I have data set from a incomplete lattice design study that I have imported into R from excel and would like to conduct a PBIB.test. However, after running the function as shown below, the output shows object Area not found, even after repeated times.
library("agricolae", lib.loc = "~/R/win-library/3.3")
Rdata2 <- PBIB.test("BlockNo", "AccNo", "Rep", Area, k = 9, c("REML"), console = TRUE)
Error in data.frame(v1 = 1, y) : object 'Area' not found
What is the problem?
See below for a sample application of PBIB.test, based on the agricolae tutorial.
First, create some sample data.
# Construct the alpha design with 30 treatments, 2 repetitions, and block size = 3
Genotype <- c(paste("gen0", 1:9, sep= ""), paste("gen", 10:30, sep= ""));
r <- 2;
k <- 3;
s <- 10;
b <- s * r;
book <- design.alpha(Genotype, k, r,seed = 5);
# Source dataframe
df <- book$book;
Create a vector of response values.
# Response variable
response <- c(
5,2,7,6,4,9,7,6,7,9,6,2,1,1,3,2,4,6,7,9,8,7,6,4,3,2,2,1,1,2,
1,1,2,4,5,6,7,8,6,5,4,3,1,1,2,5,4,2,7,6,6,5,6,4,5,7,6,5,5,4);
Run PBIB.test
model <- with(df, PBIB.test(block, Genotype, replication, response, k = 3, method="REML"))
head(model);
#$ANOVA
#Analysis of Variance Table
#
#Response: yield
# Df Sum Sq Mean Sq F value Pr(>F)
#Genotype 29 72.006 2.4830 1.2396 0.3668
#Residuals 11 22.034 2.0031
#
#$method
#[1] "Residual (restricted) maximum likelihood"
#
#$parameters
# test name.t treatments blockSize blocks r alpha
# PBIB-lsd Genotype 30 3 10 2 0.05
#
#$statistics
# Efficiency Mean CV
# 0.6170213 4.533333 31.22004
#
#$model
#Linear mixed-effects model fit by REML
# Data: NULL
# Log-restricted-likelihood: -73.82968
# Fixed: y ~ trt.adj
# (Intercept) trt.adjgen02 trt.adjgen03 trt.adjgen04 trt.adjgen05 trt.adjgen06
# 6.5047533 -3.6252940 -0.7701618 -2.5264354 -3.1633495 -1.9413054
#trt.adjgen07 trt.adjgen08 trt.adjgen09 trt.adjgen10 trt.adjgen11 trt.adjgen12
# -3.0096514 -4.0648738 -3.5051139 -2.8765561 -1.7111335 -1.6308755
#trt.adjgen13 trt.adjgen14 trt.adjgen15 trt.adjgen16 trt.adjgen17 trt.adjgen18
# -2.2187974 -2.3393290 -2.0807215 -0.3122845 -3.4526453 -1.0320169
#trt.adjgen19 trt.adjgen20 trt.adjgen21 trt.adjgen22 trt.adjgen23 trt.adjgen24
# -3.1257616 0.2101325 -1.7632411 -1.9177848 -1.0500345 -2.5612960
#trt.adjgen25 trt.adjgen26 trt.adjgen27 trt.adjgen28 trt.adjgen29 trt.adjgen30
# -4.3184716 -2.3071359 1.2239927 -1.3643068 -1.4354599 -0.4726870
#
#Random effects:
# Formula: ~1 | replication
# (Intercept)
#StdDev: 8.969587e-05
#
# Formula: ~1 | block.adj %in% replication
# (Intercept) Residual
#StdDev: 1.683459 1.415308
#
#Number of Observations: 60
#Number of Groups:
# replication block.adj %in% replication
# 2 20
#
#$Fstat
# Fit Statistics
#AIC 213.65937
#BIC 259.89888
#-2 Res Log Likelihood -73.82968
Related
I did a Ljung-Box Test for independence in r with 36 lags and stored the results in a list.
for (lag in c(1:36)){
box.test.list[[lag]] <- (Box.test(btcr, type = "Ljung", lag))
}
I want to extract the p-values as well as the test statistic (X-squared) and print them out to look something like:
X-squared = 100, p-value = 0.0001
I also want to pull it out p-value indivually but rather than just spit out numbers, I want something like:
[1] p-value = 0.001
[2] p-value = 0.0001
and so on. Can this be done?
With the test data
set.seed(7)
btcr <- rnorm(100)
you can perform all your tests with
box.test.list <- lapply(1:36, function(i) Box.test(btcr, type = "Ljung", i))
and then put all the results in a data.frame with
results <- data.frame(
lag = 1:36,
xsquared = sapply(box.test.list, "[[", "statistic"),
pvalue = sapply(box.test.list, "[[", "p.value")
)
Then you can do what you like with the results
head(results)
# lag xsquared pvalue
# 1 1 3.659102 0.05576369
# 2 2 7.868083 0.01956444
# 3 3 8.822760 0.03174261
# 4 4 9.654935 0.04665920
# 5 5 11.190969 0.04772238
# 6 6 12.607454 0.04971085
As far as I know the breakpoint should correspond to the observation which maximizes the F statistics, but I can't see any meaningful association between the F statistics and the timing of the break. What do I get wrong?
y <- c(rnorm(30), 2+rnorm(20)) # 1 breakpoint
f <- Fstats(y ~ 1) # calculate F statistics
f$breakpoint # breakpoint Fstats suggests
which(f$Fstats == max(f$Fstats)) # observation with max F statistics
order(f$Fstats) # observations ordered by F statistics
As can be seen the observation of the breakpoint is not the observation with the highest F statistics.
Your y is not class ts. So the output became a bit curious ts data and unfortunately you failed to interpret it.
set.seed(1)
y <- c(rnorm(30), 2+rnorm(20))
ts.y <- ts(y, start = 1, frequency = 1) # change `y` into class `ts`
ts.f <- Fstats(ts.y ~ 1)
ts.f$breakpoint # [1] 30
ts.f$Fstats
# Time Series:
# Start = 7 # this means ts.f$Fstats[1] is 7th
which(ts.f$Fstats == max(ts.f$Fstats)) # [1] 24 # ts.f$Fstats[24] is 30th
plot(ts.f)
lines(breakpoints(ts.f))
Let's consider the following vectors in the dataframe:
ctrl <- rnorm(50)
x1 <- rnorm(30, mean=0.2)
x2 <- rnorm(100,mean=0.1)
x3 <- rnorm(100,mean=0.4)
x <- data.frame(data=c(ctrl,x1,x2,x3),
Group=c(
rep("ctrl", length(ctrl)),
rep("x1", length(x1)),
rep("x2", length(x2)),
rep("x3", length(x3))) )
I know I could use
pairwise.t.test(x$data,
x$Group,
pool.sd=FALSE)
to get pairwise comparison like
Pairwise comparisons using t tests with non-pooled SD
data: x$data and x$Group
ctrl x1 x2
x1 0.08522 - -
x2 0.99678 0.10469 -
x3 0.00065 0.99678 2.8e-05
P value adjustment method: holm
However I am not interested in every possible combination of vectors. I am seeking a way to compare ctrl vector with every other vectors, and to take into account alpha inflation. I'd like to avoid
t.test((x$data[x$Group=='ctrl']), (x$data[x$Group=='x1']), var.equal=T)
t.test((x$data[x$Group=='ctrl']), (x$data[x$Group=='x2']), var.equal=T)
t.test((x$data[x$Group=='ctrl']), (x$data[x$Group=='x3']), var.equal=T)
And then perform manual correction for multiple comparisons. What would be the best way to do so ?
You can use p.adjust to get a Bonferroni adjustment to multiple p-values. You should not bundle thos unequal length vectors inot t adataframe but rather use a list.
ctrl <- rnorm(50)
x1 <- rnorm(30, mean=0.2)
x2 <- rnorm(100,mean=0.1)
x3 <- rnorm(100,mean=0.4)
> lapply( list(x1,x2,x3), function(x) t.test(x,ctrl)$p.value)
[[1]]
[1] 0.2464039
[[2]]
[1] 0.8576423
[[3]]
[1] 0.0144275
> p.adjust( .Last.value)
[1] 0.4928077 0.8576423 0.0432825
#BondedDust 's answer looks great. I provide a bit more complicated solution if you really need to work with dataframes.
library(dplyr)
ctrl <- rnorm(50)
x1 <- rnorm(30, mean=0.2)
x2 <- rnorm(100,mean=0.1)
x3 <- rnorm(100,mean=0.4)
x <- data.frame(data=c(ctrl,x1,x2,x3),
Group=c(
rep("ctrl", length(ctrl)),
rep("x1", length(x1)),
rep("x2", length(x2)),
rep("x3", length(x3))), stringsAsFactors = F )
# provide the combinations you want
# set1 with all from set2
set1 = c("ctrl")
set2 = c("x1","x2","x3")
dt_res =
data.frame(expand.grid(set1,set2)) %>% # create combinations
mutate(test_id = row_number()) %>% # create a test id
group_by(test_id) %>% # group by test id, so everything from now on is performed for each test separately
do({x_temp = x[(x$Group==.$Var1 | x$Group==.$Var2),] # for each test id keep groups of interest
x_temp = data.frame(x_temp)}) %>%
do(test = t.test(data~Group, data=.)) # perform the test and save it
# you create a dataset that has the test id and a column with t.tests results as elements
dt_res
# Source: local data frame [3 x 2]
# Groups: <by row>
#
# test_id test
# 1 1 <S3:htest>
# 2 2 <S3:htest>
# 3 3 <S3:htest>
# get all tests as a list
dt_res$test
# [[1]]
#
# Welch Two Sample t-test
#
# data: data by Group
# t = -1.9776, df = 58.36, p-value = 0.05271
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
# -0.894829477 0.005371207
# sample estimates:
# mean in group ctrl mean in group x1
# -0.447213560 -0.002484425
#
#
# [[2]]
#
# Welch Two Sample t-test
#
# data: data by Group
# t = -2.3549, df = 100.68, p-value = 0.02047
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
# -0.71174095 -0.06087081
# sample estimates:
# mean in group ctrl mean in group x2
# -0.44721356 -0.06090768
#
#
# [[3]]
#
# Welch Two Sample t-test
#
# data: data by Group
# t = -5.4235, df = 101.12, p-value = 4.001e-07
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
# -1.2171386 -0.5652189
# sample estimates:
# mean in group ctrl mean in group x3
# -0.4472136 0.4439652
PS : It's always interesting to work with p-values and alpha corrections. It's a bit of a philosophical issue how to approach that and some people agree and other disagree. Personally, I tend to correct alpha based on all possible comparison I can do after an experiment, because you never know when you'll come back to investigate other pairs. Imagine what happens if in the future people decide that you have to go back and compare the winning group (let's say x1) with x2 and x3. You'll focus on those pairs and you'll again correct alpha based on those compariosns. But on the whole you performed all possible comparisons, apart from x2 vs x3! You may write your reports or publish findings that should have been a bit more strict on the alpha correction.
I have written the code below to obtain a bootstrap estimate of a mean. My objective is to view the numbers selected from the data set, ideally in the order they are selected, by the function boot in the boot package.
The data set only contains three numbers: 1, 10, and 100 and I am only using two bootstrap samples.
The estimated mean is 23.5 and the R code below indicates that the six numbers included one '1', four '10' and one '100'. However, there are 30 possible combinations of those numbers that would have resulted in a mean of 23.5.
Is there a way for me to determine which of those 30 possible combinations is the combination that actually appeared in the two bootstrap samples?
library(boot)
set.seed(1234)
dat <- c(1, 10, 100)
av <- function(dat, i) { sum(dat[i])/length(dat[i]) }
av.boot <- boot(dat, av, R = 2)
av.boot
#
# ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
# Call:
# boot(data = dat, statistic = av, R = 2)
#
#
# Bootstrap Statistics :
# original bias std. error
# t1* 37 -13.5 19.09188
#
mean(dat) + -13.5
# [1] 23.5
# The two samples must have contained one '1', four '10' and one '100',
# but there are 30 possibilities.
# Which of these 30 possible sequences actual occurred?
# This code shows there must have been one '1', four '10' and one '100'
# and shows the 30 possible combinations
my.combos <- expand.grid(V1 = c(1, 10, 100),
V2 = c(1, 10, 100),
V3 = c(1, 10, 100),
V4 = c(1, 10, 100),
V5 = c(1, 10, 100),
V6 = c(1, 10, 100))
my.means <- apply(my.combos, 1, function(x) {( (x[1] + x[2] + x[3])/3 + (x[4] + x[5] + x[6])/3 ) / 2 })
possible.samples <- my.combos[my.means == 23.5,]
dim(possible.samples)
n.1 <- rowSums(possible.samples == 1)
n.10 <- rowSums(possible.samples == 10)
n.100 <- rowSums(possible.samples == 100)
n.1[1]
n.10[1]
n.100[1]
length(unique(n.1)) == 1
length(unique(n.10)) == 1
length(unique(n.100)) == 1
I think you can determine the numbers sampled and the order in which they are sampled with the code below. You have to extract the function ordinary.array from the boot package and paste that function into your R code. Then specify the values for n, R and strata, where n is the number of observations in the data set and R is the number of replicate samples you want.
I do not know how general this approach is, but it worked with a couple of simple examples I tried, including the example below.
library(boot)
set.seed(1234)
dat <- c(1, 10, 100, 1000)
av <- function(dat, i) { sum(dat[i])/length(dat[i]) }
av.boot <- boot(dat, av, R = 3)
av.boot
#
# ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
# Call:
# boot(data = dat, statistic = av, R = 3)
#
#
# Bootstrap Statistics :
# original bias std. error
# t1* 277.75 -127.5 132.2405
#
#
mean(dat) + -127.5
# [1] 150.25
# boot:::ordinary.array
ordinary.array <- function (n, R, strata)
{
inds <- as.integer(names(table(strata)))
if (length(inds) == 1L) {
output <- sample.int(n, n * R, replace = TRUE)
dim(output) <- c(R, n)
}
else {
output <- matrix(as.integer(0L), R, n)
for (is in inds) {
gp <- seq_len(n)[strata == is]
output[, gp] <- if (length(gp) == 1)
rep(gp, R)
else bsample(gp, R * length(gp))
}
}
output
}
# I think the function ordinary.array determines which elements
# of the data are sampled in each of the R samples
set.seed(1234)
ordinary.array(n=4,R=3,1)
# [,1] [,2] [,3] [,4]
# [1,] 1 3 1 3
# [2,] 3 4 1 3
# [3,] 3 3 3 3
#
# which equals:
((1+100+1+100) / 4 + (100+1000+1+100) / 4 + (100+100+100+100) / 4) / 3
# [1] 150.25
I am new to R and having a problem with printing the results of 'for' loop in R. Here is my code:
afile <- read.table(file = 'data.txt', head =T)##Has three columns Lab, Store and Batch
lab1 <- afile$Lab[afile$Batch == 1]
lab2 <- afile$Lab[afile$Batch == 2]
lab3 <- afile$Lab[afile$Batch == 3]
lab_list <- list(lab1,lab2,lab3)
for (i in 1:2){
x=lab_list[[i]]
y=lab_list[[i+1]]
t.test(x,y,alternative='two.sided',conf.level=0.95)
}
This code runs without any error but produces no output on screen. I tried taking results in a variable using 'assign' but that produces error:
for (i in 1:2){x=lab_list[[i]];y=lab_list[[i+1]];assign(paste(res,i,sep=''),t.test(x,y,alternative='two.sided',conf.level=0.95))}
Warning messages:
1: In assign(paste(res, i, sep = ""), t.test(x, y, alternative = "two.sided", :
only the first element is used as variable name
2: In assign(paste(res, i, sep = ""), t.test(x, y, alternative = "two.sided", :
only the first element is used as variable name
Please help me on how can I perform t.test in loop and get their results i.e. print on screen or save in variable.
AK
I would rewrite your code like this :
I assume your data is like this
afile <- data.frame(Batch= sample(1:3,10,rep=TRUE),lab=rnorm(10))
afile
Batch lab
1 2 0.4075675
2 1 0.3006192
3 1 -0.4824655
4 3 1.0656481
5 1 0.1741648
6 2 -1.4911526
7 2 0.2216970
8 1 -0.3862147
9 1 -0.4578520
10 1 -0.6298040
Then using lapply you can store your result in a list :
lapply(1:2,function(i){
x <- subset(afile,Batch==i)
y <- subset(afile,Batch==i+1)
t.test(x,y,alternative='two.sided',conf.level=0.95)
})
[[1]]
Welch Two Sample t-test
data: x and y
t = -0.7829, df = 6.257, p-value = 0.4623
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.964637 1.005008
sample estimates:
mean of x mean of y
0.3765373 0.8563520
[[2]]
Welch Two Sample t-test
data: x and y
t = -1.0439, df = 1.797, p-value = 0.4165
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-6.588720 4.235776
sample estimates:
mean of x mean of y
0.856352 2.032824
In a loop, you need to explicitly print your results in many cases. Try:
print(t.test(x,y,alternative='two.sided',conf.level=0.95))
or
print(summary(t.test(x,y,alternative='two.sided',conf.level=0.95)))
In addition to 'Hansons' solution of printing, results can be saved and printed like:
result <- vector("list",6)
for (i in 1:5){x=lab_list[[i]];y=lab_list[[i+1]];result[[i]] = t.test(x,y,alternative='two.sided',conf.level=0.95)}
result
AK