about the design of Part related to Association

Question

<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1]]

gives values

while

<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1;;2]]

gives

<|"a" -> 1, "b" -> 2|>

I just feel an inconsistency in this kind of Part design.

Why

<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1]]

doesn't give

"a" -> 1

as

{"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4}[[1]]

gives

"a" -> 1

or as many people point out that functions act transparently on Association's values, why doesn't

<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1;;2]]

give

{1,2}

??? Can somebody give an explanation?

I think I figured it. To obtain the whole part, i.e. (key->value), one must use [[n;;n]] syntax or [[n;;]]. But if you use [[n]] this will only give the value at part n. The reason why [[n;;m]] work for both key,values at both n through m parts, is because of the ;; usage. So to pull both key+value, use ;;, else it will default to only the value at that part. — Nasser, Dec 13 '14 at 4:55
@Nasser Yeah, it is definitely not a bug. They designed it in this way! They write these example in the help doc. But it just doesn't consistent with Part of List. — matheorem, Dec 13 '14 at 4:55
Yes, I think as well it is consistent. Observe that for an expression f[x,y,z,u,v,w], the Part function returns an argument only when it is called with a single number, in all other cases, such as 1;;2, {3,2,1}, the result has head f. With Associations, it is almost the same. Part with a number gives the value, in all other cases it gives an Association — Fred Simons, Dec 13 '14 at 8:27

Murta · Accepted Answer · 2014-12-13 12:02:03Z

7

IMO the behavior is consistent, and acts like in Lists.

For Lists we have:

{"a", "b", "c", "d"}[[1]]
{"a", "b", "c", "d"}[[{1}]]

"a"
{"a"}

and for Associations:

<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1]]
<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[{1}]]

1
<|"a" -> 1|>

For for the Spam form ;; the result is always like a list inside Part, so:

<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[{1}]]
<|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1;;1]]

are equivalent.

answered Dec 13 '14 at 9:03

Murta

22.7k4 gold badges57 silver badges136 bronze badges

1

$\begingroup$ I agree, the behaviour of Part with Association is consistent and expected. $\endgroup$ – Simon Woods Dec 13 '14 at 11:05
$\begingroup$ @SimonWoods But why <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4|>[[1]] doesn't give "a" -> 1? $\endgroup$ – matheorem Dec 13 '14 at 12:10
$\begingroup$ @matheorem, because the value is not the rule "a"->1, it is just 1. Note that FreeQ[<|"a"->1|>, Rule] is True - there is no Rule in that expression, despite appearances. $\endgroup$ – Simon Woods Dec 13 '14 at 13:05
$\begingroup$ @SimonWoods But the FullForm[<|"a"->1|>] gives Association[Rule["a",1]]. There is Rule. $\endgroup$ – matheorem Dec 13 '14 at 13:14
1

$\begingroup$ @matheorem, there's some discussion of that here. I don't claim to fully understand the situation, but I have found that associations mostly make sense if you regard the values as being what's really there at level 1, and the keys as ephemeral magic labels which are transparent to most operations and which vanish as soon as the value is no longer contained inside an association. $\endgroup$ – Simon Woods Dec 13 '14 at 13:25

| show 2 more comments

1 Answer 1