Hi Paul,
I would use something like this:
x - c(2,2,3,3,4,6)
table(x)
x
2 3 4 6
2 2 1 1
x - factor(x, levels=1:8)
table(x)
x
1 2 3 4 5 6 7 8
0 2 2 1 0 1 0 0
Sarah
On Sun, Jul 31, 2011 at 5:41 PM, Paul Menzel
paulepanter at users.sourceforge.net wrote:
Dear R folks,
I am sorry to ask
Hi:
Another possibility is
x - c(2,2,3,3,4,6)
tabulate(x, 8)
[1] 0 2 2 1 0 1 0 0
where the second argument of tabulate() allows one to specify the
number of bins, in order from 1 up to the specified value, with all
others outside that range ignored.
HTH,
Dennis
On Thu, Aug 18, 2011 at 12:29
Dear R folks,
I am sorry to ask this simple question, but my search for the right
way/command was unsuccessful.
I have a vector
x - c(2, 2, 3, 3, 4, 6)
Now the values of x should be considered the index of another vector
with possible greater length, say 8, and the value should be how often
Hi Paul,
I would use something like this:
x - c(2,2,3,3,4,6)
table(x)
x
2 3 4 6
2 2 1 1
x - factor(x, levels=1:8)
table(x)
x
1 2 3 4 5 6 7 8
0 2 2 1 0 1 0 0
Sarah
On Sun, Jul 31, 2011 at 5:41 PM, Paul Menzel
paulepan...@users.sourceforge.net wrote:
Dear R folks,
I am sorry to ask this
Here's an attempt using sapply:
x - c(2, 2, 3, 3, 4, 6)
ys - 1:8
sapply(ys, function(y) { length(which(x==y)) } )
[1] 0 2 2 1 0 1 0 0
Jeff
On Sun, Jul 31, 2011 at 2:41 PM, Paul Menzel
paulepan...@users.sourceforge.net wrote:
Dear R folks,
I am sorry to ask this simple question, but my
See also ?tabulate.
tabulate(x,8)
Hi Paul,
I would use something like this:
x - c(2,2,3,3,4,6)
table(x)
x
2 3 4 6
2 2 1 1
x - factor(x, levels=1:8)
table(x)
x
1 2 3 4 5 6 7 8
0 2 2 1 0 1 0 0
Sarah
On Sun, Jul 31, 2011 at 5:41 PM, Paul Menzel
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