Split Python list into several lists based on index

Question

So I've got a string of bytes which represents cubes in three dimensions. The coordinates are ordered like this:

[x0y0z0, x0y1z0, x0y2z0, ..., x0y127z0, x0y0z1, x0y1z1, ..., x15y127z15]

I'd like to split this into 128 lists, one for each Y coordinate. This code already does that, but I think inefficiently. Is there some way to split this list based on mod(128) of the index?

From the original code:

col.extend(izip_longest(*[iter(file["Level"]["Blocks"].value)]*128))

That takes quite a while, and I think it should be possible to make something better performing by avoiding the *128 part of this. But zipping is definitely not my strong side, and neither is binary file handling.

Your example isn't clear. Is it a list of strings or a string? Why are you showing a y127? is this the 127'th y coordinate? I thought there were only 16. — aaronasterling
– aaronasterling, Commented Oct 29, 2010 at 10:55
This is ambiguous. Please provide code showing actual input data (not pseudocode that requires interpretation), and the expected result from that data. — Glenn Maynard
– Glenn Maynard, Commented Oct 29, 2010 at 10:57
The *128 part does not take long, you are just making 128 references to the same iterator — John La Rooy
– John La Rooy, Commented Oct 29, 2010 at 12:33

Ignacio Vazquez-Abrams · Accepted Answer · 2010-10-29 10:50:52Z

3

# l = [x0y0z0, ...]
def bucketsofun(l, sp=16):
  r = [[] for x in range(sp)]
  for b, e in itertools.izip(itertools.cycle(r), l):
    b.append(e)
  return r

answered Oct 29, 2010 at 10:50

Ignacio Vazquez-Abrams

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2 Comments

l0b0 Over a year ago

Do you have more descriptive names for b and e?

Ignacio Vazquez-Abrams Over a year ago

bucket and element, if you prefer.

John La Rooy · Accepted Answer · 2010-10-29 12:47:13Z

2

Something like this might be worth trying

L = file["Level"]["Blocks"].value
col += [L[i::128] for i in range(127)]

answered Oct 29, 2010 at 12:47

John La Rooy

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Comments

Jochen Ritzel · Accepted Answer · 2010-10-29 11:27:24Z

0

Itertools can do it too:

from itertools import izip, izip_longest, chain

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

# you have something like this
# [0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4]
l = list(chain.from_iterable(range(5) for _ in range(5)))

# you want to put all the 0s, 1s (every 5th element) together
print list(izip(*grouper(5, l)))
# [(0, 0, 0, 0, 0), (1, 1, 1, 1, 1), ... , (4, 4, 4, 4, 4)]

answered Oct 29, 2010 at 11:27

Jochen Ritzel

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1000101cz · Accepted Answer · 2023-01-04 17:42:32Z

0

Faster way to do this is using reshape() function from NumPy

original_1D_array = [1, 2, 3, 4, 5, 6]
modulo = 3

new_2D_array = np.reshape(original_1D_array, (-1, modulo))

# new_2D_array = [[1, 2, 3], [4, 5, 6]]

Now we just need to transpose this matrix

new_2D_array = new_2D_array.T

# new_2D_array = [[1, 4], [2, 5], [3, 6]]

edited Jan 4, 2023 at 17:42

answered Jan 4, 2023 at 17:41

1000101cz

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