Index:
      ------------>
  0   1   2   3   4
+---+---+---+---+---+
| a | b | c | d | e |
+---+---+---+---+---+
  0  -4  -3  -2  -1
      <------------

Slice:
    <---------------|
|--------------->
:   1   2   3   4   :
+---+---+---+---+---+
| a | b | c | d | e |
+---+---+---+---+---+
:  -4  -3  -2  -1   :
|--------------->
    <---------------|

我希望这将帮助您用Python建模列表。

参考：http://wiki.python.org/moin/MovingToPythonFromOtherLanguages

还可以使用切片分配从列表中删除一个或多个元素：

r = [1, 'blah', 9, 8, 2, 3, 4]
>>> r[1:4] = []
>>> r
[1, 2, 3, 4]

在Python 2.7中

Python中的切片

[a:b:c]

len = length of string, tuple or list

c -- default is +1. The sign of c indicates forward or backward, absolute value of c indicates steps. Default is forward with step size 1. Positive means forward, negative means backward.

a --  When c is positive or blank, default is 0. When c is negative, default is -1.

b --  When c is positive or blank, default is len. When c is negative, default is -(len+1).

理解索引分配非常重要。

In forward direction, starts at 0 and ends at len-1

In backward direction, starts at -1 and ends at -len

当你说[a:b:c]时，你是说根据c的符号（向前或向后），从a开始，到b结束（不包括bth索引中的元素）。使用上面的索引规则，并记住您只能找到此范围内的元素：

-len, -len+1, -len+2, ..., 0, 1, 2,3,4 , len -1

但这一范围在两个方向上无限延伸：

...,-len -2 ,-len-1,-len, -len+1, -len+2, ..., 0, 1, 2,3,4 , len -1, len, len +1, len+2 , ....

例如：

             0    1    2   3    4   5   6   7   8   9   10   11
             a    s    t   r    i   n   g
    -9  -8  -7   -6   -5  -4   -3  -2  -1

如果在使用上面的a、b、c的规则进行遍历时，a、b和c的选择允许与上面的范围重叠，则会得到一个包含元素的列表（在遍历过程中被触摸），或者得到一个空列表。

最后一件事：如果a和b相等，那么也会得到一个空列表：

>>> l1
[2, 3, 4]

>>> l1[:]
[2, 3, 4]

>>> l1[::-1] # a default is -1 , b default is -(len+1)
[4, 3, 2]

>>> l1[:-4:-1] # a default is -1
[4, 3, 2]

>>> l1[:-3:-1] # a default is -1
[4, 3]

>>> l1[::] # c default is +1, so a default is 0, b default is len
[2, 3, 4]

>>> l1[::-1] # c is -1 , so a default is -1 and b default is -(len+1)
[4, 3, 2]


>>> l1[-100:-200:-1] # Interesting
[]

>>> l1[-1:-200:-1] # Interesting
[4, 3, 2]


>>> l1[-1:-1:1]
[]


>>> l1[-1:5:1] # Interesting
[4]


>>> l1[1:-7:1]
[]

>>> l1[1:-7:-1] # Interesting
[3, 2]

>>> l1[:-2:-2] # a default is -1, stop(b) at -2 , step(c) by 2 in reverse direction
[4]

在找到这张很棒的桌子http://wiki.python.org/moin/MovingToPythonFromOtherLanguages

Python indexes and slices for a six-element list.
Indexes enumerate the elements, slices enumerate the spaces between the elements.

Index from rear:    -6  -5  -4  -3  -2  -1      a=[0,1,2,3,4,5]    a[1:]==[1,2,3,4,5]
Index from front:    0   1   2   3   4   5      len(a)==6          a[:5]==[0,1,2,3,4]
                   +---+---+---+---+---+---+    a[0]==0            a[:-2]==[0,1,2,3]
                   | a | b | c | d | e | f |    a[5]==5            a[1:2]==[1]
                   +---+---+---+---+---+---+    a[-1]==5           a[1:-1]==[1,2,3,4]
Slice from front:  :   1   2   3   4   5   :    a[-2]==4
Slice from rear:   :  -5  -4  -3  -2  -1   :
                                                b=a[:]
                                                b==[0,1,2,3,4,5] (shallow copy of a)

我自己使用“元素之间的索引点”方法来思考它，但描述它的一种方式有时有助于其他人获得它：

mylist[X:Y]

X是所需的第一个元素的索引。Y是不需要的第一个元素的索引。

已经有很多答案了，但我想添加一个性能比较

~$ python3.8 -m timeit -s 'fun = "this is fun;slicer = slice(0, 3)"' "fun_slice = fun[slicer]" 
10000000 loops, best of 5: 29.8 nsec per loop
~$ python3.8 -m timeit -s 'fun = "this is fun"' "fun_slice = fun[0:3]" 
10000000 loops, best of 5: 37.9 nsec per loop
~$ python3.8 -m timeit -s 'fun = "this is fun"' "fun_slice = fun[slice(0, 3)]" 
5000000 loops, best of 5: 68.7 nsec per loop
~$ python3.8 -m timeit -s 'fun = "this is fun"' "slicer = slice(0, 3)" 
5000000 loops, best of 5: 42.8 nsec per loop

因此，如果您重复使用同一个切片，使用切片对象将有益并提高可读性。然而，如果您只进行了几次切片，则应首选[：]表示法。