如何拆分每个第 N 个元素的 Python 列表

Question

What I am trying to do is pretty simple, but I couldn't find how to do it.

• Starting with 1st element, put every 4th element into a new list.
• Repeat with the 2nd, 3rd, and 4th elements.

From:

list = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']

To:

list1 = ['1', '5', '9']
list2 = ['2', '6', 'a']
list3 = ['3', '7', 'b']
list4 = ['4', '9']

In other words, I need to know how to:

• Get the Nth element from a list (in a loop)
• Store it in new arrays
• Repeat

Answer 1

The specific solution is to use slicing with a stride:

source = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']
list1 = source[::4]
list2 = source[1::4]
list3 = source[2::4]
list4 = source[3::4]

source[::4] takes every 4th element, starting at index 0; the other slices only alter the starting index.

The generic solution is to use a loop to do the slicing, and store the result in an outer list; a list comprehension can do that nicely:

def slice_per(source, step):
    return [source[i::step] for i in range(step)]

Demo:

>>> source = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']
>>> source[::4]
['1', '5', '9']
>>> source[1::4]
['2', '6', 'a']
>>> def slice_per(source, step):
...     return [source[i::step] for i in range(step)]
... 
>>> slice_per(source, 4)
[['1', '5', '9'], ['2', '6', 'a'], ['3', '7', 'b'], ['4', '8']]
>>> slice_per(source, 3)
[['1', '4', '7', 'a'], ['2', '5', '8', 'b'], ['3', '6', '9']]

Answer 2

The numbered names are a bad idea, and you shouldn't name your own variable list (it shadows the built-in), but in general you can do something like:

>>> startlist = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']
>>> n = 4
>>> endlist = [[] for _ in range(n)]
>>> for index, item in enumerate(startlist):
    endlist[index % n].append(item)


>>> endlist
[['1', '5', '9'], ['2', '6', 'a'], ['3', '7', 'b'], ['4', '8']]

Answer 3

lst = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n'] s = [] n = 3 for x in range(n): s.append(lst[x::n]) print(s)

Answer 4

Alternative approach:

list = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']
lists = []
list_temp = []

for _, __ in enumerate(list, 1):
  list_temp.append(__)
  if _ % 3 == 0 or _ == len(list):
    lists.append(list_temp)
    list_temp = []

This will append the chunks in a temporary list and append the temporary list to our final list every time the index divided by the desired list length has no remainders.

It's more a mathematical approach than a pythonic one :)