on 01/13/2009 01:17 PM Doran, Harold wrote:
> Suppose I have a dataframe as follows:
>
> dat <- data.frame(id = c(1,1,2,2,2), var1 = c(10,10,20,20,25), var2 =
> c('foo', 'foo', 'foo', 'foobar', 'foo'))
>
> Now, if I were to subset by id, such as:
>
>> subset(dat, id==1)
> id var1 var2
> 1 1 10 foo
> 2 1 10 foo
>
> I can see that the elements in var1 are exactly the same and the
> elements in var2 are exactly the same. However,
>
>> subset(dat, id==2)
> id var1 var2
> 3 2 20 foo
> 4 2 20 foobar
> 5 2 25 foo
>
> Shows the elements are not the same for either variable in this
> instance. So, what I am looking to create is a data frame that would be
> like this
>
> id freq var1 var2
> 1 2 TRUE TRUE
> 2 3 FALSE FALSE
>
> Where freq is the number of times the ID is repeated in the dataframe. A
> TRUE appears in the cell if all elements in the column are the same for
> the ID and FALSE otherwise. It is insignificant which values differ for
> my problem.
>
> The way I am thinking about tackling this is to loop through the ID
> variable and compare the values in the various columns of the dataframe.
> The problem I am encountering is that I don't think all.equal or
> identical are the right functions in this case.
>
> So, say I was wanting to compare the elements of var1 for id ==1. I
> would have
>
> x <- c(10,10)
>
> Of course, the following works
>
>> all.equal(x[1], x[2])
> [1] TRUE
>
> As would a similar call to identical. However, what if I only have a
> vector of values (or if the column consists of names) that I want to
> assess for equality when I am trying to automate a process over
> thousands of cases? As in the example above, the vector may contain only
> two values or it may contain many more. The number of values in the
> vector differ by id.
>
> Any thoughts?
>
> Harold
Harold,
If we are not talking about testing floats for equivalence:
> merge(table(id = dat$id),
aggregate(dat[-1], list(id = dat$id),
function(x) length(unique(x)) == 1),
by = "id")
id Freq var1 var2
1 1 2 TRUE TRUE
2 2 3 FALSE FALSE
HTH,
Marc Schwartz
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