Carrying out a PBIB.test

Question

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?

First off: You really should provide a [minimal reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) with sample data. That aside, you're calling `PBIB.test` without giving the source and response data. Take a look at `?PBIB.test` where `PBIB.test` is called as `with(source_data_frame, PBIB.test(blocks, trt, replication, response_vector, ...))`. `blocks`, `trt` ,`replication` are columns in `source_data_frame`. `response_vector` has to be a vector of the same length as you have rows in `source_data_frame`. — Maurits Evers, Jan 23 '18 at 22:48
See below for an application of `PBIB.test` using sample data. — Maurits Evers, Jan 23 '18 at 23:03

score 0 · Answer 1 · answered Jan 23 '18 at 23:01

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

Carrying out a PBIB.test

1 Answers1