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Slicing Python List with inconsistent intervals

I have a list of stock prices of a company. Now I want to split the list with multiple intervals. we will store the price like: The first 2 elements, then next 3 elements, then 2 elements, and so on.

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]

Output

meta_stocks = [[10, 9],[11, 15, 19],[22, 25],[ 11, 15, 17]]

I am able to split the list with 5 items each but not able to split it further

>>> [meta_stocks[i:i+interval2] for i in range(0, len(meta_stocks), interval2)]
>>> [[10, 9, 11, 15, 19], [22, 25, 11, 15, 17]]
over 4 years ago · Santiago Trujillo
3 answers
Answer question

0

You can use a list comprehension with the help of itertools.cycle:

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]

from itertools import cycle

start = 0
l = [2,3]
c = cycle(l)

[meta_stocks[start:(start:=start+next(c))]
 for i in range(len(l)*len(meta_stocks)//sum(l))]

Output:

[[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]
over 4 years ago · Santiago Trujillo Report

0

You could do it like this without the aid of additional imports:

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]
meta_stocks_out = []
offset = 0
interval = 2
while offset + interval <= len(meta_stocks):
    meta_stocks_out.append(meta_stocks[offset:offset+interval])
    offset += interval
    interval = 2 if interval == 3 else 3
print(meta_stocks_out)
over 4 years ago · Santiago Trujillo Report

0

Some itertools to the rescue:

from itertools import islice, cycle, takewhile

i = iter(meta_stocks)
intervals = [2, 3]

[*takewhile(lambda _: _, ([*islice(i, n)] for n in cycle(intervals)))]
# [[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]
over 4 years ago · Santiago Trujillo Report
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