Displaying 7 results from an estimated 7 matches for "ind5".
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2011 Mar 10
2
within group sequential subtraction
...uential subtractions within a group so that I know the
time between separate observations for a group of individuals.
My data:
data <- structure(list(group = c("IND1", "IND1", "IND2",
"IND2", "IND2", "IND3", "IND4", "IND5",
"IND6", "IND6"), date_obs = structure(c(6468,
7063, 9981, 14186, 14372, 5129, 9767, 11168, 10243, 10647), class =
"Date")), .Names = c("group",
"date_obs"), row.names = c(NA, 10L), class = "data.frame")
So I start with:
group...
2012 Apr 30
3
95% confidence interval of the coefficients from a bootstrap analysis
...ysis to obtain the variation of my variables to
replacement.
I am trying to obtain the coefficients 95% confidence interval from the
bootstrap procedure.
Here is my script for the bootstrap:
N = length (data_Pb[,1])
B = 10000
stor.r2 = rep(0,B)
stor.r2 = rep(0,B)
stor.inter = rep(0,B)
stor.Ind5 = rep(0,B)
stor.LNPRI25 = rep(0,B)
stor.NPRI10 = rep(0,B)
stor.Mine = rep(0,B)
for (i in 1:B){
idx = sample(1:N, replace=T)
newdata = data_Pb[idx,]
L_NPRI_25k <- log(newdata$NPRI_25k+1)
data_Pb.boot = lm(newdata$Log_Level ~
newdata$Ind_5k + newdata$MineP_50k +
newdata$NPRI_10k +...
2011 Aug 10
2
choosing selective data with permutations
Hello,
I am a R beginner and hoping to obtain some hints or suggestions about
using permutations to sort a data set I have.
Here is an example dataset:
Ind1 11 00 12 15 28
Ind2 21 33 22 67 52
Ind3 22 45 21 22 56
Ind4 11 25 74 77 42
Ind5 41 32 67 45 22
This will be read into a variable using read.table. What I want to do
is permute these individuals and every time pick 3 individuals and
write them to a new variable. I want to do this 100 times so that in
the end I will have 100 tables containing data for 3 individuals eac...
2009 Feb 13
0
lists on a script
...nn<-length(foo[[m]]);
prinf<-median(foo[[m]][1:12]);
prsup<-median(foo[[m]][(nn-12):nn]);
if (xor((prsup/prinf)>s,(prinf/prsup)>s)) {ind4<-c(ind4,m)}};
crit3<<-ind4;
So far so good (aparently). I mean, the "crit3" set it's ok. But, the next
paragraph:
ind5<-integer();
for (k in ind4){
print(k); #to know when the error happens
dife<-numeric();
l<-length(foo[[k]]);
print(l);
print(foo[[K]]) #to know why the error happens
for (i in 9:(l-8)){
minf<-median(foo[[k]][(i-8):(i-1)]);
msup<-median(foo[[k]][(i+1):(i...
2011 Mar 09
2
Cleaning date columns
Hi Everyone,
I have the following problem:
data <- structure(list(prochi = c("IND1", "IND1", "IND1",
"IND2", "IND2", "IND2", "IND2", "IND3",
"IND4", "IND5"), date_admission = structure(c(6468,
6470, 7063, 9981, 9983, 14186, 14372, 5129, 9767, 11168), class = "Date")),
.Names = c("prochi",
"date_admission"), row.names = c("27", "28", "21", "86", "77",
"80"...
2017 Nov 18
0
Using cforest on a hierarchically structured dataset
...?? 7?? ...
???????????????? 0.55???????? ind2??????? S1??? 5 10??? 7?? ...
????????????????? 0.2???????? ind3??????? S1??? 5 10??? 7?? ...
...?????????????????????????????????????? S1??? 5 10??? 7?? ...
????????????????? 0.3???????? ind1??????? S2?? 15 7?? 50?? ...
???????????????? 0.01???????? ind5??????? S2?? 15 7?? 50?? ...
...?????????????????????????????????????? S2?? 15 7?? 50?? ...
????????????????? 0.4???????? ind1??????? S3??? 2 8??? 5?? ...
???????????????? 0.05???????? ind3??????? S3??? 2 8??? 5?? ...
????????????????? 0.1???????? ind4??????? S3??? 2 8??? 5?? ...
???????????????...
2005 Jul 25
0
lda: scaling to 'disctiminant function'
...(EW$AGR, EW$MIN, EW$MAN, EW$PS, EW$CON, EW$SER, EW$FIN,
EW$SPS, EW$TC)
Dep <- EW$GROUP
LDA <- lda(Dep ~ Ind, prior=c(1,1,1)/3)
So far so good.
I have....
> LDA$scaling
LD1 LD2
Ind1 1.3415788 3.689959
Ind2 1.5118460 5.215123
Ind3 1.4480197 3.653298
Ind4 2.1898063 3.875867
Ind5 1.2055849 3.913750
Ind6 0.8257858 3.718032
Ind7 1.3481728 3.699259
Ind8 1.2364320 3.871438
Ind9 2.0652630 2.891655
What do I do with this to convert it to the coefficients for the
discrimination functions?
cheers
Worik