[R] Quickest way to match two vectors besides %in%?
Hello list,
I have two data frames, X (48469,2) and Y (79771,5).
X[,1] contains distinct values of Y[,2].
I want to match values in X[,1] and Y[,2], then take
the corresponding value in [X,2] and place it in
Y[,4].
So far I have been doing it like so:
for(i in 1:48469) {
y[which(x[i,1]==y[,3]),4]-x[i,2]
}
But it chunks along so very slowly that I can't help
but wonder if there's a faster way, mainly because on
my box it takes R about 30 seconds to simply COUNT to
48,469 in the for loop.
I have already tried using %in%. It tells me if the
values in X[,1] are IN Y[,2], which is useful in
removing unnecessary values from X[,1]. But it does
not tell me exactly where they match. which(X[,1]
%in% Y[,2]) does but it only matches on the first
instance.
This is the slowest part of the script I'm working
on--if I could improve it I could shave off some
serious operating time. Any pointers?
Regards,
Pete
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Re: [R] Quickest way to match two vectors besides %in%?
On 11/8/2005 2:28 PM, Pete Cap wrote:
Hello list,
I have two data frames, X (48469,2) and Y (79771,5).
X[,1] contains distinct values of Y[,2].
I want to match values in X[,1] and Y[,2], then take
the corresponding value in [X,2] and place it in
Y[,4].
So far I have been doing it like so:
for(i in 1:48469) {
y[which(x[i,1]==y[,3]),4]-x[i,2]
}
But it chunks along so very slowly that I can't help
but wonder if there's a faster way, mainly because on
my box it takes R about 30 seconds to simply COUNT to
48,469 in the for loop.
I have already tried using %in%. It tells me if the
values in X[,1] are IN Y[,2], which is useful in
removing unnecessary values from X[,1]. But it does
not tell me exactly where they match. which(X[,1]
%in% Y[,2]) does but it only matches on the first
instance.
This is the slowest part of the script I'm working
on--if I could improve it I could shave off some
serious operating time. Any pointers?
Look at the merge() function to add the X and Y columns to a new
dataframe, then process that to merge the X[,2] and Y[,4] values.
It will be something like
Z - merge(X, Y, by.x=1, by.y=2, all.y=TRUE)
changes - !is.na(Z[,2])
Z[changes,5] - Z[changes,2]
but you are almost certainly better off (from a maintenance point of
view) to use the names of the columns, rather than guessing at column
numbers.
Duncan Murdoch
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Re: [R] Quickest way to match two vectors besides %in%?
?match
x
X1 X2
1 1 5
2 2 6
3 3 7
4 4 8
y
Y1 Y4
1 1 8
2 2 9
3 3 10
4 4 11
5 1 12
6 2 13
7 3 14
8 4 15
y.orig-y # backup
y$Y4-x$X2[match(y$Y1, x$X1)]
y
Y1 Y4
1 1 5
2 2 6
3 3 7
4 4 8
5 1 5
6 2 6
7 3 7
8 4 8
HTH,
Weiwei
On 11/8/05, Pete Cap [EMAIL PROTECTED] wrote:
Hello list,
I have two data frames, X (48469,2) and Y (79771,5).
X[,1] contains distinct values of Y[,2].
I want to match values in X[,1] and Y[,2], then take
the corresponding value in [X,2] and place it in
Y[,4].
So far I have been doing it like so:
for(i in 1:48469) {
y[which(x[i,1]==y[,3]),4]-x[i,2]
}
But it chunks along so very slowly that I can't help
but wonder if there's a faster way, mainly because on
my box it takes R about 30 seconds to simply COUNT to
48,469 in the for loop.
I have already tried using %in%. It tells me if the
values in X[,1] are IN Y[,2], which is useful in
removing unnecessary values from X[,1]. But it does
not tell me exactly where they match. which(X[,1]
%in% Y[,2]) does but it only matches on the first
instance.
This is the slowest part of the script I'm working
on--if I could improve it I could shave off some
serious operating time. Any pointers?
Regards,
Pete
__
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--
Weiwei Shi, Ph.D
Did you always know?
No, I did not. But I believed...
---Matrix III
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Re: [R] Quickest way to match two vectors besides %in%?
Pete Cap wrote:
Hello list,
I have two data frames, X (48469,2) and Y (79771,5).
X[,1] contains distinct values of Y[,2].
I want to match values in X[,1] and Y[,2], then take
the corresponding value in [X,2] and place it in
Y[,4].
So far I have been doing it like so:
for(i in 1:48469) {
y[which(x[i,1]==y[,3]),4]-x[i,2]
}
I'm not sure but isn't that a case where merge() can help?
cheers
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