• Python基础学习-列表的常用方法


    列表方法

    =

    Python 3.5.2 (default, Sep 14 2016, 11:27:58) 
    [GCC 6.2.1 20160901 (Red Hat 6.2.1-1)] on linux
    Type "help", "copyright", "credits" or "license" for more information.
    >>> L=['good','boy','!']
    >>> L
    ['good', 'boy', '!']
    >>> L.
    L.__add__(           L.__format__(        L.__init__(          L.__reduce__(        L.__str__(           L.insert(
    L.__class__(         L.__ge__(            L.__iter__(          L.__reduce_ex__(     L.__subclasshook__(  L.pop(
    L.__contains__(      L.__getattribute__(  L.__le__(            L.__repr__(          L.append(            L.remove(
    L.__delattr__(       L.__getitem__(       L.__len__(           L.__reversed__(      L.clear(             L.reverse(
    L.__delitem__(       L.__gt__(            L.__lt__(            L.__rmul__(          L.copy(              L.sort(
    L.__dir__(           L.__hash__           L.__mul__(           L.__setattr__(       L.count(             
    L.__doc__            L.__iadd__(          L.__ne__(            L.__setitem__(       L.extend(            
    L.__eq__(            L.__imul__(          L.__new__(           L.__sizeof__(        L.index(             

    1、append

       append方法用于在列表末尾追加新的对象:

    >>> L=['index0','index1','index2']
    >>> L
    ['index0', 'index1', 'index2']
    >>> L.append('index3')
    >>> L
    ['index0', 'index1', 'index2', 'index3']
    >>> L.append([1,2,3])
    >>> L
    ['index0', 'index1', 'index2', 'index3', [1, 2, 3]]
    >>> L.append(('Hello','World','!'))
    >>> L
    ['index0', 'index1', 'index2', 'index3', [1, 2, 3], ('Hello', 'World', '!')]
    >>> L.append('index5')
    >>> L
    ['index0', 'index1', 'index2', 'index3', [1, 2, 3], ('Hello', 'World', '!'), 'index5']
    >>> len(L)
    7
    >>> L.append()
    Traceback (most recent call last):
      File "<pyshell#11>", line 1, in <module>
        L.append()
    TypeError: append() takes exactly one argument (0 given)
    >>> L.append(None)
    >>> L
    ['index0', 'index1', 'index2', 'index3', [1, 2, 3], ('Hello', 'World', '!'), 'index5', None]
    >>> L.append([])
    >>> L
    ['index0', 'index1', 'index2', 'index3', [1, 2, 3], ('Hello', 'World', '!'), 'index5', None, []]
    >>> id(L)
    19215544
    >>> L.append(1231)
    >>> L
    ['index0', 'index1', 'index2', 'index3', [1, 2, 3], ('Hello', 'World', '!'), 'index5', None, [], 1231]
    >>> id(L)
    19215544
    >>> 

    2、count

      count方法统计某个元素在列表中出现的次数:

    >>> help(L.count)
    Help on built-in function count:
    
    count(...)
        L.count(value) -> integer -- return number of occurrences of value
    
    >>> L=['to','be','or','not','to','be']
    >>> L
    ['to', 'be', 'or', 'not', 'to', 'be']
    >>> L.count('to')
    2
    >>> x=[[1,2],1,1,[1,2,1,[1,2]]]
    >>> x.count(1)
    2>>> x.count([1,2])
    1
    >>> 

    3、extend  

      官方说明:

    >>> help(L.extend)
    Help on built-in function extend:
    
    extend(...)
        L.extend(iterable) -> None -- extend list by appending elements from the iterable

       extend方法可以在列表的末尾一次性追加别一个序列中的多个值;换句话说,可以用新列表扩展原有的列表:

    >>> a=[1,2,3,4]
    >>> id(a)
    20761256
    >>> b=[5,6,7,8,9]
    >>> id(b)
    20760536
    >>> c=a.extend(b)
    >>> c
    >>> print(c)
    None
    >>> a.extend(b)
    >>> a
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9]
    >>> b
    [5, 6, 7, 8, 9]
    
    >>> id(a)
    20761256
    >>> b.extend(a)
    >>> b
    [5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9]
    >>> 
    >>> a
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9]
    >>> b
    [5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9]
    >>> a.extend(b)
    >>> a
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9]

     4、index

      index方法用于从列表中找出某个值第一个匹配项的索引位置:

       举例:

    >>> s='He who commences many things finishes but a few'
    >>> list1=s.split()
    >>> list1
    ['He', 'who', 'commences', 'many', 'things', 'finishes', 'but', 'a', 'few']
    >>> list1.index('who')
    1
    >>> list1.index('Few')
    Traceback (most recent call last):
      File "<pyshell#16>", line 1, in <module>
        list1.index('Few')
    ValueError: 'Few' is not in list
    >>> list1.index('who',2)
    Traceback (most recent call last):
      File "<pyshell#17>", line 1, in <module>
        list1.index('who',2)
    ValueError: 'who' is not in list
    >>> list1.index('but',2)
    6
    >>> list1.index('things',4,5)
    4
    >>> list1.index('things',5,len(list1))
    Traceback (most recent call last):
      File "<pyshell#20>", line 1, in <module>
        list1.index('things',5,len(list1))
    ValueError: 'things' is not in list
    >>> list1.index('things',3,len(list1))
    4
    >>> list1.index('a',-2,-1)
    7
    >>> len(list1)
    9
    >>> 

    5、insert

      insert方法用于将对象插入到列表中;

    >>> list1
    ['He', 'commences', 'many', 'things', 'finishes', 'but', 'a', 'few']
    >>> list1.insert(1,'who')
    >>> list1
    ['He', 'who', 'commences', 'many', 'things', 'finishes', 'but', 'a', 'few']
    >>> list1.insert(0,'who')
    >>> list1
    ['who', 'He', 'who', 'commences', 'many', 'things', 'finishes', 'but', 'a', 'few']
    >>> 

    6、pop

      pop方法会移除列表中的一个元素(默认是最后一个),并且返回该元素的值:

    >>> x=[1,2,3,4]
    >>> x.pop()
    4
    >>> x
    [1, 2, 3]
    >>> x.pop(0)
    1
    >>> x
    [2, 3]
    >>> x.pop(1)
    3
    >>> x
    [2]
    >>> help(x.pop)
    Help on built-in function pop:
    
    pop(...)
        L.pop([index]) -> item -- remove and return item at index (default last).
        Raises IndexError if list is empty or index is out of range.
    
    >>> 

    注意: pop方法是唯一一个既能修改列表又返回元素(除了None)的列表方法。

    7、remove

      remove方法用于移除列表中某个值的第一个匹配项:

    >>> x
    ['to', 'be', 'or', 'not', 'to', 'be']
    >>> x.remove('be')
    >>> x
    ['to', 'or', 'not', 'to', 'be']
    >>> print(x.remove('bee'))
    Traceback (most recent call last):
      File "<pyshell#65>", line 1, in <module>
        print(x.remove('bee'))
    ValueError: list.remove(x): x not in list
    >>> print(x.remove('to'))
    None
    >>> x
    ['or', 'not', 'to', 'be']
    >>> x.insert(0,'to')
    >>> x
    ['to', 'or', 'not', 'to', 'be']
    >>> x.remove()
    Traceback (most recent call last):
      File "<pyshell#70>", line 1, in <module>
        x.remove()
    TypeError: remove() takes exactly one argument (0 given)
    >>> 

    8、reverse

      reverse方法将列表中的元素反向存放;

    >>> help(list.reverse)
    Help on method_descriptor:
    
    reverse(...)
        L.reverse() -- reverse *IN PLACE*
    
    >>> x=[1,2,3,4]
    >>> x
    [1, 2, 3, 4]
    >>> x.reverse()
    >>> x
    [4, 3, 2, 1]
    >>> x.reverse()
    >>> x
    [1, 2, 3, 4]
    >>> 

     9、sort

    sort方法用于在原位置对列表进行排序,在‘原位置排序’意味着改变原来的列表,从而让其中的元素能按一定的顺序排列,而不是简单地返回一个已经排序

     的列表的副本;

    >>> x=[2,55,123,89,-2,23]
    >>> x.sort()
    >>> x
    [-2, 2, 23, 55, 89, 123]
    >>> x.sort()
    >>> x
    [-2, 2, 23, 55, 89, 123]

     sort方法修改原列表并返回了空值(None):

    示例:

    >>> help(list.sort)
    Help on method_descriptor:
    
    sort(...)
        L.sort(key=None, reverse=False) -> None -- stable sort *IN PLACE*
    
    >>> x=[23,44,1,5,-100]
    >>> y=x.sort()  #千万不要这样做!
    >>> x
    [-100, 1, 5, 23, 44]
    >>> y
    >>> print(y)
    None
    >>> 

    那么当用户需要一个排好序的列表副本,同时又保留原来列表不变的时候,问题就出现了,为了实现这个功能的正解方法是,首先把x的副本赋值给y,然后对y进行排序,

    >>> x=[23,12,9,4,1231]
    >>> x
    [23, 12, 9, 4, 1231]
    >>> y=x[:]
    >>> y
    [23, 12, 9, 4, 1231]
    >>> id(x)
    20712184
    >>> id(y)
    20711584
    >>> z=x
    >>> id(z)
    20712184
    >>> y.sort()
    >>> x
    [23, 12, 9, 4, 1231]
    >>> y
    [4, 9, 12, 23, 1231]
    >>> 
    

      再次调用x[:]得到的是包含了x所有元素的分片,这是一种很有效率的复制整个列表的方法

    假如只是简单把x赋值给y是没用的,因为这样做让x和y都指向同一个列表了。

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  • 原文地址:https://www.cnblogs.com/me80/p/6928131.html
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