Hi,
Try either:
res1 <- apply(mydata[,1:2],2,mean)
?res2 <- colMeans(mydata[,1:2])
?identical(res1,res2)
#[1] TRUE
# Also if you need to find means for each group ("Ungrazed vs.
"Grazed")
by(mydata[,-3],mydata[,3],colMeans)
#or if column names are "V1", "V2", "V3"
aggregate(.~V3,mydata,mean)
#or
library(plyr)
?ddply(mydata,.(V3),numcolwise(mean))
A.K.
I have a data set with two columns of data that I want to find the mean of. ?
1 ? 6.225 ?59.77 Ungrazed
2 ? 6.487 ?60.98 Ungrazed
3 ? 4.919 ?14.73 Ungrazed
4 ? 5.130 ?19.28 Ungrazed
5 ? 5.417 ?34.25 Ungrazed
6 ? 5.359 ?35.53 Ungrazed
7 ? 7.614 ?87.73 Ungrazed
8 ? 6.352 ?63.21 Ungrazed
9 ? 4.975 ?24.25 Ungrazed
10 ?6.930 ?64.34 Ungrazed
11 ?6.248 ?52.92 Ungrazed
12 ?5.451 ?32.35 Ungrazed
13 ?6.013 ?53.61 Ungrazed
14 ?5.928 ?54.86 Ungrazed
15 ?6.264 ?64.81 Ungrazed
16 ?7.181 ?73.24 Ungrazed
17 ?7.001 ?80.64 Ungrazed
18 ?4.426 ?18.89 Ungrazed
19 ?7.302 ?75.49 Ungrazed
20 ?5.836 ?46.73 Ungrazed
21 10.253 116.05 Ungrazed
22 ?6.958 ?38.94 ? Grazed
23 ?8.001 ?60.77 ? Grazed
24 ?9.039 ?84.37 ? Grazed
25 ?8.910 ?70.11 ? Grazed
26 ?6.106 ?14.95 ? Grazed
27 ?7.691 ?70.70 ? Grazed
28 ?8.988 ?80.31 ? Grazed
29 ?8.975 ?82.35 ? Grazed
30 ?9.844 105.07 ? Grazed
31 ?8.508 ?73.79 ? Grazed
32 ?7.354 ?50.08 ? Grazed
33 ?8.643 ?78.28 ? Grazed
34 ?7.916 ?41.48 ? Grazed
35 ?9.351 ?98.47 ? Grazed
36 ?7.066 ?40.15 ? Grazed
37 ?8.158 ?52.26 ? Grazed
38 ?7.382 ?46.64 ? Grazed
39 ?8.515 ?71.01 ? Grazed
40 ?8.530 ?83.03 ? Grazed
This is from an introduction handout that instructs me to enter the command
>mean(mydata[,1:2])
but when I enter it, I get an error message
Warning message:
In mean.default(mydata[, 1:2]) :
? argument is not numeric or logical: returning NA
I've tried tacking on na.rm=T to the end of it, but I get the same
message. Can someone tell me what I'm doing wrong, or how to fix it?
I've tried searching the forum, but can't find a post relevant to this
problem.
