python 列表(list)增删改查及方法

                                 强大自己是唯一获得幸福的途径，这是长远的，而非当下的玩乐！

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增

append(self, p_object) 在最后追加

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.append("noway")
>>> print(l1)
['zwz', 'lp', 'problem', 'has', 'maybe', 'noway']

insert(self, index, p_object) 在指定位置插入

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.insert(2, "abc")
>>> print(l1)
['zwz', 'lp', 'abc', 'problem', 'has', 'maybe', 'noway']

extend(self, iterable) 迭代追加

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.extend("some4")   # 此处迭代追加的是一个字符串，字符串会被分解
>>> print(l1)
['zwz', 'lp', 'abc', 'problem', 'has', 'maybe', 'noway', 's', 'o', 'm', 'e', '4']
>>> l2 = ["some1","some2","some3"]
>>> l1.extend(l2)        # 追加一个列表则迭代追加列表里的值
>>> print(l1)
['zwz', 'lp', 'abc', 'problem', 'has', 'maybe', 'noway', 's', 'o', 'm', 'e', '4', 'some1', 'some2', 'some3']

删

pop(self, index=None) 按照指定索引删除

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.pop()     # pop会有返回值，返回值为被删除的值
'maybe'
>>> print(l1)
['zwz', 'lp', 'problem', 'has']

remove(self, value)  按照指定内容删除

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.remove("problem")
>>> print(l1)
['zwz', 'lp', 'has', 'maybe']

clear(清空列表)

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.clear()
>>> print(l1)
[]

del(按照索引删除)

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> del(l1[2])        # 可切片 如：删除从第2到最后的索引   del(l1[2:])
>>> print(l1)
['zwz', 'lp', 'has', 'maybe']

改

按索引

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1[2] = "haha"            # 此处也可以按切片批量修改，但值必须一一对应
>>> print(l1)
['zwz', 'lp', 'haha', 'has', 'maybe']

查

按索引

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> print(l1[2])                # 正常索引
problem
>>> print(l1[:3])                #切片
['zwz', 'lp', 'problem']
>>> print(l1[::2])                #加步长
['zwz', 'problem', 'maybe']
>>> for i in l1:                 #循环
...     print(i)
...
zwz
lp
problem
has
maybe

list其他方法

append

           上面已经说明

clear

           上面已经说明

copy

           请参见深浅copy --> https://blog.csdn.net/z_bright/article/details/84648996

count(统计元素出现次数)

>>> l1 = ["zwz", "lp", "problem", "has", "maybe", "zwz"]
>>> print(l1.count("lp"))
1
>>> print(l1.count("zwz"))
2

extend

            上面已经说明

index(self, value, start=None, stop=None) 根据值查找索引

>>> l1 = ["zwz", "lp", "problem", "has", "maybe", "zwz"]
>>> print(l1.index("zwz"))    # 找到即返回
0
>>> print(l1.index("has"))
3

insert

         上面已经说明

pop

         上面已经说明

remove

        上面已经说明

reverse(翻转列表)

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.reverse()
>>> print(l1)
['maybe', 'has', 'problem', 'lp', 'zwz']

sort(self, key=None, reverse=False) 排序

>>> l1 = ["zwz", "lp", "problem", "has", "maybe"]
>>> l1.extend("cabde")              # 追加元素
>>> print(l1)
['zwz', 'lp', 'problem', 'has', 'maybe', 'c', 'a', 'b', 'd', 'e']
>>> l1.sort()                       # 正向排序
>>> print(l1)
['a', 'b', 'c', 'd', 'e', 'has', 'lp', 'maybe', 'problem', 'zwz']
>>> l1.sort(reverse=True)           # 反向排序
>>> print(l1)
['zwz', 'problem', 'maybe', 'lp', 'has', 'e', 'd', 'c', 'b', 'a']