I need a good explanation (references are a plus) on Python’s slice notation.
To me, this notation needs a bit of picking up.
It looks extremely powerful, but I haven’t quite got my head around it.
回答 0
真的很简单:
a[start:stop]# items start through stop-1
a[start:]# items start through the rest of the array
a[:stop]# items from the beginning through stop-1
a[:]# a copy of the whole array
还有一个step值,可以与以上任何一种一起使用:
a[start:stop:step]# start through not past stop, by step
a[-1]# last item in the array
a[-2:]# last two items in the array
a[:-2]# everything except the last two items
同样,step可能为负数:
a[::-1]# all items in the array, reversed
a[1::-1]# the first two items, reversed
a[:-3:-1]# the last two items, reversed
a[-3::-1]# everything except the last two items, reversed
a[start:stop] # items start through stop-1
a[start:] # items start through the rest of the array
a[:stop] # items from the beginning through stop-1
a[:] # a copy of the whole array
There is also the step value, which can be used with any of the above:
a[start:stop:step] # start through not past stop, by step
The key point to remember is that the :stop value represents the first value that is not in the selected slice. So, the difference between stop and start is the number of elements selected (if step is 1, the default).
The other feature is that start or stop may be a negative number, which means it counts from the end of the array instead of the beginning. So:
a[-1] # last item in the array
a[-2:] # last two items in the array
a[:-2] # everything except the last two items
Similarly, step may be a negative number:
a[::-1] # all items in the array, reversed
a[1::-1] # the first two items, reversed
a[:-3:-1] # the last two items, reversed
a[-3::-1] # everything except the last two items, reversed
Python is kind to the programmer if there are fewer items than you ask for. For example, if you ask for a[:-2] and a only contains one element, you get an empty list instead of an error. Sometimes you would prefer the error, so you have to be aware that this may happen.
Relation to slice() object
The slicing operator [] is actually being used in the above code with a slice() object using the : notation (which is only valid within []), i.e.:
a[start:stop:step]
is equivalent to:
a[slice(start, stop, step)]
Slice objects also behave slightly differently depending on the number of arguments, similarly to range(), i.e. both slice(stop) and slice(start, stop[, step]) are supported.
To skip specifying a given argument, one might use None, so that e.g. a[start:] is equivalent to a[slice(start, None)] or a[::-1] is equivalent to a[slice(None, None, -1)].
While the :-based notation is very helpful for simple slicing, the explicit use of slice() objects simplifies the programmatic generation of slicing.
The Python tutorial talks about it (scroll down a bit until you get to the part about slicing).
The ASCII art diagram is helpful too for remembering how slices work:
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
0 1 2 3 4 5 6
-6 -5 -4 -3 -2 -1
One way to remember how slices work is to think of the indices as pointing between characters, with the left edge of the first character numbered 0. Then the right edge of the last character of a string of n characters has index n.
+---+---+---+---+---+---+| P | y | t | h | o | n |+---+---+---+---+---+---+Slice position:0123456Index position:012345>>> 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']
对于从零到n的切片,一种试探法是:“零是起点,从起点开始,并在列表中取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']
>>> 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,因此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
>>> 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']
习惯最棘手的部分是分配给空片。使用启发式1和2可以很容易地索引一个空片:
>>> 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,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
The answers above don’t discuss slice assignment. To understand slice assignment, it’s helpful to add another concept to the ASCII art:
+---+---+---+---+---+---+
| 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']
One heuristic is, for a slice from zero to n, think: “zero is the beginning, start at the beginning and take n items in a list”.
>>> 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']
Another heuristic is, “for any slice, replace the start by zero, apply the previous heuristic to get the end of the list, then count the first number back up to chop items off the beginning”
>>> 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.
The first rule of slice assignment is that since slicing returns a list, slice assignment requires a list (or other iterable):
>>> 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
The second rule of slice assignment, which you can also see above, is that whatever portion of the list is returned by slice indexing, that’s the same portion that is changed by slice assignment:
>>> p[2:4]
['T','h']
>>> p[2:4] = ['t','r']
>>> p
['P','y','t','r','o','n']
The third rule of slice assignment is, the assigned list (iterable) doesn’t have to have the same length; the indexed slice is simply sliced out and replaced en masse by whatever is being assigned:
>>> 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']
The trickiest part to get used to is assignment to empty slices. Using heuristic 1 and 2 it’s easy to get your head around indexing an empty slice:
And then once you’ve seen that, slice assignment to the empty slice makes sense too:
>>> 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
Note that, since we are not changing the second number of the slice (4), the inserted items always stack right up against the ‘o’, even when we’re assigning to the empty slice. So the position for the empty slice assignment is the logical extension of the positions for the non-empty slice assignments.
Backing up a little bit, what happens when you keep going with our procession of counting up the slice beginning?
With slicing, once you’re done, you’re done; it doesn’t start slicing backwards. In Python you don’t get negative strides unless you explicitly ask for them by using a negative number.
>>> p[5:3:-1]
['n','o']
There are some weird consequences to the “once you’re done, you’re done” rule:
>>> 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
In fact, compared to indexing, Python slicing is bizarrely error-proof:
>>> p[100:200]
[]
>>> p[int(2e99):int(1e99)]
[]
This can come in handy sometimes, but it can also lead to somewhat strange behavior:
In short, the colons (:) in subscript notation (subscriptable[subscriptarg]) make slice notation – which has the optional arguments, start, stop, step:
sliceable[start:stop:step]
Python slicing is a computationally fast way to methodically access parts of your data. In my opinion, to be even an intermediate Python programmer, it’s one aspect of the language that it is necessary to be familiar with.
Important Definitions
To begin with, let’s define a few terms:
start: the beginning index of the slice, it will include the element at this index unless it is the same as stop, defaults to 0, i.e. the first index. If it’s negative, it means to start n items from the end.
stop: the ending index of the slice, it does not include the element at this index, defaults to length of the sequence being sliced, that is, up to and including the end.
step: the amount by which the index increases, defaults to 1. If it’s negative, you’re slicing over the iterable in reverse.
How Indexing Works
You can make any of these positive or negative numbers. The meaning of the positive numbers is straightforward, but for negative numbers, just like indexes in Python, you count backwards from the end for the start and stop, and for the step, you simply decrement your index. This example is from the documentation’s tutorial, but I’ve modified it slightly to indicate which item in a sequence each index references:
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
0 1 2 3 4 5
-6 -5 -4 -3 -2 -1
And recall that there are defaults for start, stop, and step, so to access the defaults, simply leave out the argument.
Slice notation to get the last nine elements from a list (or any other sequence that supports it, like a string) would look like this:
my_list[-9:]
When I see this, I read the part in the brackets as “9th from the end, to the end.” (Actually, I abbreviate it mentally as “-9, on”)
Explanation:
The full notation is
my_list[-9:None:None]
and to substitute the defaults (actually when step is negative, stop‘s default is -len(my_list) - 1, so None for stop really just means it goes to whichever end step takes it to):
my_list[-9:len(my_list):1]
The colon, :, is what tells Python you’re giving it a slice and not a regular index. That’s why the idiomatic way of making a shallow copy of lists in Python 2 is
list_copy = sequence[:]
And clearing them is with:
del my_list[:]
(Python 3 gets a list.copy and list.clear method.)
When step is negative, the defaults for start and stop change
By default, when the step argument is empty (or None), it is assigned to +1.
But you can pass in a negative integer, and the list (or most other standard slicables) will be sliced from the end to the beginning.
Thus a negative slice will change the defaults for start and stop!
If so, the lower bound is -1 meaning we slice all the way up to and including the beginning, and the upper bound is the length minus 1, meaning we start at the end. (Note that the semantics of this -1 is different from a -1 that users may pass indexes in Python indicating the last item.)
if (step_is_negative) {
lower = PyLong_FromLong(-1L);
if (lower == NULL)
goto error;
upper = PyNumber_Add(length, lower);
if (upper == NULL)
goto error;
}
Otherwise step is positive, and the lower bound will be zero and the upper bound (which we go up to but not including) the length of the sliced list.
You may find it useful to separate forming the slice from passing it to the list.__getitem__ method (that’s what the square brackets do). Even if you’re not new to it, it keeps your code more readable so that others that may have to read your code can more readily understand what you’re doing.
However, you can’t just assign some integers separated by colons to a variable. You need to use the slice object:
Since slices of Python lists create new objects in memory, another important function to be aware of is itertools.islice. Typically you’ll want to iterate over a slice, not just have it created statically in memory. islice is perfect for this. A caveat, it doesn’t support negative arguments to start, stop, or step, so if that’s an issue you may need to calculate indices or reverse the iterable in advance.
The fact that list slices make a copy is a feature of lists themselves. If you’re slicing advanced objects like a Pandas DataFrame, it may return a view on the original, and not a copy.
And a couple of things that weren’t immediately obvious to me when I first saw the slicing syntax:
>>> x = [1,2,3,4,5,6]
>>> x[::-1]
[6,5,4,3,2,1]
Easy way to reverse sequences!
And if you wanted, for some reason, every second item in the reversed sequence:
>>> x = [1,2,3,4,5,6]
>>> x[::-2]
[6,4,2]
回答 6
在Python 2.7中
用Python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.Defaultis forward with step size 1.Positive means forward, negative means backward.
a --When c is positive or blank, default is0.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).
了解索引分配非常重要。
In forward direction, starts at 0and ends at len-1In backward direction, starts at -1and ends at -len
>>> 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]
[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).
Understanding index assignment is very important.
In forward direction, starts at 0 and ends at len-1
In backward direction, starts at -1 and ends at -len
When you say [a:b:c], you are saying depending on the sign of c (forward or backward), start at a and end at b (excluding element at bth index). Use the indexing rule above and remember you will only find elements in this range:
-len, -len+1, -len+2, ..., 0, 1, 2,3,4 , len -1
But this range continues in both directions infinitely:
...,-len -2 ,-len-1,-len, -len+1, -len+2, ..., 0, 1, 2,3,4 , len -1, len, len +1, len+2 , ....
For example:
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
If your choice of a, b, and c allows overlap with the range above as you traverse using rules for a,b,c above you will either get a list with elements (touched during traversal) or you will get an empty list.
One last thing: if a and b are equal, then also you get an empty list:
>>> 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]
Python indexes and slices for a six-element list.Indexes enumerate the elements, slices enumerate the spaces between the elements.Indexfrom rear:-6-5-4-3-2-1 a=[0,1,2,3,4,5] a[1:]==[1,2,3,4,5]Indexfrom front:012345 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]Slicefrom front::12345: a[-2]==4Slicefrom rear::-5-4-3-2-1:
b=a[:]
b==[0,1,2,3,4,5](shallow copy of a)
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)
I find it easier to remember how it works, and then I can figure out any specific start/stop/step combination.
It’s instructive to understand range() first:
def range(start=0, stop, step=1): # Illegal syntax, but that's the effect
i = start
while (i < stop if step > 0 else i > stop):
yield i
i += step
Begin from start, increment by step, do not reach stop. Very simple.
The thing to remember about negative step is that stop is always the excluded end, whether it’s higher or lower. If you want same slice in opposite order, it’s much cleaner to do the reversal separately: e.g. 'abcde'[1:-2][::-1] slices off one char from left, two from right, then reverses. (See also reversed().)
Sequence slicing is same, except it first normalizes negative indexes, and it can never go outside the sequence:
TODO: The code below had a bug with “never go outside the sequence” when abs(step)>1; I think I patched it to be correct, but it’s hard to understand.
def this_is_how_slicing_works(seq, start=None, stop=None, step=1):
if start is None:
start = (0 if step > 0 else len(seq)-1)
elif start < 0:
start += len(seq)
if not 0 <= start < len(seq): # clip if still outside bounds
start = (0 if step > 0 else len(seq)-1)
if stop is None:
stop = (len(seq) if step > 0 else -1) # really -1, not last element
elif stop < 0:
stop += len(seq)
for i in range(start, stop, step):
if 0 <= i < len(seq):
yield seq[i]
Don’t worry about the is None details – just remember that omitting start and/or stop always does the right thing to give you the whole sequence.
Normalizing negative indexes first allows start and/or stop to be counted from the end independently: 'abcde'[1:-2] == 'abcde'[1:3] == 'bc' despite range(1,-2) == [].
The normalization is sometimes thought of as “modulo the length”, but note it adds the length just once: e.g. 'abcde'[-53:42] is just the whole string.
I use the “an index points between elements” method of thinking about it myself, but one way of describing it which sometimes helps others get it is this:
mylist[X:Y]
X is the index of the first element you want.
Y is the index of the first element you don’t want.
回答 11
Index:------------>01234+---+---+---+---+---+| a | b | c | d | e |+---+---+---+---+---+0-4-3-2-1<------------Slice:<---------------||--------------->:1234:+---+---+---+---+---+| a | b | c | d | e |+---+---+---+---+---+:-4-3-2-1:|---------------><---------------|
In[122]: alpha =['a','b','c','d','e','f']In[123]: alpha
Out[123]:['a','b','c','d','e','f']In[124]: alpha[0]Out[124]:'a'In[127]: alpha[0]='A'In[128]: alpha
Out[128]:['A','b','c','d','e','f']In[129]: alpha[0,1]---------------------------------------------------------------------------TypeErrorTraceback(most recent call last)<ipython-input-129-c7eb16585371>in<module>()---->1 alpha[0,1]TypeError: list indices must be integers,not tuple
In[21]: alpha =['a','b','c','d','e','f']In[22]: s = slice(None,None,None)In[23]: s
Out[23]: slice(None,None,None)In[24]: s.indices(len(alpha))Out[24]:(0,6,1)In[25]: range(*s.indices(len(alpha)))Out[25]:[0,1,2,3,4,5]In[26]: s = slice(None,None,-1)In[27]: range(*s.indices(len(alpha)))Out[27]:[5,4,3,2,1,0]In[28]: s = slice(None,3,-1)In[29]: range(*s.indices(len(alpha)))Out[29]:[5,4]
Understanding the difference between indexing and slicing:
Wiki Python has this amazing picture which clearly distinguishes indexing and slicing.
It is a list with six elements in it. To understand slicing better, consider that list as a set of six boxes placed together. Each box has an alphabet in it.
Indexing is like dealing with the contents of box. You can check contents of any box. But you can’t check the contents of multiple boxes at once. You can even replace the contents of the box. But you can’t place two balls in one box or replace two balls at a time.
In [122]: alpha = ['a', 'b', 'c', 'd', 'e', 'f']
In [123]: alpha
Out[123]: ['a', 'b', 'c', 'd', 'e', 'f']
In [124]: alpha[0]
Out[124]: 'a'
In [127]: alpha[0] = 'A'
In [128]: alpha
Out[128]: ['A', 'b', 'c', 'd', 'e', 'f']
In [129]: alpha[0,1]
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-129-c7eb16585371> in <module>()
----> 1 alpha[0,1]
TypeError: list indices must be integers, not tuple
Slicing is like dealing with boxes themselves. You can pick up the first box and place it on another table. To pick up the box, all you need to know is the position of beginning and ending of the box.
You can even pick up the first three boxes or the last two boxes or all boxes between 1 and 4. So, you can pick any set of boxes if you know the beginning and ending. These positions are called start and stop positions.
The interesting thing is that you can replace multiple boxes at once. Also you can place multiple boxes wherever you like.
In [130]: alpha[0:1]
Out[130]: ['A']
In [131]: alpha[0:1] = 'a'
In [132]: alpha
Out[132]: ['a', 'b', 'c', 'd', 'e', 'f']
In [133]: alpha[0:2] = ['A', 'B']
In [134]: alpha
Out[134]: ['A', 'B', 'c', 'd', 'e', 'f']
In [135]: alpha[2:2] = ['x', 'xx']
In [136]: alpha
Out[136]: ['A', 'B', 'x', 'xx', 'c', 'd', 'e', 'f']
Slicing With Step:
Till now you have picked boxes continuously. But sometimes you need to pick up discretely. For example, you can pick up every second box. You can even pick up every third box from the end. This value is called step size. This represents the gap between your successive pickups. The step size should be positive if You are picking boxes from the beginning to end and vice versa.
In [137]: alpha = ['a', 'b', 'c', 'd', 'e', 'f']
In [142]: alpha[1:5:2]
Out[142]: ['b', 'd']
In [143]: alpha[-1:-5:-2]
Out[143]: ['f', 'd']
In [144]: alpha[1:5:-2]
Out[144]: []
In [145]: alpha[-1:-5:2]
Out[145]: []
How Python Figures Out Missing Parameters:
When slicing, if you leave out any parameter, Python tries to figure it out automatically.
If you check the source code of CPython, you will find a function called PySlice_GetIndicesEx() which figures out indices to a slice for any given parameters. Here is the logical equivalent code in Python.
This function takes a Python object and optional parameters for slicing and returns the start, stop, step, and slice length for the requested slice.
def py_slice_get_indices_ex(obj, start=None, stop=None, step=None):
length = len(obj)
if step is None:
step = 1
if step == 0:
raise Exception("Step cannot be zero.")
if start is None:
start = 0 if step > 0 else length - 1
else:
if start < 0:
start += length
if start < 0:
start = 0 if step > 0 else -1
if start >= length:
start = length if step > 0 else length - 1
if stop is None:
stop = length if step > 0 else -1
else:
if stop < 0:
stop += length
if stop < 0:
stop = 0 if step > 0 else -1
if stop >= length:
stop = length if step > 0 else length - 1
if (step < 0 and stop >= start) or (step > 0 and start >= stop):
slice_length = 0
elif step < 0:
slice_length = (stop - start + 1)/(step) + 1
else:
slice_length = (stop - start - 1)/(step) + 1
return (start, stop, step, slice_length)
This is the intelligence that is present behind slices. Since Python has an built-in function called slice, you can pass some parameters and check how smartly it calculates missing parameters.
In [21]: alpha = ['a', 'b', 'c', 'd', 'e', 'f']
In [22]: s = slice(None, None, None)
In [23]: s
Out[23]: slice(None, None, None)
In [24]: s.indices(len(alpha))
Out[24]: (0, 6, 1)
In [25]: range(*s.indices(len(alpha)))
Out[25]: [0, 1, 2, 3, 4, 5]
In [26]: s = slice(None, None, -1)
In [27]: range(*s.indices(len(alpha)))
Out[27]: [5, 4, 3, 2, 1, 0]
In [28]: s = slice(None, 3, -1)
In [29]: range(*s.indices(len(alpha)))
Out[29]: [5, 4]
As a general rule, writing code with a lot of hardcoded index values leads to a readability
and maintenance mess. For example, if you come back to the code a year later, you’ll
look at it and wonder what you were thinking when you wrote it. The solution shown
is simply a way of more clearly stating what your code is actually doing.
In general, the built-in slice() creates a slice object that can be used anywhere a slice
is allowed. For example:
If you have a slice instance s, you can get more information about it by looking at its
s.start, s.stop, and s.step attributes, respectively. For example:
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].
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?
惊奇: 当索引超出范围时,slice不会引发IndexError!
如果索引超出范围,Python将尽力将索引设置为0或len(s)根据情况。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.例子
让我们以示例结束这个答案,解释我们所讨论的一切:
# Create our array for demonstrationIn[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 indexOut[3]:[2,3,4,5,6,7,8,9]In[4]: s[:8]# From index 0 up to index 8Out[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 orderOut[9]:[9,7,5,3,1]In[11]: s[-2::-2]# All even numbers in reversed orderOut[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 listOut[14]:[]In[15]: s[11]# Access index 11 (greater than len(s)) will raise an IndexError---------------------------------------------------------------------------IndexErrorTraceback(most recent call last)<ipython-input-15-79ffc22473a3>in<module>()---->1 s[11]IndexError: list index out of range
To make it simple, remember slice has only one form:
s[start:end:step]
and here is how it works:
s: an object that can be sliced
start: first index to start iteration
end: last index, NOTE that end index will not be included in the resulted slice
step: pick element every step index
Another import thing: all start,end, step can be omitted! And if they are omitted, their default value will be used: 0,len(s),1 accordingly.
So possible variations are:
# 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[:]
NOTE: If start >= end (considering only when step>0), Python will return a empty slice [].
2. Pitfalls
The above part explains the core features on how slice works, and it will work on most occasions. However, there can be pitfalls you should watch out, and this part explains them.
Negative indexes
The very first thing that confuses Python learners is that an index can be negative!
Don’t panic: a negative index means count backwards.
For example:
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].
Negative step
Making things more confusing is that step can be negative too!
A negative step means iterate the array backwards: from the end to start, with the end index included, and the start index excluded from the result.
NOTE: when step is negative, the default value for start is len(s) (while end does not equal to 0, because s[::-1] contains s[0]). For example:
s[::-1] # Reversed slice
s[len(s)::-1] # The same as above, reversed slice
s[0:len(s):-1] # Empty list
Out of range error?
Be surprised: slice does not raise an IndexError when the index is out of range!
If the index is out of range, Python will try its best to set the index to 0 or len(s) according to the situation. For example:
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. Examples
Let’s finish this answer with examples, explaining everything we have discussed:
# 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
The previous answers don’t discuss multi-dimensional array slicing which is possible using the famous NumPy package:
Slicing can also be applied to multi-dimensional arrays.
# Here, a is a NumPy array
>>> a
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
>>> a[:2, 0:3:2]
array([[1, 3],
[5, 7]])
The “:2” before the comma operates on the first dimension and the “0:3:2” after the comma operates on the second dimension.
回答 19
#!/usr/bin/env pythondef slicegraphical(s, lista):if len(s)>9:print"""Enter a string of maximum 9 characters,
so the printig would looki nice"""return0;# 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]==Noneand triada[1]==Noneand triada[2]==None:# 000print s+'[ : : ]'+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]==Noneand triada[1]==Noneand triada[2]!=None:# 001print s+'[ : :{0:2d} ]'.format(triada[2],'','')+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]==Noneand triada[1]!=Noneand triada[2]==None:# 010print s+'[ :{0:2d} : ]'.format(triada[1])+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]==Noneand triada[1]!=Noneand triada[2]!=None:# 011print s+'[ :{0:2d} :{1:2d} ]'.format(triada[1], triada[2])+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]!=Noneand triada[1]==Noneand triada[2]==None:# 100print s+'[{0:2d} : : ]'.format(triada[0])+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]!=Noneand triada[1]==Noneand triada[2]!=None:# 101print s+'[{0:2d} : :{1:2d} ]'.format(triada[0], triada[2])+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]!=Noneand triada[1]!=Noneand triada[2]==None:# 110print s+'[{0:2d} :{1:2d} : ]'.format(triada[0], triada[1])+' = ', s[triada[0]:triada[1]:triada[2]]elif triada[0]!=Noneand triada[1]!=Noneand triada[2]!=None:# 111print 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]==Noneand triada[1]==None:# 00print s+'[ : ] '+' = ', s[triada[0]:triada[1]]elif triada[0]==Noneand triada[1]!=None:# 01print s+'[ :{0:2d} ] '.format(triada[1])+' = ', s[triada[0]:triada[1]]elif triada[0]!=Noneand triada[1]==None:# 10print s+'[{0:2d} : ] '.format(triada[0])+' = ', s[triada[0]:triada[1]]elif triada[0]!=Noneand triada[1]!=None:# 11print 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)
您可以运行此脚本并进行实验,以下是我从脚本中获得的一些示例。
+---+---+---+---+---+---+---+---+---+| C | O | M | P | U | T | E | R | S |+---+---+---+---+---+---+---+---+---+0123456789-9-8-7-6-5-4-3-2-1
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
[Finishedin0.9s]
In my opinion, you will understand and memorize better the Python string slicing notation if you look at it the following way (read on).
Let’s work with the following string …
azString = "abcdefghijklmnopqrstuvwxyz"
For those who don’t know, you can create any substring from azString using the notation azString[x:y]
Coming from other programming languages, that’s when the common sense gets compromised. What are x and y?
I had to sit down and run several scenarios in my quest for a memorization technique that will help me remember what x and y are and help me slice strings properly at the first attempt.
My conclusion is that x and y should be seen as the boundary indexes that are surrounding the strings that we want to extra. So we should see the expression as azString[index1, index2] or even more clearer as azString[index_of_first_character, index_after_the_last_character].
Here is an example visualization of that …
Letters a b c d e f g h i j ...
↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑
┊ ┊
Indexes 0 1 2 3 4 5 6 7 8 9 ...
┊ ┊
cdefgh index1 index2
So all you have to do is setting index1 and index2 to the values that will surround the desired substring. For instance, to get the substring “cdefgh”, you can use azString[2:8], because the index on the left side of “c” is 2 and the one on the right size of “h” is 8.
Remember that we are setting the boundaries. And those boundaries are the positions where you could place some brackets that will be wrapped around the substring like this …
a b [ c d e f g h ] i j
That trick works all the time and is easy to memorize.
In[7]: l[:-1]# Include all elements but the last oneOut[7]:[0,1,2,3,4,5,6,7,8]In[8]: l[-3:]# Take the last three elementsOut[8]:[7,8,9]
切片时可以提供超出范围的索引,例如:
In[9]: l[:20]# 20 is out of index bounds, and l[20] will raise an IndexError exceptionOut[9]:[0,1,2,3,4,5,6,7,8,9]In[11]: l[-20:]# -20 is out of index bounds, and l[-20] will raise an IndexError exceptionOut[11]:[0,1,2,3,4,5,6,7,8,9]
In[16]: l[2:6]= list('abc')# Assigning fewer elements than the ones contained in the sliced collection l[2:6]In[17]: l
Out[17]:[0,1,'a','b','c',6,7,8,9]In[18]: l[2:5]= list('hello')# Assigning more elements than the ones contained in the sliced collection l [2:5]In[19]: l
Out[19]:[0,1,'h','e','l','l','o',6,7,8,9]
如果省略开始索引和结束索引,则将复制该集合:
In[14]: l_copy = l[:]In[15]: l == l_copy and l isnot l_copy
Out[15]:True
如果在执行赋值操作时省略了开始索引和结束索引,则集合的全部内容将被引用的副本代替:
In[20]: l[:]= list('hello...')In[21]: l
Out[21]:['h','e','l','l','o','.','.','.']
除了基本切片之外,还可以应用以下符号:
l[start:end:step]
where l是一个集合,start是一个包含索引,end是一个排他索引,并且step是一个可用于获取第n个项目的跨度l。
In[22]: l = list(range(10))In[23]: l[::2]# Take the elements which indexes are evenOut[23]:[0,2,4,6,8]In[24]: l[1::2]# Take the elements which indexes are oddOut[24]:[1,3,5,7,9]
In Python, the most basic form for slicing is the following:
l[start:end]
where l is some collection, start is an inclusive index, and end is an exclusive index.
In [1]: l = list(range(10))
In [2]: l[:5] # First five elements
Out[2]: [0, 1, 2, 3, 4]
In [3]: l[-5:] # Last five elements
Out[3]: [5, 6, 7, 8, 9]
When slicing from the start, you can omit the zero index, and when slicing to the end, you can omit the final index since it is redundant, so do not be verbose:
In [5]: l[:3] == l[0:3]
Out[5]: True
In [6]: l[7:] == l[7:len(l)]
Out[6]: True
Negative integers are useful when doing offsets relative to the end of a collection:
In [7]: l[:-1] # Include all elements but the last one
Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8]
In [8]: l[-3:] # Take the last three elements
Out[8]: [7, 8, 9]
It is possible to provide indices that are out of bounds when slicing such as:
In [9]: l[:20] # 20 is out of index bounds, and l[20] will raise an IndexError exception
Out[9]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
In [11]: l[-20:] # -20 is out of index bounds, and l[-20] will raise an IndexError exception
Out[11]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Keep in mind that the result of slicing a collection is a whole new collection. In addition, when using slice notation in assignments, the length of the slice assignments do not need to be the same. The values before and after the assigned slice will be kept, and the collection will shrink or grow to contain the new values:
In [16]: l[2:6] = list('abc') # Assigning fewer elements than the ones contained in the sliced collection l[2:6]
In [17]: l
Out[17]: [0, 1, 'a', 'b', 'c', 6, 7, 8, 9]
In [18]: l[2:5] = list('hello') # Assigning more elements than the ones contained in the sliced collection l [2:5]
In [19]: l
Out[19]: [0, 1, 'h', 'e', 'l', 'l', 'o', 6, 7, 8, 9]
If you omit the start and end index, you will make a copy of the collection:
In [14]: l_copy = l[:]
In [15]: l == l_copy and l is not l_copy
Out[15]: True
If the start and end indexes are omitted when performing an assignment operation, the entire content of the collection will be replaced with a copy of what is referenced:
In [20]: l[:] = list('hello...')
In [21]: l
Out[21]: ['h', 'e', 'l', 'l', 'o', '.', '.', '.']
Besides basic slicing, it is also possible to apply the following notation:
l[start:end:step]
where l is a collection, start is an inclusive index, end is an exclusive index, and step is a stride that can be used to take every nth item in l.
In [22]: l = list(range(10))
In [23]: l[::2] # Take the elements which indexes are even
Out[23]: [0, 2, 4, 6, 8]
In [24]: l[1::2] # Take the elements which indexes are odd
Out[24]: [1, 3, 5, 7, 9]
Using step provides a useful trick to reverse a collection in Python:
It is also possible to use negative integers for step as the following example:
In[28]: l[::-2]
Out[28]: [9, 7, 5, 3, 1]
However, using a negative value for step could become very confusing. Moreover, in order to be Pythonic, you should avoid using start, end, and step in a single slice. In case this is required, consider doing this in two assignments (one to slice, and the other to stride).
In [29]: l = l[::2] # This step is for striding
In [30]: l
Out[30]: [0, 2, 4, 6, 8]
In [31]: l = l[1:-1] # This step is for slicing
In [32]: l
Out[32]: [2, 4, 6]
回答 24
我想添加一个世界您好!这个例子为初学者解释了切片的基础。这对我帮助很大。
我们来看一个包含六个值的列表['P', 'Y', 'T', 'H', 'O', 'N']:
+---+---+---+---+---+---+| P | Y | T | H | O | N |+---+---+---+---+---+---+012345
现在,该列表中最简单的部分就是其子列表。符号是[<index>:<index>],关键是这样阅读:
[ start cutting before this index : end cutting before this index ]
现在,如果您[2:5]从上面的列表中切出一部分,就会发生这种情况:
||+---+---|---+---+---|---+| P | Y | T | H | O | N |+---+---|---+---+---|---+01|234|5
I want to add one Hello, World! example that explains the basics of slices for the very beginners. It helped me a lot.
Let’s have a list with six values ['P', 'Y', 'T', 'H', 'O', 'N']:
+---+---+---+---+---+---+
| P | Y | T | H | O | N |
+---+---+---+---+---+---+
0 1 2 3 4 5
Now the simplest slices of that list are its sublists. The notation is [<index>:<index>] and the key is to read it like this:
[ start cutting before this index : end cutting before this index ]
Now if you make a slice [2:5] of the list above, this will happen:
| |
+---+---|---+---+---|---+
| P | Y | T | H | O | N |
+---+---|---+---+---|---+
0 1 | 2 3 4 | 5
You made a cut before the element with index 2 and another cut before the element with index 5. So the result will be a slice between those two cuts, a list ['T', 'H', 'O'].
回答 25
我个人认为它像一个for循环:
a[start:end:step]# for(i = start; i < end; i += step)
a[start:end:step]
# for(i = start; i < end; i += step)
Also, note that negative values for start and end are relative to the end of the list and computed in the example above by given_index + a.shape[0].
回答 26
以下是字符串索引的示例:
+---+---+---+---+---+| H | e | l | p | A |+---+---+---+---+---+012345-5-4-3-2-1
str="Name string"
切片示例:[开始:结束:步骤]
str[start:end]# Items start through end-1
str[start:]# Items start through the rest of the array
str[:end]# Items from the beginning through end-1
str[:]# A copy of the whole array
下面是示例用法:
print str[0]= N
print str[0:2]=Naprint str[0:7]=Name st
print str[0:7:2]=Nm t
print str[0:-1:2]=Nm ti
+---+---+---+---+---+
| H | e | l | p | A |
+---+---+---+---+---+
0 1 2 3 4 5
-5 -4 -3 -2 -1
str="Name string"
Slicing example: [start:end:step]
str[start:end] # Items start through end-1
str[start:] # Items start through the rest of the array
str[:end] # Items from the beginning through end-1
str[:] # A copy of the whole array
Below is the example usage:
print str[0] = N
print str[0:2] = Na
print str[0:7] = Name st
print str[0:7:2] = Nm t
print str[0:-1:2] = Nm ti
def slice(list, start =None, end =None, step =1):# Take care of missing start/end parameters
start =0if start isNoneelse start
end = len(list)if end isNoneelse end
# Take care of negative start/end parameters
start = len(list)+ start if start <0else start
end = len(list)+ end if end <0else end
# Now just execute a for-loop with start, end and stepreturn[list[i]for i in range(start, end, step)]
If you feel negative indices in slicing is confusing, here’s a very easy way to think about it: just replace the negative index with len - index. So for example, replace -3 with len(list) - 3.
The best way to illustrate what slicing does internally is just show it in code that implements this operation:
def slice(list, start = None, end = None, step = 1):
# Take care of missing start/end parameters
start = 0 if start is None else start
end = len(list) if end is None else end
# Take care of negative start/end parameters
start = len(list) + start if start < 0 else start
end = len(list) + end if end < 0 else end
# Now just execute a for-loop with start, end and step
return [list[i] for i in range(start, end, step)]
The basic slicing technique is to define the starting point, the stopping point, and the step size – also known as stride.
First, we will create a list of values to use in our slicing.
Create two lists to slice. The first is a numeric list from 1 to 9 (List A). The second is also a numeric list, from 0 to 9 (List B):
A = list(range(1, 10, 1)) # Start, stop, and step
B = list(range(9))
print("This is List A:", A)
print("This is List B:", B)
Index the number 3 from A and the number 6 from B.
print(A[2])
print(B[6])
Basic Slicing
Extended indexing syntax used for slicing is aList[start:stop:step]. The start argument and the step argument both default to none – the only required argument is stop. Did you notice this is similar to how range was used to define lists A and B? This is because the slice object represents the set of indices specified by range(start, stop, step). Python 3.4 documentation.
As you can see, defining only stop returns one element. Since the start defaults to none, this translates into retrieving only one element.
It is important to note, the first element is index 0, not index 1. This is why we are using 2 lists for this exercise. List A’s elements are numbered according to the ordinal position (the first element is 1, the second element is 2, etc.) while List B’s elements are the numbers that would be used to index them ([0] for the first element 0, etc.).
With extended indexing syntax, we retrieve a range of values. For example, all values are retrieved with a colon.
A[:]
To retrieve a subset of elements, the start and stop positions need to be defined.
Given the pattern aList[start:stop], retrieve the first two elements from List A.
+---+---+---+---+---+---+| P | y | t | h | o | n |+---+---+---+---+---+---+012345-6-5-4-3-2-1+---+---+---+---+---+---+---+---+---+---+---+---+| P | y | t | h | o | n | P | y | t | h | o | n |+---+---+---+---+---+---+---+---+---+---+---+---+-6-5-4-3-2-1012345
I don’t think that the Python tutorial diagram (cited in various other answers) is good as this suggestion works for positive stride, but does not for a negative stride.
This is the diagram:
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
0 1 2 3 4 5 6
-6 -5 -4 -3 -2 -1
From the diagram, I expect a[-4,-6,-1] to be yP but it is ty.
>>> a = "Python"
>>> a[2:4:1] # as expected
'th'
>>> a[-4:-6:-1] # off by 1
'ty'
What always work is to think in characters or slots and use indexing as a half-open interval — right-open if positive stride, left-open if negative stride.
This way, I can think of a[-4:-6:-1] as a(-6,-4] in interval terminology.
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
0 1 2 3 4 5
-6 -5 -4 -3 -2 -1
+---+---+---+---+---+---+---+---+---+---+---+---+
| P | y | t | h | o | n | P | y | t | h | o | n |
+---+---+---+---+---+---+---+---+---+---+---+---+
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5