Hey all. I've decided I let my Python skills (minor though they were) slip away so I started reading the new edition of Learning Python to brush up. I just read about lists again and I'm wondering if someone could explain what's going on under the hood that makes index and slice assignments behave differently when assigning an empty list.
For example:
>>> L = [1, 2, 3, 4, 5] >>> L[0:2] = [] >>> L [3, 4, 5]
>>> L = [1, 2, 3, 4, 5] >>> L[0] = [] >>> L [[], 2, 3, 4, 5]
So the question is, when you assign an empty list to an index, why does it insert an empty list, but when you assign an empty list to a slice, it simply deletes the slice?
John Salerno wrote: > So the question is, when you assign an empty list to an index, why does > it insert an empty list, but when you assign an empty list to a slice, > it simply deletes the slice?
I would say this is consistent behavior because a list slice is also a list itself. Whereas a list element is just that. A reference to an object that can be rebound. In the latter case you are rebinding a single list item to an empty list.
John Salerno wrote: > Hey all. I've decided I let my Python skills (minor though they were) > slip away so I started reading the new edition of Learning Python to > brush up. I just read about lists again and I'm wondering if someone > could explain what's going on under the hood that makes index and slice > assignments behave differently when assigning an empty list.
> So the question is, when you assign an empty list to an index, why does > it insert an empty list, but when you assign an empty list to a slice, > it simply deletes the slice?
Assignment to a list *element* rebinds the single element to the assigned value. Assignment to a list *slice* has to be of a list, and it replaces the elements in the slice by assigned elements.
regards Steve -- Steve Holden +1 571 484 6266 +1 800 494 3119 Holden Web LLC http://www.holdenweb.com/
Steve Holden <st...@holdenweb.com> wrote: > Assignment to a list *element* rebinds the single element to the > assigned value. Assignment to a list *slice* has to be of a list, and it > replaces the elements in the slice by assigned elements.
Assignment to a list *slice* just has use an iterable, it doesn't actually have to be another list.
Steve Holden wrote: > Assignment to a list *element* rebinds the single element to the > assigned value.
Ok, I understand that.
Assignment to a list *slice* has to be of a list [or iterable, as per Duncan], and it
> replaces the elements in the slice by assigned elements.
I don't understand the second part of that sentence. I'm assuming "it" refers to the list being assigned, "replaces the elements" is self-evident, but what does "by assigned elements" refer to? It seems when you assign a list to a list slice, nothing gets replaced, the slice just gets deleted.
>> Assignment to a list *element* rebinds the single element to the >> assigned value.
> Ok, I understand that.
> Assignment to a list *slice* has to be of a list [or iterable, as per > Duncan], and it >> replaces the elements in the slice by assigned elements.
> I don't understand the second part of that sentence. I'm assuming "it" > refers to the list being assigned, "replaces the elements" is > self-evident, but what does "by assigned elements" refer to? It seems > when you assign a list to a list slice, nothing gets replaced, the slice > just gets deleted.
Deletion occurs *only* in the corner case where there are no "assigned elements" i.e. only if the RHS list (sequence) is *empty*. Otherwise there would be no point at all in the language having assignment to a slice -- del L[0:2] would suffice.
Study these:
>>> L = [1, 2, 3, 4, 5] >>> L[0:2] = [] >>> L [3, 4, 5] >>> L = [1, 2, 3, 4, 5] >>> L[0:2] = ['whatever'] >>> L ['whatever', 3, 4, 5] >>> L = [1, 2, 3, 4, 5] >>> L[0:2] = tuple('foobar') >>> L ['f', 'o', 'o', 'b', 'a', 'r', 3, 4, 5] >>>
John Salerno wrote: >> replaces the elements in the slice by assigned elements.
> I don't understand the second part of that sentence. I'm assuming "it" > refers to the list being assigned, "replaces the elements" is > self-evident, but what does "by assigned elements" refer to? It seems > when you assign a list to a list slice, nothing gets replaced, the slice > just gets deleted. >>> x = range(5) >>> x[0:3] = ["a", "b"] >>> x
['a', 'b', 3, 4]
Here, '= ["a", "b"]' replaces x[0:3] with ["a", "b"]. When you do '= []', it replaces them with nothing. --
John Machin wrote: > Deletion occurs *only* in the corner case where there are no "assigned > elements" i.e. only if the RHS list (sequence) is *empty*.
Oh, it was my understanding that deletion always occurs, even when the section is being assigned a non-empty value, i.e. delete the slice and insert new value.
Otherwise
> there would be no point at all in the language having assignment to a > slice -- del L[0:2] would suffice.
Right, but I'm wondering why a statement like
L[0:2] = []
doesn't assign an empty list as the new element in L. For example:
L = [1, 2, 3, 4, 5] L[0:2] = []
Why doesn't L now equal [[], 3, 4, 5] as it does with an index assignment?
| | > Deletion occurs *only* in the corner case where there are no "assigned | > elements" i.e. only if the RHS list (sequence) is *empty*. | | Oh, it was my understanding that deletion always occurs, even when the | section is being assigned a non-empty value, i.e. delete the slice and | insert new value.
Slice replacement means replace the slice with a new slice generated from the iterable on the left. John meant that deletion only only happens when the replacement is empty. Yes, deletion always occurs, but usually addition also occurs, so the net result is replacement rather than just deletion.
| Otherwise | > there would be no point at all in the language having assignment to a | > slice -- del L[0:2] would suffice. | | Right, but I'm wondering why a statement like | L[0:2] = [] | doesn't assign an empty list as the new element in L. For example:
Because, as others already told you, slice replacement is slice replacement, not item assignment. When you say to replace the slice with nothing, the deleted slice is replaced with nothing.
L[0:2] = [[]]
says to replace the slice with a slice consisting of one item -- [] That will get you what you are expecting.
| L = [1, 2, 3, 4, 5] | L[0:2] = [] | | Why doesn't L now equal [[], 3, 4, 5] as it does with an index assignment?
See above.
| | > >>> L[0:2] = tuple('foobar')
L[0:2] = 'foobar' has same effect because s string is an iterable.
| > >>> L | > ['f', 'o', 'o', 'b', 'a', 'r', 3, 4, 5] | | Hmm...why doesn't L equal [('f', 'o', 'o', 'b', 'a', 'r'), 3, 4, 5] ? | Shouldn't L be a 4 item list instead of 9?
Because you replaced 2 items with 6.
L[0:2] = ['foobar'] will replace 2 with 1, leaving 4
