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]

Solution 1:

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)

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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[:])