Как разделить список на количество n подсписков(на n-частей)?

Question

[13_nastya_13]

Мне нужно например разделить список [1, 2 , 2, 3, 4, 3] на [1, 2], [2, 3], [4, 3]

Answer 1

[Miit]

Как вариант:

data = [1, 2, 2, 3, 4, 3]
offset = 0
length = len(data)

while offset + 2 <= length:
    print(data[offset:offset + 2])
    offset += 2

UPD: Краткий вариант записи

data = [1, 2, 2, 3, 4, 3]
out = list(zip(*[iter(data)] * 2))  # [(1, 2), (2, 3), (4, 3)]

Answer 2

[Andrey Dugin]

Классику надо знать:

a = [1, 2, 2, 3, 4, 3]

n = 2
list(zip(*[iter(a)]*n))

Или через numpy:

import numpy as np

np.array(a).reshape(-1, n)

Comments

[13_nastya_13]

первый вариант делает список с таплами

Answer 3

[ScriptKiddo]

https://pypi.org/project/more-itertools/

from more_itertools import divide

for n in range(2, 7):
    print(f'n: {n} - {[list(x) for x in divide(n=n, iterable=[1, 2, 2, 3, 4, 3])]}\n')

OUT

n: 2 - [[1, 2, 2], [3, 4, 3]]

n: 3 - [[1, 2], [2, 3], [4, 3]]

n: 4 - [[1, 2], [2, 3], [4], [3]]

n: 5 - [[1, 2], [2], [3], [4], [3]]

n: 6 - [[1], [2], [2], [3], [4], [3]]


Process finished with exit code 0

Comments

[13_nastya_13]

А можно как-то без импорта сторонних библиотек?

[ScriptKiddo]

13_nastya_13, можно. Но если задача для практики - то лучше написать свой костыльный вариант

https://github.com/more-itertools/more-itertools/b...

Code

def divide(n, iterable):
    """Divide the elements from *iterable* into *n* parts, maintaining
    order.
        >>> group_1, group_2 = divide(2, [1, 2, 3, 4, 5, 6])
        >>> list(group_1)
        [1, 2, 3]
        >>> list(group_2)
        [4, 5, 6]
    If the length of *iterable* is not evenly divisible by *n*, then the
    length of the returned iterables will not be identical:
        >>> children = divide(3, [1, 2, 3, 4, 5, 6, 7])
        >>> [list(c) for c in children]
        [[1, 2, 3], [4, 5], [6, 7]]
    If the length of the iterable is smaller than n, then the last returned
    iterables will be empty:
        >>> children = divide(5, [1, 2, 3])
        >>> [list(c) for c in children]
        [[1], [2], [3], [], []]
    This function will exhaust the iterable before returning and may require
    significant storage. If order is not important, see :func:`distribute`,
    which does not first pull the iterable into memory.
    """
    if n < 1:
        raise ValueError('n must be at least 1')

    try:
        iterable[:0]
    except TypeError:
        seq = tuple(iterable)
    else:
        seq = iterable

    q, r = divmod(len(seq), n)

    ret = []
    stop = 0
    for i in range(1, n + 1):
        start = stop
        stop += q + 1 if i <= r else q
        ret.append(iter(seq[start:stop]))

    return ret