Read ?mean.
Look at the argument list.
The mean function only applies to the first argument.
On July 21, 2018 9:39:42 AM PDT, John Kane via R-help
wrote:
>Either I am doing something very stupid or my R installation has a
>glitch. What am I missing?
>dd1 <- 50
>dd2 <- 54
>
>mean(dd1, dd2)
>[
On 21/07/2018 12:39 PM, John Kane via R-help wrote:
Either I am doing something very stupid or my R installation has a glitch. What
am I missing?
dd1 <- 50
dd2 <- 54
mean(dd1, dd2)
[1] 50 # wrong
Read the help page ?mean. You are specifying the parameters x and trim.
Duncan Murdoch
(d
Hello,
The first argument of mean is a vector, the dots argument is to be
"passed to or from other methods." (from ?mean)
Try instead
mean(c(dd1, dd2))
Hope this helps,
Rui Barradas
Às 17:39 de 21-07-2018, John Kane via R-help escreveu:
Either I am doing something very stupid or my R ins
Either I am doing something very stupid or my R installation has a glitch. What
am I missing?
dd1 <- 50
dd2 <- 54
mean(dd1, dd2)
[1] 50 # wrong
(dd1 + dd2)/2
[1] 52 # correct
aa <- c(48, 52, 56, 54, 52)
mean(aa)
[1] 52.4 # correct
[[alternative HTML version deleted]]
__
round(mean(val), digits=16)
-Ista
On Tue, Mar 9, 2010 at 10:23 AM, venkata kirankumar
wrote:
> Hi all,
> I got one intresting problem with caliculating mean that is
>
> while i am taking mean of values
>
> *0.6, -0.8, 4, -3.8*
>
> using
> *val<-c(0.6, -0.8, 4, -3.8)*
> *mean(val)*
> it given res
See FAQ 7.31
venkata kirankumar wrote:
Hi all,
I got one intresting problem with caliculating mean that is
while i am taking mean of values
*0.6, -0.8, 4, -3.8*
using
*val<-c(0.6, -0.8, 4, -3.8)*
*mean(val)*
it given result as
*2.775558e-17*
but the actual result is *"0"*
can any one sugge
Hi all,
I got one intresting problem with caliculating mean that is
while i am taking mean of values
*0.6, -0.8, 4, -3.8*
using
*val<-c(0.6, -0.8, 4, -3.8)*
*mean(val)*
it given result as
*2.775558e-17*
but the actual result is *"0"*
can any one suggest how can I get correct mean result in th
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