R devel - Sep 2000 - all.equal.list() sometimes fails with unnamed and named components (PR#674)

examples:

1) Fails to report that components 2 and 3 differ

all.equal(list(1,2,3,zap=1),list(1,3,4,zap=2))
[1] "Component zap: Mean relative  difference: 1


2) Incorrectly asserts all are equal when components 2 and 3 differ
> all.equal(list(1,2,3,zap=1),list(1,3,4,zap=1))
[1] TRUE

3) Removing named component reveals differences:
> all.equal(list(1,2,3,1),list(1,3,4,1))
[1] "Component 2: Mean relative  difference: 0.5"      
[2] "Component 3: Mean relative  difference: 0.3333333"

4) Unequal first component reveals differences
> all.equal(list(2,2,3,zap=1),list(1,3,4,zap=1))
[1] "Component : Mean relative  difference: 0.5"
[2] "Component : Mean relative  difference: 0.5"
[3] "Component : Mean relative  difference:
0.5"
> all.equal(list(2,2,3,zap=1),list(1,3,4,zap=2))
[1] "Component : Mean relative  difference: 0.5" 
[2] "Component : Mean relative  difference: 0.5" 
[3] "Component : Mean relative  difference: 0.5" 
[4] "Component zap: Mean relative  difference:
1"
> 
5) Second component IS equal, but:
> all.equal(list(2,2,3,zap=1),list(1,2,4,zap=2))
[1] "Component : Mean relative  difference: 0.5" 
[2] "Component : Mean relative  difference: 0.5" 
[3] "Component : Mean relative  difference: 0.5" 
[4] "Component zap: Mean relative  difference: 1"

--please do not edit the information below--

Version:
 platform = sparc-sun-solaris2.7
 arch = sparc
 os = solaris2.7
 system = sparc, solaris2.7
 status = 
 major = 1
 minor = 1.0
 year = 2000
 month = June
 day = 15
 language = R

Search Path:
 .GlobalEnv, Autoloads, package:base

Chuck Berry

-- 
Charles C. Berry                        (858) 534-2098 
                                         Dept of Family/Preventive Medicine
E mailto:cberry@tajo.ucsd.edu	         UC San Diego
http://hacuna.ucsd.edu/members/ccb.html  La Jolla, San Diego 92093-0645

 

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>>>>> cberry  writes:
> examples:
> 1) Fails to report that components 2 and 3 differ
> all.equal(list(1,2,3,zap=1),list(1,3,4,zap=2))
> [1] "Component zap: Mean relative  difference: 1
> 2) Incorrectly asserts all are equal when components 2 and 3 differ
>> all.equal(list(1,2,3,zap=1),list(1,3,4,zap=1))
> [1] TRUE
> 3) Removing named component reveals differences:
>> all.equal(list(1,2,3,1),list(1,3,4,1))
> [1] "Component 2: Mean relative  difference: 0.5"      
> [2] "Component 3: Mean relative  difference: 0.3333333"
> 4) Unequal first component reveals differences
>> all.equal(list(2,2,3,zap=1),list(1,3,4,zap=1))
> [1] "Component : Mean relative  difference: 0.5"
> [2] "Component : Mean relative  difference: 0.5"
> [3] "Component : Mean relative  difference: 0.5"
>> all.equal(list(2,2,3,zap=1),list(1,3,4,zap=2))
> [1] "Component : Mean relative  difference: 0.5" 
> [2] "Component : Mean relative  difference: 0.5" 
> [3] "Component : Mean relative  difference: 0.5" 
> [4] "Component zap: Mean relative  difference: 1"
>> 
> 5) Second component IS equal, but:
>> all.equal(list(2,2,3,zap=1),list(1,2,4,zap=2))
> [1] "Component : Mean relative  difference: 0.5" 
> [2] "Component : Mean relative  difference: 0.5" 
> [3] "Component : Mean relative  difference: 0.5" 
> [4] "Component zap: Mean relative  difference: 1"
> --please do not edit the information below--
> Version:
>  platform = sparc-sun-solaris2.7
>  arch = sparc
>  os = solaris2.7
>  system = sparc, solaris2.7
>  status = 
>  major = 1
>  minor = 1.0
>  year = 2000
>  month = June
>  day = 15
>  language = R
> Search Path:
>  .GlobalEnv, Autoloads, package:base
> Chuck Berry
I think the above happens because all.equal.list() is broken.  The fix
is not entirely clear, because I am not sure what we really want.

Anyway, in all of the above examples we compare lists which are named in
the sense that names() is not NULL, but not ``fully'' named in the sense
that any(nchar(names()) == NULL) is TRUE.  In this case, iseq gets set
to nt[c.in.t] which in the above example gives

Browse[1]> nt[c.in.t]
[1] ""    ""    ""    "zap"

Hence in the for(i in iseq) loop we do three comparisons of the first
component, due to subscripting with "".

The obvious idea is that we must use positional indices where we do not
have named ones.

Now for the problem I see in the current logic.  We currently do

   if both lists are named (at least partially) then ... else fi

Maybe we should change this as follows: if either of the two lists has
names, work though the named components.  Warn about the ones not
present in both.  Compare the ones present in both.  Then get rid of all
named components and compare what is left in positional order.

As I said, I am not sure that this is really what we want.

Comments?

-k
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> To: Kurt.Hornik@ci.tuwien.ac.at
> Cc: cberry@tajo.ucsd.edu, r-devel@stat.math.ethz.ch
> Subject: Re: [Rd] all.equal.list() sometimes fails with unnamed and named 
components (PR#674)
> From: Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk>
> Date: 03 Oct 2000 18:05:50 +0200
> 
> Kurt Hornik <Kurt.Hornik@ci.tuwien.ac.at> writes:
> 
> > Maybe we should change this as follows: if either of the two lists has
> > names, work though the named components.  Warn about the ones not
> > present in both.  Compare the ones present in both.  Then get rid of
all
> > named components and compare what is left in positional order.
> > 
> > As I said, I am not sure that this is really what we want.
> > 
> > Comments?
> 
> I think you might be right, and also that this is an easy thing to
> implement. Then we'd have
> 
> all.equal(list(a=1,b=2,3,4),list(3,b=2,4,a=1)) == TRUE
> 
> Right?
Probably not. Lists do have orderings: they are not sets but generic
vectors.
> However, BigBrother has
> 
> > all.equal(list(a=1,b=2,3,4),list(3,b=2,4,a=1)) 
> [1] "Names: 2 string mismatches"
> attr(, "continue"):
> [1] T
> 
> > all.equal(list(a=1,b=2,3,4), list(a=1,b=2,4,3))
> [1] T
That's not what current versions of S-PLUS give, as one might hope.
> ..which does look like a "compatible bug"
> 
> Hmm. Maybe one wants positional matching in any case? But then, what
> is the named-component extraction about??
I think that both the names and components should match exactly (the
components recursively).  Unfortunately the named-component extraction
is partial matching (at least, sometimes) so the ordering of the names
always matters.  (There's an S/R difference here I keep forgetting to 
write down. I think it is 

x <- list(aa=1, bb=2)
x["a"]

which gives in S
$aa:
[1] 1
and in R
$"NA"
NULL
so S always partial matches, but R does not always.)

-- 
Brian D. Ripley,                  ripley@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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