I am using aregImpute from the Hmisc package to impute missing values in a dataset. Since the results are not invariant to the order of variables in the formula, I have used the reformM function to generate a number of random formula permutations:
reformM(~ x1 + x2 + x3..., data = d, nperm = Z)
The function returns a list.
I have reviewed the help file, but am unclear on how the results can be passed to aregImpute.
Can the object be passed such that aregImpute will iterate over the different combinations to assess any variability? That is, can the more than one permutation be passed to aregImpute? If not, do I simply create a new formula using the new order of variables?
You can certainly use map to pass reformM to aregImpute. Below, I provide a minimal example. That said, I do think it is a better idea to reorder variables such that variables with higher missing data rates are first imputed.
library(missForest)
library(tidyverse)
iris_miss <- prodNA(iris,0.10) ## make every variable missing at 10%
## factorial(ncol(iris_miss)) ## number of possible permutations
permlist <- reformM( as.formula(paste(" ~ ", names(iris_miss) %>% str_flatten( " + "))), data=iris_miss, nperm=10 ) ## get 10 permutations
map(permlist, aregImpute, nk=3, n.impute=5, data= iris_miss)
Related
I have a tricky problem. I have a dataframe with more than 1000 variables and want to fit each variable to age using fp smoothing function. I know how to use gamlss() for a specific variable (vari), but that's not practical to repeat this explicitly for more than 1000 times. Moreover, I want to plot the fitting for all 1000 variable in a single figure. What I did is:
variables <- colnames(data)[7:dim(data)[2]]
for(vari in variables) {
print("ROI is:")
print(vari)
model_fem <- gamlss(vari ~ fp(age), family=GG, data=females)
But I got errors:
Error in model.frame.default(formula = vari ~ fp(age), data = females) :
variable lengths differ (found for 'fp(age)')
I think the tricky part is from fp(). I have tried to use as.formula, it didn't work. Also because females$vari return NULL, that's why we got this error.
Do you have any solution for this?
Thank you
Character values are very different from formuals. Formulas contain symbols and you need to properly rebuild them to make them dynamic. There are lots of different ways to do that, but here's one that uses reformulate to turn characters into formulas and update() to modify a base formula.
variables <- colnames(data)[7:dim(data)[2]]
form_resp <- ~ fp(age)
for(vari in variables) {
print("ROI is:")
form_model <- update(form, reformulate(".", response=vari))
print(form_model)
model_fem <- gamlss(form_model, family=GG, data=females)
}
I am trying to input matrix data into the brm() function to run a signal regression. brm is from the brms package, which provides an interface to fit Bayesian models using Stan. Signal regression is when you model one covariate using another within the bigger model, and you use the by parameter like this: model <- brm(response ~ s(matrix1, by = matrix2) + ..., data = Data). The problem is, I cannot input my matrices using the 'data' parameter because it only allows one data.frame object to be inputted.
Here are my code and the errors I obtained from trying to get around that constraint...
First off, my reproducible code leading up to the model-building:
library(brms)
#100 rows, 4 columns. Each cell contains a number between 1 and 10
Data <- data.frame(runif(100,1,10),runif(100,1,10),runif(100,1,10),runif(100,1,10))
#Assign names to the columns
names(Data) <- c("d0_10","d0_100","d0_1000","d0_10000")
Data$Density <- as.matrix(Data)%*%c(-1,10,5,1)
#the coefficients we are modelling
d <- c(-1,10,5,1)
#Made a matrix with 4 columns with values 10, 100, 1000, 10000 which are evaluation points. Rows are repeats of the same column numbers
Bins <- 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T)
Bins
As mentioned above, since 'data' only allows one data.frame object to be inputted, I've tried other ways of inputting my matrix data. These methods include:
1) making the matrix within the brm() function using as.matrix()
signalregression.brms <- brm(Density ~ s(Bins,by=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])),data = Data)
#Error in is(sexpr, "try-error") :
argument "sexpr" is missing, with no default
2) making the matrix outside the formula, storing it in a variable, then calling that variable inside the brm() function
Donuts <- as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])
signalregression.brms <- brm(Density ~ s(Bins,by=Donuts),data = Data)
#Error: The following variables can neither be found in 'data' nor in 'data2':
'Bins', 'Donuts'
3) inputting a list containing the matrix using the 'data2' parameter
signalregression.brms <- brm(Density ~ s(Bins,by=donuts),data = Data,data2=list(Bins = 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T),donuts=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])))
#Error in names(dat) <- object$term :
'names' attribute [1] must be the same length as the vector [0]
None of the above worked; each had their own errors and it was difficult troubleshooting them because I couldn't find answers or examples online that were of a similar nature in the context of brms.
I was able to use the above techniques just fine for gam(), in the mgcv package - you don't have to define a data.frame using 'data', you can call on variables defined outside of the gam() formula, and you can make matrices inside the gam() function itself. See below:
library(mgcv)
signalregression2 <- gam(Data$Density ~ s(Bins,by = as.matrix(Data[,c("d0_10","d0_100","d0_1000","d0_10000")]),k=3))
#Works!
It seems like brms is less flexible... :(
My question: does anyone have any suggestions on how to make my brm() function run?
Thank you very much!
My understanding of signal regression is limited enough that I'm not convinced this is correct, but I think it's at least a step in the right direction. The problem seems to be that brm() expects everything in its formula to be a column in data. So we can get the model to compile by ensuring all the things we want are present in data:
library(tidyverse)
signalregression.brms = brm(Density ~
s(cbind(d0_10_bin, d0_100_bin, d0_1000_bin, d0_10000_bin),
by = cbind(d0_10, d0_100, d0_1000, d0_10000),
k = 3),
data = Data %>%
mutate(d0_10_bin = 10,
d0_100_bin = 100,
d0_1000_bin = 1000,
d0_10000_bin = 10000))
Writing out each column by hand is a little annoying; I'm sure there are more general solutions.
For reference, here are my installed package versions:
map_chr(unname(unlist(pacman::p_depends(brms)[c("Depends", "Imports")])), ~ paste(., ": ", pacman::p_version(.), sep = ""))
[1] "Rcpp: 1.0.6" "methods: 4.0.3" "rstan: 2.21.2" "ggplot2: 3.3.3"
[5] "loo: 2.4.1" "Matrix: 1.2.18" "mgcv: 1.8.33" "rstantools: 2.1.1"
[9] "bayesplot: 1.8.0" "shinystan: 2.5.0" "projpred: 2.0.2" "bridgesampling: 1.1.2"
[13] "glue: 1.4.2" "future: 1.21.0" "matrixStats: 0.58.0" "nleqslv: 3.3.2"
[17] "nlme: 3.1.149" "coda: 0.19.4" "abind: 1.4.5" "stats: 4.0.3"
[21] "utils: 4.0.3" "parallel: 4.0.3" "grDevices: 4.0.3" "backports: 1.2.1"
I'm trying to write a function to plot ROC curves based on different scoring systems I have to predict an outcome.
I have a dataframe data_all, with columns "score_1" and "Threshold.2000". I generate a ROC curve as desired with the following:
plot.roc(data_all$Threshold.2000, data_all$score_1)
My goal is to generate a ROC curve for a number of different outcomes (e.g. Threshold.1000) and scores (score_1, score_2 etc), but am initially trying to set it up just for different scores. My function is as follows:
roc_plot <- function(dataframe_of_interest, score_of_interest) {
plot.roc(dataframe_of_interest$Threshold.2000, dataframe_of_interest$score_of_interest)}
I get the following error: Error in roc.default(x, predictor, plot =
TRUE, ...) : No valid data provided.
I'd be very grateful if someone can spot why my function doesn't work! I'm a python coder and new-ish to R, and haven't had much luck trying a number of different things. Thanks very much.
EDIT:
Here is the same example with mtcars so it's reproducible:
data(mtcars)
plot.roc(mtcars$vs, mtcars$mpg) # --> makes correct graph
roc_plot <- function(dataframe_of_interest, score_of_interest) {
plot.roc(dataframe_of_interest$mpg, dataframe_of_interest$score_of_interest)}
Outcome:
Error in roc.default(x, predictor, plot = TRUE, ...) : No valid data provided.
roc_plot(mtcars, vs)
Here's one solution that works as desired (i.e. lets the user specify different values for score_of_interest):
library(pROC)
data(mtcars)
plot.roc(mtcars$vs, mtcars$mpg) # --> makes correct graph
# expects `score_of_interest` to be a string!!!
roc_plot <- function(dataframe_of_interest, score_of_interest) {
plot.roc(dataframe_of_interest$vs, dataframe_of_interest[, score_of_interest])
}
roc_plot(mtcars, 'mpg')
roc_plot(mtcars, 'cyl')
Note that your error was not resulting from an incorrect column name, it was resulting from an incorrect use of the data.frame class. Notice what happens with a simpler function:
foo <- function(x, col_name) {
head(x$col_name)
}
foo(mtcars, mpg)
## NULL
This returns NULL. So in your original function when you tried to supply plot.roc with dataframe_of_interest$score_of_interest you were actually feeding plot.roc a NULL.
There are several ways to extract a column from a data.frame by the column name when that name is stored in an object (which is what you're doing when you pass it as an argument in a function). Perhaps the easiest way is to remember that a data.frame is like a 2D array-type object and so we can use familiar object[i, j] syntax, but we ask for all rows and we specify the column by name, e.g., mtcars[, 'mpg']. This still works if we assign the string 'mpg' to an object:
x <- 'mpg'
mtcars[, x]
So that's how I produced my solution. Going a step further, it's not hard to imagine how you would be able to supply both a score_of_interest and a threshold_of_interest:
roc_plot2 <- function(dataframe_of_interest, threshold_of_interest, score_of_interest) {
plot.roc(dataframe_of_interest[, threshold_of_interest],
dataframe_of_interest[, score_of_interest])
}
roc_plot2(mtcars, 'vs', 'mpg')
I am trying to perform a simple RDA using the vegan package to test the effects of depth, basin and sector on genetic population structure using the following data frame.
datafile.
The "ALL" variable is the genetic population assignment (structure).
In case the link to my data doesn't work well, I'll paste a snippet of my data frame here.
I read in the data this way:
RDAmorph_Oct6 <- read.csv("RDAmorph_Oct6.csv")
My problems are two-fold:
1) I can't seem to get my genetic variable to read correctly. I have tried three things to fix this.
gen=rda(ALL ~ Depth + Basin + Sector, data=RDAmorph_Oct6, na.action="na.exclude")
Error in eval(specdata, environment(formula), enclos = globalenv()) :
object 'ALL' not found
In addition: There were 12 warnings (use warnings() to see them)
so, I tried things like:
> gen=rda("ALL ~ Depth + Basin + Sector", data=RDAmorph_Oct6, na.action="na.exclude")
Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric
so I specified numeric
> RDAmorph_Oct6$ALL = as.numeric(RDAmorph_Oct6$ALL)
> gen=rda("ALL ~ Depth + Basin + Sector", data=RDAmorph_Oct6, na.action="na.exclude")
Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric
I am really baffled. I've also tried specifying each variable with dataset$variable, but this doesn't work either.
The strange thing is, I can get an rda to work if I look the effects of the environmental variables on a different, composite, variable
MC = RDAmorph_Oct6[,5:6]
H_morph_var=rda(MC ~ Depth + Basin + Sector, data=RDAmorph_Oct6, na.action="na.exclude")
Note that I did try to just extract the ALL column for the genetic rda above. This didn't work either.
Regardless, this leads to my second problem.
When I try to plot the rda I get a super weird plot. Note the five dots in three places. I have no idea where these come from.
I will have to graph the genetic rda, and I figure I'll come up with the same issue, so I thought I'd ask now.
I've been though several tutorials and tried many iterations of each issue. What I have provided here is I think the best summary. If anyone can give me some clues, I would much appreciate it.
The documentation, ?rda, says that the left-hand side of the formula specifying your model needs to be a data matrix. You can't pass it the name of a variable in the data object as the left-hand side (or at least if this was ever anticipated, doing so exposes bugs in how we parse the formula which is what leads to further errors).
What you want is a data frame containing a variable ALL for the left-hand side of the formula.
This works:
library('vegan')
df <- read.csv('~/Downloads/RDAmorph_Oct6.csv')
ALL <- df[, 'ALL', drop = FALSE]
Notice the drop = FALSE, which stops R from dropping the empty dimension (i.e. converting the single column data frame to a vector.
Then your original call works:
ord <- rda(ALL ~ Basin + Depth + Sector, data = df, na.action = 'na.exclude')
The problem is that rda expects a separate df for the first part of the formula (ALL in your code), and does not use the one in the data = argument.
As mentioned above, you can create a new df with the variable needed for analysis, but here's a oneline solution that should also work:
gen <- rda(RDAmorph_Oct6$ALL ~ Depth + Basin + Sector, data = RDAmorph_Oct6, na.action = na.exclude)
This is partly similar to Gavin simpson's answer. There is also a problem with the categorical vectors in your data frame. You can either use library(data.table) and the rowid function to set the categorical variables to unique integers. Most preferably, not use them. I also wanted to set the ID vector as site names, but I am too lazy now.
library(data.table)
RDAmorph_Oct6 <- read.csv("C:/........../RDAmorph_Oct6.csv")
#remove NAs before. I like looking at my dataframes before I analyze them.
RDAmorph_Oct6 <- na.omit(RDAmorph_Oct6)
#I removed one duplicate
RDAmorph_Oct6 <- RDAmorph_Oct6[!duplicated(RDAmorph_Oct6$ID),]
#Create vector with only ALL
ALL <- RDAmorph_Oct6$ALL
#Create data frame with only numeric vectors and remove ALL
dfn <- RDAmorph_Oct6[,-c(1,4,11,12)]
#Select all categorical vectors.
dfc <- RDAmorph_Oct6[,c(1,11,12)]
#Give the categorical vectors unique integers doesn't do this for ID (Why?).
dfc2 <- as.data.frame(apply(dfc, 2, function(x) rowid(x)))
#Bind back with numeric data frame
dfnc <- cbind.data.frame(dfn, dfc2)
#Select only what you need
df <- dfnc[c("Depth", "Basin", "Sector")]
#The rest you know
rda.out <- rda(ALL ~ ., data=df, scale=T)
plot(rda.out, scaling = 2, xlim=c(-3,2), ylim=c(-1,1))
#Also plot correlations
plot(cbind.data.frame(ALL, df))
Sector and depth have the highest variation. Almost logical, since there are only three vectors used. The assignment of integers to the categorical vector has probably no meaning at all. The function assigns from top to bottom unique integers to the following unique character string. I am also not really sure which question you want to answer. Based on this you can organize the data frame.
I received a comment from a reviewer who wanted to have all the p-values for each line of specific variables levels in a demographic characteristic table (Table 1). Even though the request appears quite strange (and inexact) to me, I would like to agree with his suggestion.
library(tableone)
## Load data
library(survival); data(pbc)
# drop ID from variable list
vars <- names(pbc)[-1]
## Create Table 1 stratified by trt (can add more stratifying variables)
tableOne <- CreateTableOne(vars = vars, strata = c("trt"), data = pbc, factorVars = c("status","edema","stage"))
print(tableOne, nonnormal = c("bili","chol","copper","alk.phos","trig"), exact = c("status","stage"), smd = TRUE)
the output:
I need to have the p-values for each level of the variables status, edema and stage, with Bonferroni correction. I went through the documentation without success.
In addition, is it correct to use chi-squared to compare sample sizes across rows?
UPDATE:
I'm not sure if my approach is correct, however I would like to share it with you. I generated for the variable status a dummy variable for each strata, than I calculated the chisq .
library(tableone)
## Load data
library(survival); data(pbc)
d <- pbc[,c("status", "trt")]
# Convert dummy variables
d$status.0 <- ifelse(d$status==0, 1,0)
d$status.1 <- ifelse(d$status==1, 1,0)
d$status.2 <- ifelse(d$status==2, 1,0)
t <- rbind(
chisq.test(d$status.0, d$trt),
# p-value = 0.7202
chisq.test(d$status.1, d$trt),
# p-value = 1
chisq.test(d$status.2, d$trt)
#p-value = 0.7818
)
t
BONFERRONI ADJ FOR MULTIPLE COMPARISONS:
p <- t[,"p.value"]
p.adjust(p, method = "bonferroni")
This question was posted some time ago, so I supose you already answered the reviewer.
I don't really understand why computing adjusted p values for just three varibles. In fact, adjusting p values depends on the number of comparisons made. If you use p.adjust() with a vector of 3 p values, results will not really be "adjusted" by the amount of comparison made (you really did more than a dozen and a half!)
I show how to extract all p-values so you can compute the adjusted ones.
To extract pValues from package tableOne there is a way calling object attributes (explained first), and two quick and dirty ways (at the bottom part).
To extract them, first I copy your code to create your tableOne:
library(tableone)
## Load data
library(survival); data(pbc)
# drop ID from variable list
vars <- names(pbc)[-1]
## Create Table 1 stratified by trt (can add more stratifying variables)
tableOne <- CreateTableOne(vars = vars, strata = c("trt"), data = pbc, factorVars = c("status","edema","stage"))
You can see what your "tableOne" object has via attributes()
attributes(tableOne)
You can see a tableOne usually has a table for continuous and categorical variables. You can use attributes() in them too
attributes(tableOne$CatTable)
# you can notice $pValues
Now you know "where" the pValues are, you can extract them with attr()
attr(tableOne$CatTable, "pValues")
Something similar with numerical variables:
attributes(tableOne$ContTable)
# $pValues are there
attr(tableOne$ContTable, "pValues")
You have pValues for Normal and NonNormal variables.
As you set them before, you can extract both
mypCont <- attr(tableOne$ContTable, "pValues") # put them in an object
nonnormal = c("bili","chol","copper","alk.phos","trig") # copied from your code
mypCont[rownames(mypCont) %in% c(nonnormal), "pNonNormal"] # extract NonNormal
"%!in%" <- Negate("%in%")
mypCont[rownames(mypCont) %!in% c(nonnormal), "pNonNormal"] # extract Normal
All that said, and your pValues extracted, I think there are two much more convenient quick and dirty ways to accomplish the same:
Quick and dirty way A: using dput() with your printed tableOne. Then search in the console where the pValues are and copy-paste them to the script, to store them in an object
Quick and dirty way B: If you look in tableOne vignette there is an "Exporting" section, you can use print(tableOne, quote = TRUE) and then just copy and paste to a spreadsheet (like LibreOffice, Excel...).
Then I would select the column with pValue, transpose it, and get it back to R, to compute adjusted p values with p.adjust() and copy them back to the spreadsheet for journal submission