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Calculate Closest 5 Cases?

4 messages · dsheuman@rogers.com, Tom Blackwell, Peter Dalgaard +1 more

#
I've only begun investigating R as a substitute for SPSS.

I have a need to identify for each CASE the closest (or most similar) 5 
other CASES (not including itself as it is automatically the closest).  I 
have a fairly large matrix (50000 cases by 50 vars).  In SPSS, I can use Correlate > Distances to generate a matrix of similarity, but only on a small sample.  The entire matrix can not be processed at once due to memory limitations.

The data are all percents, so they are easy comparable.  

Is there any way to do this in R?

Below is a small sample of the data (from SPSS) and the desired output.

Thanks,

Danny




*Sample Data.
DATA LIST LIST /id(F8) var1(F8.2) var2(F8.2) var3(F8.2) var4(F8.2) var5
(F8.2) var6(F8.2) var7(F8.2) var8(F8.2) var9(F8.2) var10(F8.2) var11(F8.2).
BEGIN DATA.
10170069	3.51	4.02	6.53	11.05	6.53	8.04	13.57	20.10	11.05	8.55
	7.04
10190229	1.89	5.66	4.61	7.62	8.45	13.21	9.50	20.82	16.07	9.36
	3.77
10540023	3.40	5.08	3.39	4.52	10.18	14.71	13.56	16.38	9.60	7.89
	11.85
10650413	6.64	6.64	3.73	4.70	3.78	13.23	19.82	15.98	12.26	8.48
	3.78
10662074	5.11	5.81	4.37	5.11	6.55	14.60	18.97	11.68	10.25	8.75
	8.79
10770041	6.43	4.17	6.34	4.26	6.34	4.26	19.11	19.20	14.95	12.77
	4.35
11010422	3.14	4.71	6.81	7.85	5.75	6.81	15.18	15.18	13.61	11.00
	9.44
11060762	7.03	5.03	6.95	5.99	5.92	12.94	15.01	12.06	11.98	8.06
	9.02
11070078	4.61	9.22	4.61	7.94	6.27	12.75	14.02	20.49	7.75	7.75
	4.61
11180646	4.48	5.35	6.29	5.42	4.55	11.71	20.74	15.32	14.45	8.09
	3.61
11460001	5.71	7.34	6.48	5.68	4.07	10.55	13.83	18.69	12.15	9.76
	4.87
11650133	6.00	3.72	6.72	6.00	7.50	17.94	13.44	16.37	13.51	5.15
	3.65
11650275	4.02	8.06	6.06	8.10	5.06	8.10	17.16	14.12	12.14	14.12
	4.02
11780034	4.25	4.28	5.30	5.33	6.38	14.88	15.96	18.08	14.85	7.48
	3.20
11790016	4.40	4.40	5.54	4.40	4.40	10.93	17.67	19.72	13.20	12.13
	4.33
12660338	6.60	7.54	5.66	8.49	10.38	11.31	16.06	12.26	8.49	8.49
	4.73
12660644	5.51	3.14	3.95	7.09	7.11	14.98	15.72	18.90	9.44	5.50
	8.65
12661667	5.44	4.50	5.44	4.50	5.44	12.69	13.63	11.81	9.07	13.68
	13.79
END DATA.

*Output should be:.
*.
*	ID1	CLOSEID1	CLOSEID2	CLOSEID3	CLOSEID4	CLOSEID5.
*	ID2	CLOSEID1	CLOSEID2	CLOSEID3	CLOSEID4	CLOSEID5.
*	ID3	CLOSEID1	CLOSEID2	CLOSEID3	CLOSEID4	CLOSEID5.
*	ID4	CLOSEID1	CLOSEID2	CLOSEID3	CLOSEID4	CLOSEID5.
*	ID5	CLOSEID1	CLOSEID2	CLOSEID3	CLOSEID4	CLOSEID5.
#
Danny  -

The flip answer is, it depends on the size of your computer.
One can readily calculate the number of entries in the pairwise
distance matrix that you would like to calculate, and ask whether
it will fit inside the physical memory installed in your computer.
It is  50,000 x 50,000 x 8 bytes per floating point number, for
a total of 20,000,000,000 bytes or 20 gigabytes.  The critical
information that's still missing is that R needs enough space
for 10 or 20 copies of the largest object in its workspace, in
order to turn around and assign that object to a new name, or
do any summaries on it, etc.  So,  . . .  if you have a computer
with between 200 and 400 gigabytes of random access memory, yes,
you can calculate and summarize the matrix of pairwise distances.
But that requires more memory slots than any ordinary motherboard
provides.  (It would be a mother of a motherboard !)

So, failing that, you could always use Adrian Raftery and Chris
Fraley's 'mclust' package to cluster your data into 50 or more
clusters of very similar cases (instructions for running mclust()
on large data sets are found in the manual which comes with the
package), then calculate all pairwise distances only between the
cases within each cluster.  That's a bit of work to code up.
You wouldn't want to work interactively for each of 50 clusters.
But it certainly can be done in R.  Depends how much effort you
want to put into it.

-  tom blackwell  -  u mighigan medical school  -  ann arbor  -
On Fri, 13 Feb 2004 dsheuman at rogers.com wrote:

            
#
<dsheuman at rogers.com> writes:
This seems to be close:

d <- read.table("tempfile") # needed to edit to get 12 items per line.
close6 <- function(r)
  d$V1[order(apply(d[-1],1,
                   function(r2)dist(rbind(r,r2))))][1:6]
t(apply(d[-1],1,close6))

       [,1]     [,2]     [,3]     [,4]     [,5]     [,6]
1  10170069 11010422 11460001 11070078 12660644 11790016
2  10190229 11780034 11460001 10170069 11650133 11070078
3  10540023 12660644 10662074 11060762 12661667 11070078
4  10650413 11180646 11780034 11790016 10662074 11460001
5  10662074 11060762 10650413 12660338 11180646 10540023
6  10770041 11790016 11650275 11010422 11460001 11180646
7  11010422 10170069 11650275 11460001 11060762 11790016
8  11060762 10662074 12660338 12661667 11010422 11460001
9  11070078 11460001 12660644 10170069 11780034 12660338
10 11180646 10650413 11780034 11790016 11460001 10662074
11 11460001 11790016 11070078 11780034 10650413 10170069
12 11650133 11780034 12660644 11060762 10650413 11460001
13 11650275 11010422 11460001 11790016 11180646 10770041
14 11780034 11650133 11180646 10650413 11790016 11460001
15 11790016 11460001 11180646 11780034 10650413 10770041
16 12660338 11060762 10662074 11650275 11070078 10650413
17 12660644 10540023 11650133 11780034 11070078 11060762
18 12661667 11060762 10662074 10540023 11010422 12660644


Notice that I use a function to get the closest *6* ID's because the
method will include the row itself. If multiple rows have distance
zero, this might be a problem since you're not guaranteed to get the
ID of the "self" row sorted first.

Here's another try:

close5 <- function(i)
  d$V1[-i][order(apply(d[-i,-1],1,function(r)dist(rbind(d[i,-1],r))))[1:5]]

do.call("rbind",lapply(1,nrow(d),close5))

However, for some reason this is much slower. Getting rid of the more
obvious inefficiencies (some of which would really kill you on a large
data set since they involve copying the entire data frame!) doesn't
really help:

dd <- d[-1]
close5 <- function(i) {r1 <- dd[i,];
d$V1[-i][order(apply(dd,1,function(r)dist(rbind(r1,r)))[-i])[1:5]]}
[1] 1.67 0.00 1.67 0.00 0.00

whereas
[1] 0.23 0.00 0.23 0.00 0.00

Anyone have a better idea, or just an explanation of the slowness?
#
A quick and dirty clustering method (I think its due to Hartigan, at
least I recall first seeing it in his book on clustering) is to pick a
random set of seed cases, and then make one pass through the data,
assigning each case to the seed closest to it.  Then you can compute
your distance matrices within the resulting clusters.  You could do
this within the resulting clusters again to reduce the size of the
distance matrix computation, and you would need to check neighboring
"clusters" for close points.

albyn
On Fri, Feb 13, 2004 at 01:40:25PM -0500, Tom Blackwell wrote: