Wow! Jim, this is really impressive. I can't wrap my head around how
you figured this out.
Thank you,
AC
On Sun, Feb 21, 2010 at 12:02 AM, jim holtman wrote:
> This will do it. You can see two different values for id=1:
>
>> x <- with(datas, aggregate(list(r = r), by = list(id = id, mod1 =
This will do it. You can see two different values for id=1:
> x <- with(datas, aggregate(list(r = r), by = list(id = id, mod1 =
mod1),mean))
> x
id mod1 r
1 11 0.980
2 41 0.640
3 71 0.490
4 101 0.180
5 12 0.295
6 52 0.490
7 82 0.330
8 11
OK, this is great, Jim. Last question: How about if I want the 1 copy
of each id to be selected randomly versus taking the first one?
Thank you,
AC
> On Sat, Feb 20, 2010 at 8:37 PM, jim holtman wrote:
>> I am not sure what you mean by eliminating a row. Now if you want only one
>> copy of ea
I am not sure what you mean by eliminating a row. Now if you want only one
copy of each 'id', and it is the first one, the you can use 'duplicated':
> x <- with(datas, aggregate(list(r = r), by = list(id = id, mod1 =
mod1),mean))
> x
id mod1 r
1 11 0.980
2 41 0.640
3 7
Perfect! Thanks Jim.
Do you know how I could then reduce the data even further?
Specifically, reducing it to 1 id per row? In this dataset, id 1 would
have one row eliminated.
Assume the data is much larger and cannot be deleted by visual
inspection and elimination one row at a time.
Thank you,
This seems to work fine (notice the missing 'c(...)'; why did you think you
needed it);
> with(datas, aggregate(list(r = r), by = list(id = id, mod1 =
mod1),mean))
id mod1 r
1 11 0.980
2 41 0.640
3 71 0.490
4 101 0.180
5 12 0.295
6 52 0.490
7 8
Hi All,
I am interested in aggregating a data frame based on 2
categories--mean effect size (r) for each 'id's' 'mod1'. The
'with' function works well when aggregating on one category (e.g.,
based on 'id' below) but doesnt work if I try 2 categories. How can
this be accomplished?
# sample data
i
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