I'm working on a Bayesian Network Meta-Analysis using the gemtc package on a dataset similar to the following:
df <- data.frame(study = c("A", "A", "B", "B", "C", "C", "D", "D", "E", "E", "F", "F",
"G", "G", "H", "H", "I", "I", "J", "J", "K", "K"),
treatment = c("A", "B", "B", "C", "B", "C", "A", "B", "B",
"C", "B", "C", "A", "B", "B", "C", "B", "C",
"A", "C", "B", "C"),
responders = c(1, 5, 0, 0, 3, 1, 0, 2, 0, 2, 0, 2, 0,
0, 1, 2, 0, 0, 2, 9, 1, 1),
sampleSize = c(32, 33, 30, 30, 18, 20, 15, 15, 20,
20, 30, 30, 36, 32, 15, 15, 23, 22, 24, 23, 18, 16))
While I have been able to set up the network model and run the analysis just fine, I have been struggling with specifying the order in which I would like the treatments to be compared in the node-splitting consistency analysis. For example, I want the odds ratios and 95% credible intervals to be calculated using the "B" treatment as the reference group when comparing "B" with "A" and "C" as the reference group when comparing "A" with "C" and "B" with "C". Below is the code I have tried:
library(gemtc)
library(rjags)
# Create mtc.network element to be used in modeling ------
network <- mtc.network(data.ab = df)
# Compile model ------
network.mod <- mtc.model(network,
linearModel = "random", # random effects model
n.chain = 4) # 4 Markov chains
# Assess network consistency using nodesplit method ------
nodesplit <- mtc.nodesplit(network.mod,
linearModel = "random", # random effects model
n.adapt = 5000, # burn-in iterations
n.iter = 100000, # actual simulation iterations
thin = 10) # extract values of every 10th iteration
summary(nodesplit) # High p-values indicate consistent results
plot(summary(nodesplit))
My results provide ORs (95% CrIs) for:
"A" vs. "C"
"B" vs. "C"
"B" vs. "A"
I have created a separate data frame specifying that I want "A" vs. "B" comparisons via:
# Specify desired comparisons ------
comparisons = data.frame(t1 = "A", t2 = "B")
# Assess network consistency using nodesplit method, adding comparisons argument ------
nodesplit <- mtc.nodesplit(network.mod,
comparisons = comparisons,
linearModel = "random", # random effects model
n.adapt = 5000, # burn-in iterations
n.iter = 100000, # actual simulation iterations
thin = 10) # extract values of every 10th iteration
summary(nodesplit) # High p-values indicate consistent results
But I still get "B" vs. "A" results. I have also tried specify t1="B", t2="A", and I get the same results. Any assistance with this would be greatly appreciated. Thanks in advance.
Related
I have a data set as I've shown below:
It shows which book is sold by which shop.
df <- tribble(
~shop, ~book_id,
"A", 1,
"B", 1,
"C", 2,
"D", 3,
"E", 3,
"A", 3,
"B", 4,
"C", 5,
"D", 1,
)
In the data set,
shop A sells 1, 3
shop B sells 1, 4
shop C sells 2, 5
shop D sells 3, 1
shop E sells only 3
So now, I want to calculate the Jaccard index here. For instance, let's take shop A and shop B. There are three different books that are sold by A and B (book 1, book 3, book 4). However, only one product is sold by both shops (this is product 1). So, the Jaccard index here should be 33.3% (1/3).
Here is the sample of the desired data:
df <- tribble(
~shop_1, ~shop_2, ~similarity,
"A", "B", 33.3,
"B", "A", 33.33,
"A", "C", 0,
"C", "A", 0,
"A", "D", 100,
"D", "A", 100,
"A", "E", 50,
"E", "A", 50,
)
Any comments/assistance really appreciated! Thanks in advance.
I don't know about a package but you can write your own function. I guess by similarity you mean something like this:
similarity <- function(x, y) {
k <- length(intersect(x, y))
n <- length(union(x, y))
k / n
}
Then you can use tidyr::crossing to merge the same data frame with itself
dfg <- df %>% group_by(shop) %>% summarise(books = list(book_id))
crossing(dfg %>% set_names(paste0, "_A"), dfg %>% set_names(paste0, "_B")) %>%
filter(shop_A != shop_B) %>%
mutate(similarity = map2_dbl(books_A, books_B, similarity))
I am trying to create a graph where the x axis (a factor) is reordered by descending order of the y axis (numerical values), but only for one of two levels of another factor.
Originally, I tried using the code below:
reorder(factor1, desc(value1))
However, this code only reorganizes the graph (in a descending order) by the sum of the two values under each factor2 (I presume); while I am only interested in reorganizing the data for one level (i.e. "A") under factor2.
Here is some sample data to illustrate better.
sampledata <- data.frame(factor1 = c("A", "A", "B", "B", "C", "C", "D", "D", "E", "E",
"F", "F", "G", "G", "H", "H", "I", "I", "J", "J"),
factor2 = c("A", "H", "A", "H", "A", "H", "A", "H", "A", "H",
"A", "H", "A", "H", "A", "H", "A", "H", "A", "H"),
value1 = c(1, 5, 6, 2, 6, 8, 10, 21, 30, 5,
3, 5, 4, 50, 4, 7, 15, 48, 20, 21))
Here is what I used previously:
sampledata %>%
ggplot(aes(x=reorder(factor1, desc(value1)), y=value1, group=factor2, color=factor2)) +
geom_point()
The reason why I would like to reorder by a specific level (say factor2=="A") is that I can view any deviance of the values for factor2=="H" away from "A" points.
I would appreciate using tidyverse or dplyr as means to solve this problem.
library(ggplto2)
library(dplyr)
sampledata %>%
mutate(value2 = +(factor2=="A")*value1) %>%
ggplot(aes(x=reorder(factor1, desc(value2 + value1/max(value1))), y=value1,
group=factor2, color=factor2)) +
geom_point() +
xlab("factor1")
I am new to igraph and it seems to be a very powerful (and therefore also complex) package.
I tried to convert the following lists into an igraph object.
graph <- list(s = c("a", "b"),
a = c("s", "b", "c", "d"),
b = c("s", "a", "c", "d"),
c = c("a", "b", "d", "e", "f"),
d = c("a", "b", "c", "e", "f"),
e = c("c", "d", "f", "z"),
f = c("c", "d", "e", "z"),
z = c("e", "f"))
weights <- list(s = c(3, 5),
a = c(3, 1, 10, 11),
b = c(5, 3, 2, 3),
c = c(10, 2, 3, 7, 12),
d = c(15, 7, 2, 11, 2),
e = c(7, 11, 3, 2),
f = c(12, 2, 3, 2),
z = c(2, 2))
Interpretation is as follows: s is the starting node, it links to nodes a and b. The edges are weighted 3 for s to a and 5 for s to b and so on.
I tried all kinds of functions from igraph but only got all kinds of errors. What is the most elegant and easy way to convert the above into an igraph object for plotting the graph?
Create an edgelist and then a graph from that. Assign the weights and plot it.
set.seed(123)
e <- as.matrix(stack(graph))
g <- graph_from_edgelist(e)
E(g)$weight <- stack(weights)[[1]]
plot(g, edge.label = E(g)$weight)
I have a data set of values (val) grouped by multiple categories (distance & phase). I would like to test each category by Kruskal-Wallis test, where val is dependent variable, distance is a factor, and phase split my data in 3 groups.
As such, I need to specify the subset of the data within Kruskal-Wallis test and then apply the test to each of groups. BUT, I can not get my subsetting to work!
In R help, it is specified that the subset is an optional vector specifying a subset of observations to be used. But how to correctly put this to my lapply function?
My dummy data:
# create data
val<-runif(60, min = 0, max = 100)
distance<-floor(runif(60, min=1, max=3))
phase<-rep(c("a", "b", "c"), 20)
df<-data.frame(val, distance, phase)
# get unique groups
ii<-unique(df$phase)
# get basic statistics per group
aggregate(val ~ distance + phase, df, mean)
# run Kruskal test, specify the subset
kruskal.test(df$val ~df$distance,
subset = phase == "c")
This works well, so my subset should be correctly set as a vector.
But how to use this in a lapply function?
# DOES not work!!
lapply(ii, kruskal.test(df$val ~ df$distance,
subset = df$phase == as.character(ii)))
My overall goal is to create a function from kruskal.test, and save all statistics for each group into one table.
All help is highly appreciated.
Usually you would start by splitting, and then lapplying.
Something like
lapply(split(df, df$phase), function(d) { kruskal.test(val ~ distance, data=d) })
would yield a list, indexed by the phase, of the results of kruskal.test.
Your final expression does not work because lapply expects a function, and applying kruskal.test does not result in a function, it results in the result of running that test. If you surround it with a function definition with the index, then it would work, just be a little less idiomatic.
lapply(ii, function(i) { kruskal.test(df$val ~ df$distance, subset=df$phase==i )})
Though it is late, it might help someone having the same problem. So, I am putting an answer implemented using tidyverse and rstatix packages. The rstatix package which "provides a simple and intuitive pipe friendly framework, coherent with the 'tidyverse' design philosophy for performing basic statistical tests".
library(rstatix)
library(tidyverse)
df %>%
group_by(phase) %>%
kruskal_test(val ~ distance)
Output
# A tibble: 3 x 7
phase .y. n statistic df p method
* <chr> <chr> <int> <dbl> <int> <dbl> <chr>
1 a val 20 0.230 1 0.631 Kruskal-Wallis
2 b val 20 0.0229 1 0.88 Kruskal-Wallis
3 c val 20 0.322 1 0.570 Kruskal-Wallis
which is same as provided by #user295691.
Data
df = structure(list(val = c(93.8056977232918, 31.0681172646582, 40.5262873973697,
47.6368983509019, 65.23181500379, 64.4571609096602, 10.3301600087434,
90.4661140637472, 41.2359046051279, 28.3357713604346, 49.8977075796574,
10.8744730940089, 5.31001624185592, 71.9248640118167, 99.0267782937735,
73.7928744405508, 3.31214582547545, 40.2693636715412, 27.6980920461938,
79.501334275119, 60.5167196830735, 89.9171086261049, 87.4633299885318,
43.1893823202699, 91.1248738644645, 99.755659350194, 7.25280269980431,
96.957387868315, 75.0860505970195, 52.3794749286026, 26.6221587313339,
52.5518182432279, 24.1361060412601, 49.5364486705512, 65.5214034719393,
38.9469220302999, 0.687191751785576, 19.3090825574473, 19.6511475136504,
25.5966754630208, 7.33999472577125, 33.9820940745994, 50.3751677693799,
10.811762069352, 17.2359711956233, 53.958406439051, 64.2723652534187,
92.7404976682737, 26.824192632921, 30.0975760444999, 52.0105463219807,
74.4495407678187, 56.0636054025963, 91.891074879095, 14.0827904455364,
59.3607738381252, 66.5170294465497, 24.1726311156526, 83.0881901318207,
35.5380675755441), distance = c(2, 1, 1, 1, 1, 2, 1, 2, 2, 1,
2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1,
1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 2, 1,
1, 2, 1, 1, 2, 2, 2, 2), phase = c("a", "b", "c", "a", "b", "c",
"a", "b", "c", "a", "b", "c", "a", "b", "c", "a", "b", "c", "a",
"b", "c", "a", "b", "c", "a", "b", "c", "a", "b", "c", "a", "b",
"c", "a", "b", "c", "a", "b", "c", "a", "b", "c", "a", "b", "c",
"a", "b", "c", "a", "b", "c", "a", "b", "c", "a", "b", "c", "a",
"b", "c")), class = "data.frame", row.names = c(NA, -60L))
I am trying to split a dataset in 80/20 - training and testing sets. I am trying to split by location, which is a factor with 4 levels, however each level has not been sampled equally. Out of 1892 samples -
Location1: 172
Location2: 615
Location3: 603
Location4: 502
I am trying to split the whole dataset 80/20, as mentioned above, but I also want each location to be split 80/20 so that I get an even proportion from each location in the training and testing set. I've seen one post about this using stratified function from the splitstackshape package but it doesn't seem to want to split my factors up.
Here is a simplified reproducible example -
x <- c(1, 2, 3, 4, 1, 3, 7, 4, 5, 7, 8, 9, 4, 6, 7, 9, 7, 1, 5, 6)
xx <- c("A", "A", "B", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C", "C", "D", "D", "D", "D", "D")
df <- data.frame(x, xx)
validIndex <- stratified(df, "xx", size=16/nrow(df))
valid <- df[-validIndex,]
train <- df[validIndex,]
where A, B, C, D correspond to the factors in the approximate proportions as the actual dataset (~ 10, 32, 32, and 26%, respectively)
Using bothSets should return you a list containing the split of the original data frame into validation and training set (whose union should be the original data frame):
splt <- stratified(df, "xx", size=16/nrow(df), replace=FALSE, bothSets=TRUE)
valid <- splt[[1]]
train <- splt[[2]]
## check
df2 <- as.data.frame(do.call("rbind",splt))
all.equal(df[with(df, order(xx, x)), ],
df2[with(df2, order(xx, x)), ],
check.names=FALSE)