[R] all.equal and use.names

[Bert Gunter]

Nope. You misread I think. It says that use.names = TRUE causes mismatches
to be **reported** by name rather than index, not that it is recursing by
name. It still recurses by component indices.

However, I still think that is wrong. It is not reporting mismatches **by**
name -- it is reporting mismatches **in** names as well as in value.


Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, May 27, 2020 at 8:23 PM John Harrold <john.m.harrold using gmail.com>
wrote:

> Howdy Folks,
>
> I believe I'm having trouble understanding the documentation for all.equal.
> If I have two lists like this:
>
> t1 = list(a = c(1,2,3),
>           b = c("1", "2", "3"))
> t2 = list( b = c("1", "2", "3"),
>            a = c(1,2,3))
>
> If I read the documentation correctly, by setting use.names equal to TRUE I
> believe this comparison should evaluate as true:
>
> all.equal(t1,t2, use.names=TRUE)
>
> However, I get the following output:
>
> which appears as though it is performing the comparison based on walking
> through indices and comparing that way.
>
> [1] "Names: 2 string mismatches"
> [2] "Component 1: Modes: numeric, character"
> [3] "Component 1: target is numeric, current is character"
> [4] "Component 2: Modes: character, numeric"
> [5] "Component 2: target is character, current is numeric"
>
> Can someone tell me what I'm doing wrong here?
> --
> John
> :wq
>
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>
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