CODEWITHSUNDEEP
pythonpython-3.xslice
SaukratesK
SaukratesK

Reputation: 51

Multiple slicing assignement with Python

I was reading a post called "Duplicate each member in a list". I found an answer to solve this problem that I can't understand:

>>> a = [3, 1, 4, 1, 5]
>>> a[:0] = a[::2] = a[1::2] = a[:]
>>> a
[3, 3, 1, 1, 4, 4, 1, 1, 5, 5]

Could you explain to me, step by step, how the following line works?

>>> a[:0] = a[::2] = a[1::2] = a[:]

Upvotes: 3

Views: 232

Answers (1)

xjcl
xjcl

Reputation: 15309

The key here is understanding that the sequence is executed left-to-right, all with reference to the rightmost value.

The rightmost slice a[:] gets assigned to each of the 3 leftmost slices. Note the use of a[:] instead of a, this ensures we make one copy at the beginning and our operations on a don't change this value. A clearer way to write this would be to store the copy a[:]:

>>> a = [3, 1, 4, 1, 5]
>>> b = a[:]
>>> b
[3, 1, 4, 1, 5]

>>> a[:0] = b
>>> a
[3, 1, 4, 1, 5, 3, 1, 4, 1, 5]
#^^^^^^^^^^^^^ changed

>>> a[::2] = b
>>> a
[3, 1, 1, 1, 4, 3, 1, 4, 5, 5]
#^     ^     ^     ^     ^ changed 

>>> a[1::2] = b
>>> a
[3, 3, 1, 1, 4, 4, 1, 1, 5, 5]
#   ^     ^     ^     ^     ^ changed

Let us discuss the individual steps

  1. b = a[:] makes a temp copy of a
  2. a[:0] = b inserts b at the front, but is actually just used for padding a to twice the length, so that the next two statements work
  3. a[::2] = b sets the values at even indices (i.e. every second index starting from the front)
  4. a[1::2] = b sets the values at odd indices (i.e. every second index excluding the front)

Another way to get the same result is the following tuple assignment, which also evaluates the right-hand-side first, and then the left-hand-side from left-to-right:

(a[:0], a[::2], a[1::2]) = (a[:], a[:], a[:])

Upvotes: 4

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