Howdy Folks,
Eric - I'm just using this as a toy example to illustrate what I'm trying
to do. The actual lists I'm trying to compare are much larger and more
complex. So sorting them like that is something I'd have to do recursively.
I wanted to take advantage of all.equal because I want to be abl
Note: all.equal() with all its S3 methods is implemented entirely in R
code, so it should not be hard to find out where things happen
and how.
> John Harrold
> on Wed, 27 May 2020 21:52:16 -0700 writes:
> Is there a way to compare t1 and t2 above such that the
Why not just reorder the elements of the list so they should match?
t1 <- t1[ names(t1)[order(names(t1))]]
t2 <- t2[ names(t2)[order(names(t2))]]
identical(t1,t2)
On Thu, May 28, 2020 at 7:52 AM John Harrold
wrote:
> Is there a way to compare t1 and t2 above such that the name is used
> ins
Is there a way to compare t1 and t2 above such that the name is used
instead of the index?
On Wed, May 27, 2020 at 9:14 PM Bert Gunter wrote:
> 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
>
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 mis
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