3

this might look like a simple question but I could not figure out the solution.

Assuming I have this list

my_list = [1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99]

I am attempting to slice item in this list and increment it by index. In short, I want to get the result like this

[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
[4, 5, 6]
[5, 6, 7]
[6, 7, 8]
[7, 8, 9]
[8, 9, 11]
[9, 11, 22]
[11, 22, 33]
[22, 33, 44]
[33, 44, 55]
[44, 55, 66]
[55, 66, 77]
[66, 77, 88]
[77, 88, 99]

But what I got is like this

[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
[4, 5, 6]
[5, 6, 7]
[6, 7, 8]
[7, 8, 9]
[8, 9, 11]
[9, 11, 22]
[11, 22, 33]
[22, 33, 44]
[33, 44, 55]
[44, 55, 66]
[55, 66, 77]
[66, 77, 88]
[77, 88, 99]
[88, 99]
[99]

I want to slice my_list to several smaller set each with length of 3. But the result I got also included the list that have length below than 3.

This is the python script I use

my_list = [1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99]

m = 0

def incr():
    global m
    for n in range(len(my_list)):
        if n < len(my_list):
            n = m + 3
            new_list = my_list[m:n]
            print new_list
            m+=1
        else:
            pass

incr()

I figured that I could add if len(new_list) < 3 but I do not quite sure how should I modify the script for that.

Thank you for your help.

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4 Answers 4

Reset to default
4 
[my_list[i:i+3] for i in range(len(my_list)-2)]

Little explanation:

  • I used List Comprehensions

  • The number of items the resulting list have is len(my_list)-2)

  • Each result item is built slicing three consecutive items from the original list [i:i+3]

  • i started at 0 to get the first item of the result list and then shift one position (being incremented by one) to get the second item and so on..

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3 Comments

thank you. other answers is great but your answer looks much simpler that I can re-use it inside more complicated script.
Please consider adding some explanation as well. That would be helpful to OP as well as future viewers.
@AshwiniChaudhary Done ;)
3

You want to iterate in chunks, with an overlap. It's a common requirement and in many applications that's called a rolling window. You may use tee from itertools:

>>> def thinger(iterable, n=3):
...     iterators = tee(iterable, n)
...     for i, iterator in enumerate(iterators):
...         for skip in range(i):
...             next(iterator, None)
...     return zip(*iterators)
... 
>>> list(thinger([1,2,3,4,5,6]))
[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]
>>> list(thinger([1,2,3,4,5,6], n=4))
[(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6)]

The zip actually returns tuples, which is likely be no problem if you're duck-typing, but you can convert them to lists if necessary: 

>>> [list(x) for x in thinger([1,2,3], n=2)]
[[1, 2], [2, 3]]
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Comments

1 
my_list = [1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99]

def temp(lst,no):
    result = []
    for i in range(len(lst)-no+1):
        result.append(lst[i:i+no])
    return result

print temp(my_list,3)    
print temp(my_list,4)

Output

[[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [5, 6, 7], [6, 7, 8], [7, 8, 9], [8, 9, 11], [9, 11, 22], [11, 22, 33], [22, 33, 44], [33, 44, 55], [44, 55, 66], [55, 66, 77], [66, 77, 88], [77, 88, 99]]                                        

[[1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6], [4, 5, 6, 7], [5, 6, 7, 8], [6, 7, 8, 9], [7, 8, 9, 11], [8, 9, 11, 22], [9, 11, 22, 33], [11, 22, 33, 44], [22, 33, 44, 55], [33, 44, 55, 66], [44, 55, 66, 77], [55, 66, 77, 88], [66, 77, 88, 99]]
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0

like this:

my_list = [1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99]

m = 0

def incr():
    global m
    for n in range(len(my_list)):
        if n+3 <= len(my_list):
            n = m + 3
            new_list = my_list[m:n]
            print new_list
            m +=1  
        else:
            pass

incr()
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2 Comments

It is not I want to achieve. as you can see my expected result is like this --> [1,2,3] [2,3,4]... while your answer gave me this result --> [1,2,3] [4,5,6]
sorry， answer updated， try again.

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