3
[123, 456, 123, 456, 123, 456]
>>> list3 *= 5
>>> list3
[123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456]

>>> 123 in list3
True
>>> '小甲鱼' not in list3
True
>>> 123 not in list3
False

>>> list5 = [123,['小甲鱼','牡丹'],456]
>>>
>>>
>>> list5
[123, ['小甲鱼', '牡丹'], 456]
>>> '小甲鱼' in list5
False
>>> '小甲鱼' in list5[1]
True
>>> list5[1][1]
'牡丹'

列表的一些其他方法

>>> dir(list)
[...,'append', 'clear', 'copy', 'count', 'extend', 'index', 'insert', 'pop', 'remove', 'reverse', 'sort']

count

>>> list3
[123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456]

>>> list3.count(123)
15

index

>>> a = [1,2,3,4,5,6,7,'a','v',1,3]
>>> a.index(1)
0
>>> a.index(1,5,10)
9

reverse

>>> list3.reverse()
>>> list3
[456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123, 456, 123]

sort

>>> list3.sort()
>>> list3
[123, 123, 123, 123, 123, 123, 123, 123, 123, 123, 123, 123, 123, 123, 123, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456]

>>> list6 = [4,2,5,1,9,23,32,0]
>>> list6.sort()
>>> list6
[0, 1, 2, 4, 5, 9, 23, 32]
>>>  list6 = [4,2,5,1,9,23,32,0]
SyntaxError: unexpected indent
>>> list6 = [4,2,5,1,9,23,32,0]
>>> (list6.sort()).reverse()
Traceback (most recent call last):
File "<pyshell#54>", line 1, in <module>
(list6.sort()).reverse()
AttributeError: 'NoneType' object has no attribute 'reverse'

>>> list6 = list6.sort()
>>> list6
>>> list6 = [4,2,5,1,9,23,32,0]

>>> list6.sort()
>>> list6
[0, 1, 2, 4, 5, 9, 23, 32]
>>> list6.reverse()
>>> list6
[32, 23, 9, 5, 4, 2, 1, 0]

>>> list6 = [4,2,5,1,9,23,32,0]
>>>
>>>
>>> list6.sort(reverse=True)
>>> list6
[32, 23, 9, 5, 4, 2, 1, 0]

课后作业

测试题

>>> old = [1, 2, 3, 4, 5]
>>> new = old
>>> old = [6]
>>> print(new)

结果：
[1,2,3,4,5]

小甲鱼

小鱿鱼

list1 = [1, [1, 2, ['小甲鱼']], 3, 5, 8, 13, 18]

方法：
>>> list1[1][2][0] = '小鱿鱼'
>>> list1
[1, [1, 2, ['小鱿鱼']], 3, 5, 8, 13, 18]

['小甲鱼','小鱿鱼']

>>> list1[1][2] = ['小甲鱼','小鱿鱼']
>>> list1
[1, [1, 2, ['小甲鱼', '小鱿鱼']], 3, 5, 8, 13, 18]
>>>

列表名.sort()

列表名.sort(reverse=True)

列表名.sort()

列表名.reverse()

copy()

clear()

copy()方法跟使用切片拷贝是一样的

>>> list1 = [1,5,89,2,6,21,6,5,4]
>>> list2 = list1.copy()
>>> list2
[1, 5, 89, 2, 6, 21, 6, 5, 4]
>>>
>>> list1.sort()

>>> list1
[1, 2, 4, 5, 5, 6, 6, 21, 89]
>>> list2
[1, 5, 89, 2, 6, 21, 6, 5, 4]
>>>

clear()方法用于清空列表的元素，但要注意，清空后列表仍然存在，只是变成一个空列表。
>>> list1.clear()
>>> list1
[]
>>> list2
[1, 5, 89, 2, 6, 21, 6, 5, 4]
>>> list2.clear(1)
Traceback (most recent call last):
File "<pyshell#15>", line 1, in <module>
list2.clear(1)
TypeError: clear() takes no arguments (1 given)

>>> [ i*i for i in range(10) ]

输出内容：
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
打印了0到9各个数的平方
列表推导式（List comprehensions）也叫列表解析，灵感取自函数式编程语言Haskell。它是一个非常有用和灵活的工具，可以用来动态的创建列表，语法如:

[有关A的表达式for A in B]

例如：
>>> list1 = [x**2 for x in range(10)]
>>> list1
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

相当于
list1 = []
for x in range(10):
list1.append(x**2)

>>> list1 = [(x, y) for x in range(10) for y in range(10) if x%2==0 if y%2!=0]

输出结果：

>>> list1
[(0, 1), (0, 3), (0, 5), (0, 7), (0, 9), (2, 1), (2, 3), (2, 5), (2, 7), (2, 9), (4, 1), (4, 3), (4, 5), (4, 7), (4, 9), (6, 1), (6, 3), (6, 5), (6, 7), (6, 9), (8, 1), (8, 3), (8, 5), (8, 7), (8, 9)]
>>>

list1= []
for x in range(10):
for y in range(10):
if x%2 ==0 and y%2 !=0:
list1.append((x,y))
print(list1)
================== RESTART: C:/Users/ThinkPad/Desktop/5.py ==================
[(0, 1), (0, 3), (0, 5), (0, 7), (0, 9), (2, 1), (2, 3), (2, 5), (2, 7), (2, 9), (4, 1), (4, 3), (4, 5), (4, 7), (4, 9), (6, 1), (6, 3), (6, 5), (6, 7), (6, 9), (8, 1), (8, 3), (8, 5), (8, 7), (8, 9)]

list1 = ['1.just do it','2.一切皆有可能','3.让编程改变世界','4.Impossible is Nothing']
list2 = ['4.阿迪达斯','2.李宁','3.鱼c工作室','1.耐克']
list3 = [name + '：' + slogan[2:] for slogan in list1 for name in list2 if slogan[0] == name[0]]

>>> list3
['1.耐克：just do it', '2.李宁：一切皆有可能', '3.鱼c工作室：让编程改变世界', '4.阿迪达斯：Impossible is Nothing']

>>> for each in list3:
print(each)

1.耐克：just do it
2.李宁：一切皆有可能
3.鱼c工作室：让编程改变世界
4.阿迪达斯：Impossible is Nothing

list1 = ['1.just do it','2.一切皆有可能','3.让编程改变世界','4.Impossible is Nothing']
list2 = ['4.阿迪达斯','2.李宁','3.鱼c工作室','1.耐克']
list3 = []
for slogan in list1:
for name in list2:
if slogan[0] == name[0]:
list3.append((name + ':'+ slogan[2:]))
for each in list3:
print(each)

================== RESTART: C:/Users/ThinkPad/Desktop/5.py ==================
1.耐克:just do it
2.李宁:一切皆有可能
3.鱼c工作室:让编程改变世界
4.阿迪达斯:Impossible is Nothing
>>>