Python的切片表示法是如何工作的？也就是说：当我编写[x:y:z]、a[：]、a]：：2]等代码时，我如何理解哪些元素最终会出现在切片中？请在适当的地方附上参考资料。
另请参见：为什么切片和范围上限是互斥的？
当前回答
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
其他回答
我的大脑似乎很乐意接受lst[开始：结束]包含开始项。我甚至可以说这是一个“自然的假设”。
但偶尔会有一种怀疑悄悄出现，我的大脑会要求我保证它不包含结尾元素。
在这些时刻，我依靠这个简单的定理：
for any n, lst = lst[:n] + lst[n:]
这个漂亮的属性告诉我，lst[start:end]不包含end-th项，因为它位于lst[end:]中。
注意，这个定理对任何n都是正确的。例如，您可以检查
lst = range(10)
lst[:-42] + lst[-42:] == lst
返回True。
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]
上面的答案不讨论切片分配。为了理解切片分配，可以在ASCII艺术中添加另一个概念：
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
Slice position: 0 1 2 3 4 5 6
Index position: 0 1 2 3 4 5
>>> p = ['P','y','t','h','o','n']
# Why the two sets of numbers:
# indexing gives items, not lists
>>> p[0]
'P'
>>> p[5]
'n'
# Slicing gives lists
>>> p[0:1]
['P']
>>> p[0:2]
['P','y']
一种启发式方法是，对于从零到n的切片，思考：“零是开始，从开始开始，在列表中取n个项目”。
>>> p[5] # the last of six items, indexed from zero
'n'
>>> p[0:5] # does NOT include the last item!
['P','y','t','h','o']
>>> p[0:6] # not p[0:5]!!!
['P','y','t','h','o','n']
另一种启发式方法是，“对于任何一个切片，用零替换开头，应用前面的启发式方法获得列表的结尾，然后将第一个数字向后计数，以从开头删除项目”
>>> p[0:4] # Start at the beginning and count out 4 items
['P','y','t','h']
>>> p[1:4] # Take one item off the front
['y','t','h']
>>> p[2:4] # Take two items off the front
['t','h']
# etc.
切片分配的第一个规则是，由于切片返回一个列表，所以切片分配需要一个列表（或其他可迭代的）：
>>> p[2:3]
['t']
>>> p[2:3] = ['T']
>>> p
['P','y','T','h','o','n']
>>> p[2:3] = 't'
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable
切片分配的第二个规则（您也可以在上面看到）是，无论切片索引返回列表的哪个部分，都是由切片分配更改的相同部分：
>>> p[2:4]
['T','h']
>>> p[2:4] = ['t','r']
>>> p
['P','y','t','r','o','n']
切片分配的第三条规则是，分配的列表（可迭代）不必具有相同的长度；索引切片被简单地切片，并被分配的任何内容整体替换：
>>> p = ['P','y','t','h','o','n'] # Start over
>>> p[2:4] = ['s','p','a','m']
>>> p
['P','y','s','p','a','m','o','n']
最难习惯的部分是分配给空切片。使用启发式1和2，很容易让你的头脑围绕空切片进行索引：
>>> p = ['P','y','t','h','o','n']
>>> p[0:4]
['P','y','t','h']
>>> p[1:4]
['y','t','h']
>>> p[2:4]
['t','h']
>>> p[3:4]
['h']
>>> p[4:4]
[]
然后，一旦您看到了这一点，将切片分配给空切片也是有意义的：
>>> p = ['P','y','t','h','o','n']
>>> p[2:4] = ['x','y'] # Assigned list is same length as slice
>>> p
['P','y','x','y','o','n'] # Result is same length
>>> p = ['P','y','t','h','o','n']
>>> p[3:4] = ['x','y'] # Assigned list is longer than slice
>>> p
['P','y','t','x','y','o','n'] # The result is longer
>>> p = ['P','y','t','h','o','n']
>>> p[4:4] = ['x','y']
>>> p
['P','y','t','h','x','y','o','n'] # The result is longer still
请注意，因为我们没有更改切片的第二个编号（4），所以插入的项目总是紧靠“o”堆叠，即使我们分配给空切片也是如此。因此，空切片分配的位置是非空切片分配位置的逻辑扩展。
稍微后退一点，当你继续进行我们的切片开始计数过程时会发生什么？
>>> p = ['P','y','t','h','o','n']
>>> p[0:4]
['P','y','t','h']
>>> p[1:4]
['y','t','h']
>>> p[2:4]
['t','h']
>>> p[3:4]
['h']
>>> p[4:4]
[]
>>> p[5:4]
[]
>>> p[6:4]
[]
通过切片，一旦你完成，你就完成了；它不会开始向后倾斜。在Python中，除非使用负数明确要求，否则不会获得负的步幅。
>>> p[5:3:-1]
['n','o']
“一旦你完成了，你就完成了”规则会产生一些奇怪的后果：
>>> p[4:4]
[]
>>> p[5:4]
[]
>>> p[6:4]
[]
>>> p[6]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: list index out of range
事实上，与索引相比，Python切片具有奇怪的防错误性：
>>> p[100:200]
[]
>>> p[int(2e99):int(1e99)]
[]
这有时会派上用场，但也会导致一些奇怪的行为：
>>> p
['P', 'y', 't', 'h', 'o', 'n']
>>> p[int(2e99):int(1e99)] = ['p','o','w','e','r']
>>> p
['P', 'y', 't', 'h', 'o', 'n', 'p', 'o', 'w', 'e', 'r']
根据您的应用程序，这可能。。。或者可能不。。。成为你在那里所希望的！
以下是我的原始答案。它对很多人都很有用，所以我不想删除它。
>>> r=[1,2,3,4]
>>> r[1:1]
[]
>>> r[1:1]=[9,8]
>>> r
[1, 9, 8, 2, 3, 4]
>>> r[1:1]=['blah']
>>> r
[1, 'blah', 9, 8, 2, 3, 4]
这也可以澄清切片和索引之间的区别。
1.切片符号
为了简单起见，请记住切片只有一种形式：
s[start:end:step]
下面是它的工作原理：
s： 可以切片的物体start：开始迭代的第一个索引end：最后一个索引，请注意，最终的切片中不会包含结束索引step：每个步骤索引拾取元素
另一个重要的事情：所有开始、结束和步骤都可以省略！如果省略它们，则将使用它们的默认值：0，len（s），1。
因此，可能的变化如下：
# Mostly used variations
s[start:end]
s[start:]
s[:end]
# Step-related variations
s[:end:step]
s[start::step]
s[::step]
# Make a copy
s[:]
注意：如果start>=end（仅在步骤>0时考虑），Python将返回一个空切片[]。
2.陷阱
上面的部分解释了切片如何工作的核心特性，它将在大多数情况下工作。然而，可能会有陷阱，您应该注意，本部分将对它们进行解释。
负面指数
让Python学习者困惑的第一件事是索引可以是负数！不要惊慌：负指数意味着倒数。
例如：
s[-5:] # Start at the 5th index from the end of array,
# thus returning the last 5 elements.
s[:-5] # Start at index 0, and end until the 5th index from end of array,
# thus returning s[0:len(s)-5].
负阶跃
让事情更令人困惑的是，这一步也可能是消极的！
负步骤意味着向后迭代数组：从结束到开始，包括结束索引，从结果中排除开始索引。
注意：当step为负值时，start的默认值为len（s）（而end不等于0，因为s[：：-1]包含s[0]）。例如：
s[::-1] # Reversed slice
s[len(s)::-1] # The same as above, reversed slice
s[0:len(s):-1] # Empty list
超出范围错误？
请注意：当索引超出范围时，切片不会引发IndexError！
如果索引超出范围，Python将根据情况尽量将索引设置为0或len。例如：
s[:len(s)+5] # The same as s[:len(s)]
s[-len(s)-5::] # The same as s[0:]
s[len(s)+5::-1] # The same as s[len(s)::-1], and the same as s[::-1]
3.示例
让我们用例子来完成这个回答，解释我们讨论的所有内容：
# Create our array for demonstration
In [1]: s = [i for i in range(10)]
In [2]: s
Out[2]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
In [3]: s[2:] # From index 2 to last index
Out[3]: [2, 3, 4, 5, 6, 7, 8, 9]
In [4]: s[:8] # From index 0 up to index 8
Out[4]: [0, 1, 2, 3, 4, 5, 6, 7]
In [5]: s[4:7] # From index 4 (included) up to index 7(excluded)
Out[5]: [4, 5, 6]
In [6]: s[:-2] # Up to second last index (negative index)
Out[6]: [0, 1, 2, 3, 4, 5, 6, 7]
In [7]: s[-2:] # From second last index (negative index)
Out[7]: [8, 9]
In [8]: s[::-1] # From last to first in reverse order (negative step)
Out[8]: [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
In [9]: s[::-2] # All odd numbers in reversed order
Out[9]: [9, 7, 5, 3, 1]
In [11]: s[-2::-2] # All even numbers in reversed order
Out[11]: [8, 6, 4, 2, 0]
In [12]: s[3:15] # End is out of range, and Python will set it to len(s).
Out[12]: [3, 4, 5, 6, 7, 8, 9]
In [14]: s[5:1] # Start > end; return empty list
Out[14]: []
In [15]: s[11] # Access index 11 (greater than len(s)) will raise an IndexError
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
<ipython-input-15-79ffc22473a3> in <module>()
----> 1 s[11]
IndexError: list index out of range
#!/usr/bin/env python
def slicegraphical(s, lista):
if len(s) > 9:
print """Enter a string of maximum 9 characters,
so the printig would looki nice"""
return 0;
# print " ",
print ' '+'+---' * len(s) +'+'
print ' ',
for letter in s:
print '| {}'.format(letter),
print '|'
print " ",; print '+---' * len(s) +'+'
print " ",
for letter in range(len(s) +1):
print '{} '.format(letter),
print ""
for letter in range(-1*(len(s)), 0):
print ' {}'.format(letter),
print ''
print ''
for triada in lista:
if len(triada) == 3:
if triada[0]==None and triada[1] == None and triada[2] == None:
# 000
print s+'[ : : ]' +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] == None and triada[1] == None and triada[2] != None:
# 001
print s+'[ : :{0:2d} ]'.format(triada[2], '','') +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] == None and triada[1] != None and triada[2] == None:
# 010
print s+'[ :{0:2d} : ]'.format(triada[1]) +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] == None and triada[1] != None and triada[2] != None:
# 011
print s+'[ :{0:2d} :{1:2d} ]'.format(triada[1], triada[2]) +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] != None and triada[1] == None and triada[2] == None:
# 100
print s+'[{0:2d} : : ]'.format(triada[0]) +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] != None and triada[1] == None and triada[2] != None:
# 101
print s+'[{0:2d} : :{1:2d} ]'.format(triada[0], triada[2]) +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] != None and triada[1] != None and triada[2] == None:
# 110
print s+'[{0:2d} :{1:2d} : ]'.format(triada[0], triada[1]) +' = ', s[triada[0]:triada[1]:triada[2]]
elif triada[0] != None and triada[1] != None and triada[2] != None:
# 111
print s+'[{0:2d} :{1:2d} :{2:2d} ]'.format(triada[0], triada[1], triada[2]) +' = ', s[triada[0]:triada[1]:triada[2]]
elif len(triada) == 2:
if triada[0] == None and triada[1] == None:
# 00
print s+'[ : ] ' + ' = ', s[triada[0]:triada[1]]
elif triada[0] == None and triada[1] != None:
# 01
print s+'[ :{0:2d} ] '.format(triada[1]) + ' = ', s[triada[0]:triada[1]]
elif triada[0] != None and triada[1] == None:
# 10
print s+'[{0:2d} : ] '.format(triada[0]) + ' = ', s[triada[0]:triada[1]]
elif triada[0] != None and triada[1] != None:
# 11
print s+'[{0:2d} :{1:2d} ] '.format(triada[0],triada[1]) + ' = ', s[triada[0]:triada[1]]
elif len(triada) == 1:
print s+'[{0:2d} ] '.format(triada[0]) + ' = ', s[triada[0]]
if __name__ == '__main__':
# Change "s" to what ever string you like, make it 9 characters for
# better representation.
s = 'COMPUTERS'
# add to this list different lists to experement with indexes
# to represent ex. s[::], use s[None, None,None], otherwise you get an error
# for s[2:] use s[2:None]
lista = [[4,7],[2,5,2],[-5,1,-1],[4],[-4,-6,-1], [2,-3,1],[2,-3,-1], [None,None,-1],[-5,None],[-5,0,-1],[-5,None,-1],[-1,1,-2]]
slicegraphical(s, lista)
你可以运行这个脚本并进行实验，下面是我从脚本中获得的一些示例。
+---+---+---+---+---+---+---+---+---+
| C | O | M | P | U | T | E | R | S |
+---+---+---+---+---+---+---+---+---+
0 1 2 3 4 5 6 7 8 9
-9 -8 -7 -6 -5 -4 -3 -2 -1
COMPUTERS[ 4 : 7 ] = UTE
COMPUTERS[ 2 : 5 : 2 ] = MU
COMPUTERS[-5 : 1 :-1 ] = UPM
COMPUTERS[ 4 ] = U
COMPUTERS[-4 :-6 :-1 ] = TU
COMPUTERS[ 2 :-3 : 1 ] = MPUT
COMPUTERS[ 2 :-3 :-1 ] =
COMPUTERS[ : :-1 ] = SRETUPMOC
COMPUTERS[-5 : ] = UTERS
COMPUTERS[-5 : 0 :-1 ] = UPMO
COMPUTERS[-5 : :-1 ] = UPMOC
COMPUTERS[-1 : 1 :-2 ] = SEUM
[Finished in 0.9s]
当使用否定步骤时，请注意答案向右移动1。
