List getting extended when assigned to itself

K

Krishnan Shankar

Hi Python Friends,

I came across an example which is as below,
var = [1, 12, 123, 1234]
var [1, 12, 123, 1234]
var[:0] []
var[:0] = var
var [1, 12, 123, 1234, 1, 12, 123, 1234]

Here in var[:0] = var we are assigning an entire list to the beginning of
itself. So shouldn't it be something like,

[[1, 12, 123, 1234], 1, 12, 123, 1234]

It happens when we do the below,
var = [1, 12, 123, 1234]
var[0] = var
var [[...], 12, 123, 1234]

Literally var[0] = var and var[:0] = var almost meens the same. But why is
the difference in output? Can anyone explain what happens when
slicing assignment and direct assignment.

Regards,
Krishnan
 
S

Steven D'Aprano

Hi Python Friends,

I came across an example which is as below,
var = [1, 12, 123, 1234]
var [1, 12, 123, 1234]
var[:0] []
var[:0] = var
var [1, 12, 123, 1234, 1, 12, 123, 1234]
Here in var[:0] = var we are assigning an entire list to the beginning
of itself. So shouldn't it be something like,

[[1, 12, 123, 1234], 1, 12, 123, 1234]

No, you have misunderstood how slicing works. When you assign to a slice,
you *replace* the existing slice with items from the other list:


py> L = [1, 2, 3, 4, 5]
py> L[2:4] = [None, None, None]
py> L
[1, 2, None, None, None, 5]


Notice that it does not insert the list [None, None, None] as a single
item.

If the slice you replace is empty, this is equivalent to inserting the
items:

py> L = [1, 2, 3, 4, 5]
py> L[2:2] = [None, None, None]
py> L
[1, 2, None, None, None, 3, 4, 5]

The beginning of the list is just an empty slice: L[:0] means "all the
items, starting at the beginning of the list, and finishing *before*
index zero". So that makes it an empty slice, and items are inserted
after the start of the list but before index zero:


py> L = [1, 2, 3, 4, 5]
py> L[:0] = [None, None, None]
py> L
[None, None, None, 1, 2, 3, 4, 5]


So assigning to a slice *always* extends. If you try to assign a single
value, not inside a list or other iterable, it fails:

py> L = [1, 2, 3, 4, 5]
py> L[:0] = None
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable


Remember that indexes in Python are *before* the item, so when slicing,
Python cuts at the vertical lines | as shown below:

| a | b | c | d | e | f | g |

The vertical lines are numbered 0 through 7, and the slice [2:5] cuts as
shown:

| a | b | c | d | e | f | g |
0---1---2---3---4---5---6---7

[2:5] => | c | d | e |


It happens when we do the below,
var = [1, 12, 123, 1234]
var[0] = var
var
[[...], 12, 123, 1234]


Here you are *not* assigning to a slice, but setting a single item. The
item at position 0 is replaced with the contents of var, which happens to
be the same list, but that is just a distraction. It is more clear when
you use a different value:

py> L = [1, 2, 3, 4, 5]
py> L[0] = "something"
py> L
['something', 2, 3, 4, 5]

Literally var[0] = var and var[:0] = var almost meens the same.

No, they are very different.

Even though the difference is a single character, var[:index] and
var[index] are very different, no matter what the value of index. The
first is a slice ending at index, the second is a single item at index.

The same naturally applies to var[index:] as well, which is a slice
starting at index.

If you wish to insert a sequence as a single object, you have two
choices: you can use the list insert() method, or you can wrap the
sequence in a list, and then use slice assignment:


py> L = [1, 2, 3, 4, 5]
py> L[2:4] = [L]
py> L
[1, 2, [...], 5]
 

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