Multiple Unpacking Assignment in Python when you don’t know the sequence length

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Question :

Multiple Unpacking Assignment in Python when you don’t know the sequence length

The textbook examples of multiple unpacking assignment are something like:

import numpy as NP
M = NP.arange(5)
a, b, c, d, e = M
# so of course, a = 0, b = 1, etc.

M = NP.arange(20).reshape(5, 4)     # numpy 5x4 array
a, b, c, d, e = M
# here, a = M[0,:], b = M[1,:], etc. (ie, a single row of M is assigned each to a through e)

(My question is not numpy specific. Indeed, I would prefer a pure Python solution.)

For the piece of code I’m looking at now, I see two complications on that straightforward scenario:

  • I usually won’t know the shape of M; and

  • I want to unpack a certain number of
    items (definitely less than all items), and
    I want to put the remainder into a single
    container

So back to the 5×4 array above, what I would very much like to do is assign the first three rows of M to a, b, and c respectively (exactly as above), and the rest of the rows (I have no idea how many there will be, just some positive integer) to a single container, all_the_rest = [].

Asked By: doug

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Answer #1:

Python 3.x can do this easily:

a, b, *c = someseq

Python 2.x needs a bit more work:

(a, b), c = someseq[:2], someseq[2:]

Answer #2:

Syntax for this is added to Python 3

>>> # Python 3.x only
>>> a, b, *c = range(10)
>>> a
0
>>> b
1
>>> c
[2, 3, 4, 5, 6, 7, 8, 9]

but no similar solution exists in Python 2.

You can of course do

>>> s = range(10)
>>> s
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> (a, b, c), rest = s[0:3], s[3:]
>>> a
0
>>> b
1
>>> c
2
>>> rest
[3, 4, 5, 6, 7, 8, 9]

or other similar solutions.

Answered By: Mike Graham

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