Displaying 5 results from an estimated 5 matches for "meanss".
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2009 Jun 08
2
help to speed up loops in r
...r is very slow (2days). Any suggestions to speed it up.
NB I have also tried using rowMeans rather than adding the 2 values and dividing by 2. (same problem)
#SCRIPT STARTS
for (i in 1:length(averagedreplicates[,1]))
#for (i in 1:dim(averagedreplicates)[1])
{
cat(i,'\n')
#calculates Meanss
#Sample A
averagedreplicates[i,2] <- (zz[i,2] + zz[i,3])/2
averagedreplicates[i,3] <- (zz[i,4] + zz[i,5])/2
averagedreplicates[i,4] <- (zz[i,6] + zz[i,7])/2
averagedreplicates[i,5] <- (zz[i,8] + zz[i,9])/2
averagedreplicates[i,6] <- (zz[i,10] + zz[i,11])/2
#Sample B
averagedreplicat...
2009 May 15
2
Help with loops
...am trying to average each set of values (i.e. zz[1,2:3] averaged and placed in average_value[1,2] and so on.
below is my script but it seems to be stuck in an endless loop
Any suggestions??
for (i in 1:length(average_value[,1])) {
average_value[i] <- i^100; print(average_value[i])
#calculates Meanss
#Sample A
average_value[i,2] <- rowMeans(zz[i,2:3])
average_value[i,3] <- rowMeans(zz[i,4:5])
average_value[i,4] <- rowMeans(zz[i,6:7])
average_value[i,5] <- rowMeans(zz[i,8:9])
average_value[i,6] <- rowMeans(zz[i,10:11])
#Sample B
average_value[i,7] <- rowMeans(zz[i,12:13])
aver...
2009 May 15
1
Fw: Help with loops(corrected question)
...eadings in zz[,1]
> I am trying to average each set of values (i.e. zz[1,2:3]
> averaged and placed in average_value[1,2] and so on.
> below is my script but it seems to be stuck in an endless
> loop
> Any suggestions??
>
> for (i in 1:length(zz[,1])) {
>
> #calculates Meanss
> #Sample A
> average_value[i,2] <- rowMeans(zz[i,2:3])
> average_value[i,3] <- rowMeans(zz[i,4:5])
> average_value[i,4] <- rowMeans(zz[i,6:7])
> average_value[i,5] <- rowMeans(zz[i,8:9])
> average_value[i,6] <- rowMeans(zz[i,10:11])
>
> #Sample B
> averag...
2010 Apr 24
2
multiple paired t-tests without loops
I am new to R and I suspect my problem is easily solved, but I haven't
been able to figure it out without using loops. I am trying to
implement Blair & Karniski's (1993) permutation test. I've included a
sample data frame below. This data frame represents the conditional
means (C1, C2) for 3 subjects in 2 consecutive samples of a continuous
data set (e.g. ERP waveform).
2002 May 02
2
problem with lme in nlme package
...2.9421 0.15122 7022 19.456 <.0001
sectorCatholic 1.2245 0.30611 157 4.000 1e-04
meanses:cses 1.0444 0.29107 7022 3.588 3e-04
cses:sectorCatholic -1.6421 0.23312 7022 -7.044 <.0001
Correlation:
(Intr) meanss cses sctrCt mnss:c
meanses 0.256
cses 0.000 0.000
sectorCatholic -0.699 -0.356 0.000
meanses:cses 0.000 0.000 0.295 0.000
cses:sectorCatholic 0.000 0.000 -0.696 0.000 -0.351
Standardized Within-Group Residuals:...