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Arbitrary number of nested loops dependent on the previous loop in Python

I'm trying to figure out how to iterate over an arbitrary number of loops where each loop depends on the most recent outer loop. The following code is an example of what I want to do:
def function(z):
n = int(log(z))
tupes = []
for i_1 in range(1, n):
for i_2 in range(1, i_1):
...
...
...
for i_n in range(1, i_{n - 1}):
if i_1*i_2*...*i_n > z:
tupes.append((i_1, i_2,..., i_n))
return tupes
While I'd like this to work for any z > e**2, it's sufficient for it to work for zs up to e**100. I know that if I take the Cartesian product of the appropriate ranges that I'll end up with a superset of the tuples I desire, but I'd like to obtain only the tuples I seek.
If anyone can help me with this, I'd greatly appreciate it. Thanks in advance.

Combinations can be listed in ascending order; in fact, this is the default behavior of itertools.combinations.
The code:
for i1 in range(1,6):
for i2 in range(1,i1):
for i3 in range(1,i2):
print (i3, i2, i1)
# (1, 2, 3)
# (1, 2, 4)
# ...
# (3, 4, 5)
Is equivalent to the code:
from itertools import combinations
for combination in combinations(range(1,6), 3):
print combination
# (1, 2, 3)
# (1, 2, 4)
# ...
# (3, 4, 5)
Using the combinations instead of the Cartesian product culls the sample space down to what you want.

The logic in your question implemented recursively (note that this allows for duplicate tuples):
import functools
def f(n, z, max_depth, factors=(), depth=0):
res = []
if depth == max_depth:
product = functools.reduce(lambda x, y: x*y, factors, 1)
if product > z:
res.append(factors)
else:
for i in range(1, n):
new_factors = factors + (i,)
res.extend(f(i, z, factors=new_factors, depth=depth+1, max_depth=max_depth))
return res
z = np.e ** 10
n = int(np.log(z))
print(f(n, z, max_depth=8))
yields
[(8, 7, 6, 5, 4, 3, 2, 1),
(9, 7, 6, 5, 4, 3, 2, 1),
(9, 8, 6, 5, 4, 3, 2, 1),
(9, 8, 7, 5, 4, 3, 2, 1),
(9, 8, 7, 6, 4, 3, 2, 1),
(9, 8, 7, 6, 5, 3, 2, 1),
(9, 8, 7, 6, 5, 4, 2, 1),
(9, 8, 7, 6, 5, 4, 3, 1),
(9, 8, 7, 6, 5, 4, 3, 2)]

As zondo suggested, you'll need to use a function and recursion to accomplish this task. Something along the lines of the following should work:
def recurse(tuplesList, potentialTupleAsList, rangeEnd, z):
# No range to iterate over, check if tuple sum is large enough
if rangeEnd = 1 and sum(potentialTupleAsList) > z:
tuplesList.append(tuple(potentialTupeAsList))
return
for i in range(1, rangeEnd):
potentialTupleAsList.append(i)
recurse(tuplesList, potentialTupleAsList, rangeEnd - 1, z)
# Need to remove item you used to make room for new value
potentialTupleAsList.pop(-1)
Then you could call it as such to get the results:
l = []
recurse(l, [], int(log(z)), z)
print l

Your innermost loop can (if reached at all) only go over range(1, 1). Since the endpoint is not included, the loop will not iterate over any values. The shortest implementation of your function is thus:
def function(z):
return []
If you are content with tuples of length smaller than n, then I propose the following solution:
import math
def function(z):
def f(tuples, loop_variables, product, end):
if product > z:
tuples.append(loop_variables)
for i in range(end - 1, 0, -1):
f(tuples, loop_variables + (i,), product * i, i)
n = int(math.log(z))
tuples = []
f(tuples, (), 1, n)
return tuples
The time complexity is not good though: With n nested loops over O(n) elements, we are on the order of n**n steps.

Pythonic way to add a list of vectors

I am trying to create a method (sum) that takes a variable number of vectors and adds them in. For educational purposes, I have written my own Vector class, and the underlying data is stored in an instance variable named data.
My code for the #classmethod sum works (for each of the vectors passed in, loop through each element in the data variable and add it to a result list), but it seems non-Pythonic, and wondering if there is a better way?
class Vector(object):
def __init__(self, data):
self.data = data
#classmethod
def sum(cls, *args):
result = [0 for _ in range(len(args[0].data))]
for v in args:
if len(v.data) != len(result): raise
for i, element in enumerate(v.data):
result[i] += element
return cls(result)

itertools.izip_longest may come very handy in your situation:
a = [1, 2, 3, 4]
b = [1, 2, 3, 4, 5, 6]
c = [1, 2]
lists = (a, b, c)
result = [sum(el) for el in itertools.izip_longest(*lists, fillvalue=0)]
And here you got what you wanted:
>>> result
[3, 6, 6, 8, 5, 6]
What it does is simply zips up your lists together, by filling empty value with 0. e.g. izip_longest(a, b) would be [(1, 1), (2, 2), (3, 0), (4, 0)]. Then just sums up all the values in each tuple element of the intermediate list.
So here you go step by step:
>>> lists
([1, 2, 3, 4], [1, 2, 3, 4, 5, 6], [1, 2])
>>> list(itertools.izip_longest(*lists, fillvalue=0))
[(1, 1, 1), (2, 2, 2), (3, 3, 0), (4, 4, 0), (0, 5, 0), (0, 6, 0)]
So if you run a list comprehension, summing up all sub-elements, you get your result.

Another thing that you could do (and that might be more "pythonic") would be to implement the __add__ magic method, so you can use + and sum directly on vectors.
class Vector(object):
def __init__(self, data):
self.data = data
def __add__(self, other):
if isinstance(other, Vector):
return Vector([s + o for s, o in zip(self.data, other.data)])
if isinstance(other, int):
return Vector([s + other for s in self.data])
raise TypeError("can not add %s to vector" % other)
def __radd__(self, other):
return self.__add__(other)
def __repr__(self):
return "Vector(%r)" % self.data
Here, I also implemented addition of Vector and int, adding the number on each of the Vector's data elements, and the "reverse addition" __radd__, to make sum work properly.
Example:
>>> v1 = Vector([1,2,3])
>>> v2 = Vector([4,5,6])
>>> v3 = Vector([7,8,9])
>>> v1 + v2 + v3
Vector([12, 15, 18])
>>> sum([v1,v2,v3])
Vector([12, 15, 18])

args = [[1, 2, 3],
[10, 20, 30],
[7, 3, 15]]
result = [sum(data) for data in zip(*args)]
# [18, 25, 48]
Is this what you want?

Accessing consecutive items when using a generator

Lets say I have a tuple generator, which I simulate as follows:
g = (x for x in (1,2,3,97,98,99))
For this specific generator, I wish to write a function to output the following:
(1,2,3)
(2,3,97)
(3,97,98)
(97,98,99)
(98,99)
(99)
So I'm iterating over three consecutive items at a time and printing them, except when I approach the end.
Should the first line in my function be:
t = tuple(g)
In other words, is it best to work on a tuple directly or might it be beneficial to work with a generator. If it is possible to approach this problem using both methods, please state the benefits and disadvantages for both approaches. Also, if it might be wise to use the generator approach, how might such a solution look?
Here's what I currently do:
def f(data, l):
t = tuple(data)
for j in range(len(t)):
print(t[j:j+l])
data = (x for x in (1,2,3,4,5))
f(data,3)
UPDATE:
Note that I've updated my function to take a second argument specifying the length of the window.

A specific example for returning three items could read
def yield3(gen):
b, c = gen.next(), gen.next()
try:
while True:
a, b, c = b, c, gen.next()
yield (a, b, c)
except StopIteration:
yield (b, c)
yield (c,)
g = (x for x in (1,2,3,97,98,99))
for l in yield3(g):
print l

Actually there're functions for this in itertools module - tee() and izip_longest():
>>> from itertools import izip_longest, tee
>>> g = (x for x in (1,2,3,97,98,99))
>>> a, b, c = tee(g, 3)
>>> next(b, None)
>>> next(c, None)
>>> next(c, None)
>>> [[x for x in l if x is not None] for l in izip_longest(a, b, c)]
[(1, 2, 3), (2, 3, 97), (3, 97, 98), (97, 98, 99), (98, 99), (99)]
from documentation:
Return n independent iterators from a single iterable. Equivalent to:
def tee(iterable, n=2):
it = iter(iterable)
deques = [collections.deque() for i in range(n)]
def gen(mydeque):
while True:
if not mydeque: # when the local deque is empty
newval = next(it) # fetch a new value and
for d in deques: # load it to all the deques
d.append(newval)
yield mydeque.popleft()
return tuple(gen(d) for d in deques)

If you might need to take more than three elements at a time, and you don't want to load the whole generator into memory, I suggest using a deque from the collections module in the standard library to store the current set of items. A deque (pronounced "deck" and meaning "double-ended queue") can have values pushed and popped efficiently from both ends.
from collections import deque
from itertools import islice
def get_tuples(gen, n):
q = deque(islice(gen, n)) # pre-load the queue with `n` values
while q: # run until the queue is empty
yield tuple(q) # yield a tuple copied from the current queue
q.popleft() # remove the oldest value from the queue
try:
q.append(next(gen)) # try to add a new value from the generator
except StopIteration:
pass # but we don't care if there are none left

actually it depends.
A generator might be useful in case of very large collections, where you dont really need to store them all in memory to achieve the result you want.
On the other hand, you have to print it is seems safe to guess that the collection isn't huge, so it doesn make a difference.
However, this is a generator that achieve what you were looking for
def part(gen, size):
t = tuple()
try:
while True:
l = gen.next()
if len(t) < size:
t = t + (l,)
if len(t) == size:
yield t
continue
if len(t) == size:
t = t[1:] + (l,)
yield t
continue
except StopIteration:
while len(t) > 1:
t = t[1:]
yield t
>>> a = (x for x in range(10))
>>> list(part(a, 3))
[(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 9), (8, 9), (9,)]
>>> a = (x for x in range(10))
>>> list(part(a, 5))
[(0, 1, 2, 3, 4), (1, 2, 3, 4, 5), (2, 3, 4, 5, 6), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8), (5, 6, 7, 8, 9), (6, 7, 8, 9), (7, 8, 9), (8, 9), (9,)]
>>>
note: the code actually isn't very elegant but it works also when you have to split in, say, 5 pieces

It's definitely best to work with the generator because you don't want to have to hold everything in memory.
It can be done very simply with a deque.
from collections import deque
from itertools import islice
def overlapping_chunks(size, iterable, *, head=False, tail=False):
"""
Get overlapping subsections of an iterable of a specified size.
print(*overlapping_chunks(3, (1,2,3,97,98,99)))
#>>> [1, 2, 3] [2, 3, 97] [3, 97, 98] [97, 98, 99]
If head is given, the "warm up" before the specified maximum
number of items is included.
print(*overlapping_chunks(3, (1,2,3,97,98,99), head=True))
#>>> [1] [1, 2] [1, 2, 3] [2, 3, 97] [3, 97, 98] [97, 98, 99]
If head is truthy, the "warm up" before the specified maximum
number of items is included.
print(*overlapping_chunks(3, (1,2,3,97,98,99), head=True))
#>>> [1] [1, 2] [1, 2, 3] [2, 3, 97] [3, 97, 98] [97, 98, 99]
If tail is truthy, the "cool down" after the iterable is exhausted
is included.
print(*overlapping_chunks(3, (1,2,3,97,98,99), tail=True))
#>>> [1, 2, 3] [2, 3, 97] [3, 97, 98] [97, 98, 99] [98, 99] [99]
"""
chunker = deque(maxlen=size)
iterator = iter(iterable)
for item in islice(iterator, size-1):
chunker.append(item)
if head:
yield list(chunker)
for item in iterator:
chunker.append(item)
yield list(chunker)
if tail:
while len(chunker) > 1:
chunker.popleft()
yield list(chunker)

I think what you currently do seems a lot easier than any of the above. If there isn't any particular need to make it more complicated, my opinion would be to keep it simple. In other words, it is best to work on a tuple directly.

Here's a generator that works in both Python 2.7.17 and 3.8.1. Internally it uses iterators and generators whenever possible, so it should be relatively memory efficient.
try:
from itertools import izip, izip_longest, takewhile
except ImportError: # Python 3
izip = zip
from itertools import zip_longest as izip_longest, takewhile
def tuple_window(n, iterable):
iterators = [iter(iterable) for _ in range(n)]
for n, iterator in enumerate(iterators):
for _ in range(n):
next(iterator)
_NULL = object() # Unique singleton object.
for t in izip_longest(*iterators, fillvalue=_NULL):
yield tuple(takewhile(lambda v: v is not _NULL, t))
if __name__ == '__main__':
data = (1, 2, 3, 97, 98, 99)
for t in tuple_window(3, data):
print(t)
Output:
(1, 2, 3)
(2, 3, 97)
(3, 97, 98)
(97, 98, 99)
(98, 99)
(99,)

Traversing a sequence of generators

I have a sequence of generators: (gen_0, gen_1, ... gen_n)
These generators will create their values lazily but are finite and will have potentially different lengths.
I need to be able to construct another generator that yields the first element of each generator in order, followed by the second and so forth, skipping values from generators that have been exhausted.
I think this problem is analogous to taking the tuple
((1, 4, 7, 10, 13, 16), (2, 5, 8, 11, 14), (3, 6, 9, 12, 15, 17, 18))
and traversing it so that it would yield the numbers from 1 through 18 in order.
I'm working on solving this simple example using (genA, genB, genC) with genA yielding values from (1, 4, 7, 10, 13, 16), genB yielding (2, 5, 8, 11, 14) and genC yielding (3, 6, 9, 12, 15, 17, 18).
To solve the simpler problem with the tuple of tuples the answer is fairly simple if the
elements of the tuple were the same length. If the variable 'a' referred to the tuple, you could use
[i for t in zip(*a) for i in t]
Unfortunately the items are not necessarily the same length and the zip trick doesn't seem to work for generators anyway.
So far my code is horribly ugly and I'm failing to find anything approaching a clean solution. Help?

I think you need itertools.izip_longest
>>> list([e for e in t if e is not None] for t in itertools.izip_longest(*some_gen,
fillvalue=None))
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12], [13, 14, 15], [16, 17], [18]]
>>>

If you look at the documentation for itertools.izip_longest, you'll see that it gives a pure-Python implementation. It's easy to modify this implementation so that it produces the results you need instead (that is, just like izip_longest, but without any fillvalue):
class ZipExhausted(Exception):
pass
def izip_longest_nofill(*args):
"""
Return a generator whose .next() method returns a tuple where the
i-th element comes from the i-th iterable argument that has not
yet been exhausted. The .next() method continues until all
iterables in the argument sequence have been exhausted and then it
raises StopIteration.
>>> list(izip_longest_nofill(*[xrange(i,2*i) for i in 2,3,5]))
[(2, 3, 5), (3, 4, 6), (5, 7), (8,), (9,)]
"""
iterators = map(iter, args)
def zip_next():
i = 0
while i < len(iterators):
try:
yield next(iterators[i])
i += 1
except StopIteration:
del iterators[i]
if i == 0:
raise ZipExhausted
try:
while iterators:
yield tuple(zip_next())
except ZipExhausted:
pass
This avoids the need to re-filter the output of izip_longest to discard the fillvalues. Alternatively, if you want a "flattened" output:
def iter_round_robin(*args):
"""
Return a generator whose .next() method cycles round the iterable
arguments in turn (ignoring ones that have been exhausted). The
.next() method continues until all iterables in the argument
sequence have been exhausted and then it raises StopIteration.
>>> list(iter_round_robin(*[xrange(i) for i in 2,3,5]))
[0, 0, 0, 1, 1, 1, 2, 2, 3, 4]
"""
iterators = map(iter, args)
while iterators:
i = 0
while i < len(iterators):
try:
yield next(iterators[i])
i += 1
except StopIteration:
del iterators[i]

Another itertools option if you want them all collapsed in a single list; this (as #gg.kaspersky already pointed out in another thread) does not handle generated None values.
g = (generator1, generator2, generator3)
res = [e for e in itertools.chain(*itertools.izip_longest(*g)) if e is not None]
print res
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]

You might consider itertools.izip_longest, but in case None is a valid value, that solution will fail. Here is a sample "another generator", which does exactly what you asked for, and is pretty clean:
def my_gen(generators):
while True:
rez = ()
for gen in generators:
try:
rez = rez + (gen.next(),)
except StopIteration:
pass
if rez:
yield rez
else:
break
print [x for x in my_gen((iter(xrange(2)), iter(xrange(3)), iter(xrange(1))))]
[(0, 0, 0), (1, 1), (2,)] #output

Iteration over list slices

I want an algorithm to iterate over list slices. Slices size is set outside the function and can differ.
In my mind it is something like:
for list_of_x_items in fatherList:
foo(list_of_x_items)
Is there a way to properly define list_of_x_items or some other way of doing this using python 2.5?
edit1: Clarification Both "partitioning" and "sliding window" terms sound applicable to my task, but I am no expert. So I will explain the problem a bit deeper and add to the question:
The fatherList is a multilevel numpy.array I am getting from a file. Function has to find averages of series (user provides the length of series) For averaging I am using the mean() function. Now for question expansion:
edit2: How to modify the function you have provided to store the extra items and use them when the next fatherList is fed to the function?
for example if the list is lenght 10 and size of a chunk is 3, then the 10th member of the list is stored and appended to the beginning of the next list.
Related:
What is the most “pythonic” way to iterate over a list in chunks?

If you want to divide a list into slices you can use this trick:
list_of_slices = zip(*(iter(the_list),) * slice_size)
For example
>>> zip(*(iter(range(10)),) * 3)
[(0, 1, 2), (3, 4, 5), (6, 7, 8)]
If the number of items is not dividable by the slice size and you want to pad the list with None you can do this:
>>> map(None, *(iter(range(10)),) * 3)
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]
It is a dirty little trick
OK, I'll explain how it works. It'll be tricky to explain but I'll try my best.
First a little background:
In Python you can multiply a list by a number like this:
[1, 2, 3] * 3 -> [1, 2, 3, 1, 2, 3, 1, 2, 3]
([1, 2, 3],) * 3 -> ([1, 2, 3], [1, 2, 3], [1, 2, 3])
And an iterator object can be consumed once like this:
>>> l=iter([1, 2, 3])
>>> l.next()
1
>>> l.next()
2
>>> l.next()
3
The zip function returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables. For example:
zip([1, 2, 3], [20, 30, 40]) -> [(1, 20), (2, 30), (3, 40)]
zip(*[(1, 20), (2, 30), (3, 40)]) -> [[1, 2, 3], [20, 30, 40]]
The * in front of zip used to unpack arguments. You can find more details here.
So
zip(*[(1, 20), (2, 30), (3, 40)])
is actually equivalent to
zip((1, 20), (2, 30), (3, 40))
but works with a variable number of arguments
Now back to the trick:
list_of_slices = zip(*(iter(the_list),) * slice_size)
iter(the_list) -> convert the list into an iterator
(iter(the_list),) * N -> will generate an N reference to the_list iterator.
zip(*(iter(the_list),) * N) -> will feed those list of iterators into zip. Which in turn will group them into N sized tuples. But since all N items are in fact references to the same iterator iter(the_list) the result will be repeated calls to next() on the original iterator
I hope that explains it. I advice you to go with an easier to understand solution. I was only tempted to mention this trick because I like it.

If you want to be able to consume any iterable you can use these functions:
from itertools import chain, islice
def ichunked(seq, chunksize):
"""Yields items from an iterator in iterable chunks."""
it = iter(seq)
while True:
yield chain([it.next()], islice(it, chunksize-1))
def chunked(seq, chunksize):
"""Yields items from an iterator in list chunks."""
for chunk in ichunked(seq, chunksize):
yield list(chunk)

Use a generator:
big_list = [1,2,3,4,5,6,7,8,9]
slice_length = 3
def sliceIterator(lst, sliceLen):
for i in range(len(lst) - sliceLen + 1):
yield lst[i:i + sliceLen]
for slice in sliceIterator(big_list, slice_length):
foo(slice)
sliceIterator implements a "sliding window" of width sliceLen over the squence lst, i.e. it produces overlapping slices: [1,2,3], [2,3,4], [3,4,5], ... Not sure if that is the OP's intention, though.

Do you mean something like:
def callonslices(size, fatherList, foo):
for i in xrange(0, len(fatherList), size):
foo(fatherList[i:i+size])
If this is roughly the functionality you want you might, if you desire, dress it up a bit in a generator:
def sliceup(size, fatherList):
for i in xrange(0, len(fatherList), size):
yield fatherList[i:i+size]
and then:
def callonslices(size, fatherList, foo):
for sli in sliceup(size, fatherList):
foo(sli)

Answer to the last part of the question:
question update: How to modify the
function you have provided to store
the extra items and use them when the
next fatherList is fed to the
function?
If you need to store state then you can use an object for that.
class Chunker(object):
"""Split `iterable` on evenly sized chunks.
Leftovers are remembered and yielded at the next call.
"""
def __init__(self, chunksize):
assert chunksize > 0
self.chunksize = chunksize
self.chunk = []
def __call__(self, iterable):
"""Yield items from `iterable` `self.chunksize` at the time."""
assert len(self.chunk) < self.chunksize
for item in iterable:
self.chunk.append(item)
if len(self.chunk) == self.chunksize:
# yield collected full chunk
yield self.chunk
self.chunk = []
Example:
chunker = Chunker(3)
for s in "abcd", "efgh":
for chunk in chunker(s):
print ''.join(chunk)
if chunker.chunk: # is there anything left?
print ''.join(chunker.chunk)
Output:
abc
def
gh

I am not sure, but it seems you want to do what is called a moving average. numpy provides facilities for this (the convolve function).
>>> x = numpy.array(range(20))
>>> x
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19])
>>> n = 2 # moving average window
>>> numpy.convolve(numpy.ones(n)/n, x)[n-1:-n+1]
array([ 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5,
9.5, 10.5, 11.5, 12.5, 13.5, 14.5, 15.5, 16.5, 17.5, 18.5])
The nice thing is that it accomodates different weighting schemes nicely (just change numpy.ones(n) / n to something else).
You can find a complete material here:
http://www.scipy.org/Cookbook/SignalSmooth

Expanding on the answer of #Ants Aasma: In Python 3.7 the handling of the StopIteration exception changed (according to PEP-479). A compatible version would be:
from itertools import chain, islice
def ichunked(seq, chunksize):
it = iter(seq)
while True:
try:
yield chain([next(it)], islice(it, chunksize - 1))
except StopIteration:
return

Your question could use some more detail, but how about:
def iterate_over_slices(the_list, slice_size):
for start in range(0, len(the_list)-slice_size):
slice = the_list[start:start+slice_size]
foo(slice)

For a near-one liner (after itertools import) in the vein of Nadia's answer dealing with non-chunk divisible sizes without padding:
>>> import itertools as itt
>>> chunksize = 5
>>> myseq = range(18)
>>> cnt = itt.count()
>>> print [ tuple(grp) for k,grp in itt.groupby(myseq, key=lambda x: cnt.next()//chunksize%2)]
[(0, 1, 2, 3, 4), (5, 6, 7, 8, 9), (10, 11, 12, 13, 14), (15, 16, 17)]
If you want, you can get rid of the itertools.count() requirement using enumerate(), with a rather uglier:
[ [e[1] for e in grp] for k,grp in itt.groupby(enumerate(myseq), key=lambda x: x[0]//chunksize%2) ]
(In this example the enumerate() would be superfluous, but not all sequences are neat ranges like this, obviously)
Nowhere near as neat as some other answers, but useful in a pinch, especially if already importing itertools.

A function that slices a list or an iterator into chunks of a given size. Also handles the case correctly if the last chunk is smaller:
def slice_iterator(data, slice_len):
it = iter(data)
while True:
items = []
for index in range(slice_len):
try:
item = next(it)
except StopIteration:
if items == []:
return # we are done
else:
break # exits the "for" loop
items.append(item)
yield items
Usage example:
for slice in slice_iterator([1,2,3,4,5,6,7,8,9,10],3):
print(slice)
Result:
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
[10]