scatterplot of 100000 points and pdf file format

From: Ted.Harding at nessie.mcc.ac.uk

On 25-Nov-04 Ted Harding wrote:

'unique' will eat x for breakfast, indeed, but will have some
trouble chewing (x,y).

I still can't think of a neat way of doing that.

Best wishes,
Ted.

Sorry, I don't want to be misunderstood.
I didn't mean that 'unique' won't work for arrays.
What I meant was:

X<-round(rnorm(1e6),3);Y<-round(rnorm(1e6),3)
system.time(unique(X))

[1] 0.74 0.07 0.81 0.00 0.00

system.time(unique(cbind(X,Y)))

[1] 350.81   4.56 356.54   0.00   0.00

X<-round(rnorm(1e6),3);Y<-round(rnorm(1e6),3)
system.time(unique(X), gcFirst=TRUE)

system.time(unique(cbind(X,Y)), gcFirst=TRUE)

system.time(dat <- data.frame(x=X, y=Y), gcFirst=TRUE)

system.time(unique(dat), gcFirst=TRUE)

However, still rounding to 3 d.p. we can try packing:

Z<-100000000*X + 1000*Y
system.time(W<-unique(Z))

[1] 0.83 0.05 0.88 0.00 0.00

length(W)

[1] 961523

Though the runtime is small we don't get much reduction
and still W has to be unpacked.

With rounding to 2 d.p.

X<-round(rnorm(1e6),2);Y<-round(rnorm(1e6),2)
Z<-100000000*X + 1000*Y
system.time(W<-unique(Z))

[1] 1.31 0.01 1.32 0.00 0.00

length(W)

[1] 209882

so now it's about 1/5, but visible discretisation must be
getting close.

With 1 d.p.

X<-round(rnorm(1e6),1);Y<-round(rnorm(1e6),1)
Z<-100000000*X + 1000*Y
system.time(W<-unique(Z))

[1] 0.92 0.01 0.93 0.00 0.00

length(W)

[1] 4953

there's a good reduction (about 1/200) but the discretisation
would definitely now be visible. However, as I suggested before,
there's an issue of choice of constant (i.e. of the resolution
of the discretisation so that there's a useful reduction and
also the plot is acceptable).

I'd still like to learn of a method which avoids the
above method of packing, which strikes me as clumsy
(but maybe it's the best way after all).

Ted.


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Date: 25-Nov-04                                       Time: 01:45:48
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