Creating a sequentially joined list using a for loop in Python

Question

I need help creating a new list z, where I will have list values as y0, x0, y1, x1, y2, x2, y3 in vertical order as an array.

output: z = [1, 1, 1, 0, 0, 2, 2, 2, 2, 4, 4, 3, 3, 3, 8, 8, 4, 4, 4]

I tried this for loop iteration, but instead of the desired list z, I get only list values as y2, x2.

I really appreciate any help you can provide.

x = [[0, 0], [4, 4], [8, 8]]
y = [[1, 1, 1], [2, 2, 2], [3, 3, 3 ], [4, 4, 4]

for i in range (0, 3):
    z = [*y[i], *x[i]]

Asked By: phoenix_dj

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Source

Answer 1

You can use zip and list comprehension.

x = [[0, 0], [4, 4], [8, 8]]
y = [[1, 1, 1], [2, 2, 2], [3, 3, 3 ]]

res = [[*j, *i] for i,j in zip(x,y)]
print(res)

[[1, 1, 1, 0, 0], [2, 2, 2, 4, 4], [3, 3, 3, 8, 8]]

Answered By: I'mahdi

Answer 2

I found the solution as:

from heapq import merge
from itertools import count

x = [[1, 1, 1], [2, 2, 2], [3, 3, 3 ], [4, 4, 4]]
y = [[0, 0], [4, 4], [8, 8]]
counter = count()
z = np.hstack(list(merge(x, y, key=lambda x: next(counter))))

Answered By: phoenix_dj

Answer 3

Solution without for loop, first use zip_longest zipped the two lists, chain the results of per loop with chain.from_iterable, and then flattened them with list(chain.from_iterable(iterable_of_list)):

>>> from itertools import chain, zip_longest
>>> list(chain.from_iterable(map(chain.from_iterable, zip_longest(y, x, fillvalue=()))))
[1, 1, 1, 0, 0, 2, 2, 2, 4, 4, 3, 3, 3, 8, 8, 4, 4, 4]

Or use reduce and iconcat to flatten:

>>> from functools import reduce
>>> from operator import iconcat
>>> reduce(iconcat, map(chain.from_iterable, zip_longest(y, x, fillvalue=())), [])
[1, 1, 1, 0, 0, 2, 2, 2, 4, 4, 3, 3, 3, 8, 8, 4, 4, 4]

Answered By: Mechanic Pig

Creating a sequentially joined list using a for loop in Python

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