Hello, R experts,
Could I use the command suggested by Chunk to get the estimates for
treat-interaction contrasts (non-orthognal) which indicates below:
summary(lm(score ~ A*B))
or is there any other ways to get the estimates?
Again, Chunk, thank you for your help!
Joshua
Quoting Chuck Cleland
Hello, Chuck,
Thank you very much for your help! But the contrasts I want to do
simutaneously is
contrasts(B)
[,1] [,2] [,3] [,4]
b1 -4 -3 -2 -1
b21 -3 -2 -1
b312 -2 -1
b4123 -1
b51234
Could you please show me how to calculate
Hello, R experts,
Sorry for asking this question again again since I really want a help!
I have a two-factor experiment data and like to calculate estimates of
interation contrasts say factor A has levels of a1, a2, and B has
levels of b1, b2, b3, b4, and b5 with 3 replicates. I am not sure the
Hello, R experts,
Sorry for asking this question again since I really want a help!
I have a two-factor experiment data and like to calculate estimates of
interation contrasts say factor A has levels of a1, a2, and B has
levels of b1, b2, b3, b4, and b5 with 3 replicates. I am not sure the
Hello, R experts,
Sorry for asking this question again since I really want a help!
I have a two-factor experiment data and like to calculate estimates of
interation contrasts say factor A has levels of a1, a2, and B has
levels of b1, b2, b3, b4, and b5 with 3 replicates. I am not sure the
Hello, R experts,
I have a two-factor experiment data and like to calculate estimates of
interation contrasts say factor A has levels of a1, a2, and B has
levels of b1, b2, b3, b4, and b5 with 3 replicates. I am not sure the
constrast estimate I got is right using the script below:
Hello, R experts,
I have a list called dataHP which has 30 elements (m1, m2, ..., m30).
Each element is a 7x6 matrix holding yield data from two factors
experimental design, with treatment in column, position in row. For
instance, the element 20 is:
dataHP[[20]]
col1 col2
Hello, R experts,
If I understand Ted's anwser correctly, then I can not contrast the
mean yields between sections 1-8 and 9-11 under Trt but I can
contrast mean yields for sections 1-3 and 6-11 because there exists
significant interaction between two factors (Trt:section4,
Trt:section5).
Hello, Ted,
Thank you for your help!
So I can not contrast the mean yields between sections 1-8 and 9-11
under Trt but I can contrast mean yields for sections 1-3 and 6-11
because there exists significant interaction between two factors
(Trt:section4, Trt:section5). Could I use the commands
Hello, R users,
I have two factors (treat, section) anova design experiment where
there are 3 replicates. The objective of the experiment is to test if
there is significant difference of yield between top (section 9 to 11)
and bottom (section 9 to 11) of the fruit tree under treatment. I
Hello, Jacques,
Thank you very much for your help!
It's clever code and works fine. But I don't understand what the last line code
is for. Should I really need the last line?
Thanks again.
Josh
Quoting Jacques VESLOT [EMAIL PROTECTED]:
you could try :
trt1 - do.call(rbind, lapply(listexp,
Hello, R Users,
I have a list (say listexp) of 10,000 elements, each of which consists of a
matrix (5X6). It likes:
$a
trt1rep1trt1rep2trt2rep1trt2rep2ctlrep1ctlrep2
[1,] 5054 98 8940 45
[2,] 6065 76 79
Hello, Everyone,
I am sorry that my message got truncated.
I resend it again as below:
Hello, R Users,
I have a list (say listexp) of 10,000 elements, each of which consists of a
matrix (5X6). It likes:
$a
trt1rep1trt1rep2trt2rep1trt2rep2ctlrep1ctlrep2
[1,] 5054
Hello, Everyone,
I am sorry that my message got truncated due to wrong format.
I hope it works now:
Hello, R Users,
I have a list (say listexp) of 10,000 elements, each of which consists of a
matrix (5X6). It likes:
$a
trt1rep1trt1rep2trt2rep1trt2rep2ctlrep1ctlrep2
[1,]
Hello, R users,
I have a long list with over 20,000 elements (say foo), now I want to make a sub
list of 30 elements from this list using the component names such as subname-c(
a, b, d, f, g, h, o, q,...). One way to do it is to make rest of
elements in the list NULL, that is too silly. Is there
Hello, Uwe and Robert,
Thank you for your help! I fixed the problem as your suggestion. I run into a
new issue: I can't sink the output to a file in the script within loops as
bellows, but I can sink it in the command line( I run R1.9.1 on WinXP).
The scrip:
sink(C:/level3BPterm.txt)
Hello, R experts,
I tried to retrieve all biological process GO terms at level 3 starting
biological process as level 1 using the code as bellows:
1 library(GO)
2 library(GOstats)
3 level2-getGOChildren(GO:0008150)$GO:0008150$Children
4 for ( i in 1:length(level2)) {
5level3 -
Dear R experts:
I tried two ways to install Package:XML on windows xp for R 1.9.1, all failed.
Messages are given as bellows:
1 download from CRAN
install.packages(XML, CRAN = getOption(CRAN),
+ contriburl = contrib.url(http://cran.r-project.org;),
+ available = NULL, destdir = NULL,
+
Hello, R experts:
I got data like this:
group duplicate treatment
A Y 5
A Y 3
A N 6
B Y 2
B N 4
B Y 1
How to sort the data and calculate the average treatment value for each group
in two
Dear R-users:
I am runing R 1.6.2 with Windows XP. I try to calculate Pearson correlation
and Spearman correlation of any pairwise columns of 8000 x 80 data matrix with
missing values and randomize the matix 1000 times and calculate this two
correlations again. The code bellow for Pearson is
Hello, there:
I got data matix with missing values. I want to calculate any possible
pairwise Spearman correlation rho for each column. Is there a function just
like cor(x, y, use=complete.obs) for Pearson correlation?
Thanks in advance!
Josh
__
Dear R users:
Is there a function or easier way to randomize the data between rows
(considering a row as a whole unit)? For example,
original data:
A1 A2 A3 A4
B1 1 2 3 4
B2 5 6 7 8
B3 9 10 11 12
B4 13 14 15 16
randomized data:
A1 A2 A3 A4
B4 13 14 15 16
B2 5 6
Hello, There:
I read data from tab-delimted text file(1888.txt) in C:/temp, which has
thousands rows and 80 colulms, using read.delim(C:/temp/1888.txt). when I
retrieved the numeric coulum which has real zero values and applies R buid-in
function such as mean( ), sum() and got a result of NA.
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