12

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
octosquidopus
  • 3,517
  • 8
  • 35
  • 53

4 Answers4

23

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']]
Martijn Pieters
  • 1,048,767
  • 296
  • 4,058
  • 3,343
  • Why not make it a generator? It will handle the infinite case too then. – Edgar Klerks Nov 15 '14 at 11:46
  • @EdgarKlerks: You'd have to make `n` `islice` iterators then. – Martijn Pieters Nov 15 '14 at 11:47
  • @EdgarKlerks: which means you'd have to tee off the input iterable, and that could potentially become very expensive, memory wise. – Martijn Pieters Nov 15 '14 at 11:48
  • 3
    @EdgarKlerks: it appears you were thinking of [How do you split a list into evenly sized chunks in Python?](http://stackoverflow.com/q/312443).. – Martijn Pieters Nov 15 '14 at 11:52
  • You are right, I didn't read the original question too well. My fault. – Edgar Klerks Nov 15 '14 at 11:52
  • Do note that **this will change the order of the items in the original list**. If you want `[[0, 1, 2], [3, 4, 5]]` style output, this will not work. (Yes, for this question the above answer is correct, but I was looking for this kind of output and stumbled upon this on google.) Ended up using [this](https://stackoverflow.com/a/1624988) instead. – zertap Oct 25 '17 at 22:29
  • 1
    @zertap the order is not altered, we select every 4th element *in order*. What you want is something different, you want to take fixed-size chunks, see [What is the most "pythonic" way to iterate over a list in chunks?](https://stackoverflow.com/q/434287). – Martijn Pieters Oct 25 '17 at 22:35
5

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']]
jonrsharpe
  • 115,751
  • 26
  • 228
  • 437
2
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)
Mohmmad Qudah
  • 31
  • 2
  • 3
    An explanation is required for such an answer even if the explanation is one line long and consists roughly of one sentence. Consider updating your answer. – Jenea Vranceanu Sep 05 '20 at 17:47
0

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 :)

Tibor
  • 129
  • 1
  • 10