Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector of IDs :
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
I want the function returns the following vector where each term is the
number of
Laetitia Marisa wrote:
Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector of IDs :
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
I want the function returns the following vector where
Laetitia Marisa [EMAIL PROTECTED] writes:
Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector of IDs :
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
I want the function returns the
Laetitia Marisa wrote:
Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector of IDs :
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
I want the function returns the following vector where
This should work:
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
table(ids)[ids]
ids
ID1 ID2 ID2 ID3 ID3 ID3 ID5
1 2 2 3 3 3 1
Andy
From: Laetitia Marisa
Hello,
Is there a simple and fast function that returns a vector of
the number
of replications for each object of a
Try this:
ave(as.numeric(factor(ds)), ds, FUN = length)
See ?ave for more info.
On 1/24/06, Laetitia Marisa [EMAIL PROTECTED] wrote:
Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector
table()
-thomas
On Tue, 24 Jan 2006, Laetitia Marisa wrote:
Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector of IDs :
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
I want the
?table
ids - c( ID1, ID2, ID2, ID3, ID3,ID3, ID5)
x - table(ids)
x
ids
ID1 ID2 ID3 ID5
1 2 3 1
count - x[ids] # index using the names in the string
count
ids
ID1 ID2 ID2 ID3 ID3 ID3 ID5
1 2 2 3 3 3 1
On 1/24/06, Laetitia Marisa [EMAIL PROTECTED] wrote:
Hello,
Ah. It's a bit more complicated than just table(), because you want the
result to be the same length.
tt - table(id)
tt[match(id,names(tt))]
-thomas
On Tue, 24 Jan 2006, Laetitia Marisa wrote:
Hello,
Is there a simple and fast function that returns a vector of the number
of
It is great! It takes now less than 1 second (with the table function
(0.34'), 2 sec with the ave function (1.91') ) with my big data and
only two lines of code ;).
Thanks a lot every one.
Regards,
Laetitia.
Laetitia Marisa [EMAIL PROTECTED] writes:
Hello,
Is there a simple and fast
Nice. I timed it and its much faster than mine too.
On 1/24/06, Barry Rowlingson [EMAIL PROTECTED] wrote:
Laetitia Marisa wrote:
Hello,
Is there a simple and fast function that returns a vector of the number
of replications for each object of a vector ?
For example :
I have a vector
There's an even faster one, which nobody seems to have mentioned yet:
rep(l - rle(ids)$lengths, l)
Timing on my 2.8GHz NetBSD system shows:
length(ids)
[1] 45150
# Gabor:
system.time(for (i in 1:100) ave(as.numeric(factor(ids)), ids, FUN =
length))
[1] 3.45 0.06 3.54 0.00 0.00
# Barry (and
Note that that assumes that all occurrences of a value
are contiguous.
On 1/24/06, Ray Brownrigg [EMAIL PROTECTED] wrote:
There's an even faster one, which nobody seems to have mentioned yet:
rep(l - rle(ids)$lengths, l)
Timing on my 2.8GHz NetBSD system shows:
length(ids)
[1] 45150
#
On Wed, 25 Jan 2006, Ray Brownrigg wrote:
There's an even faster one, which nobody seems to have mentioned yet:
rep(l - rle(ids)$lengths, l)
I considered this but it wasn't clear to me from the initial post that
each ID occupied a contiguous section of the vector.
Also, lazy evaluation
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