I am confused as to why the two pieces of code are returning different results.
In one the only difference between the loops is the use of Wage$age.cut1 vs. age.cut1. What is the significance of the difference?
DATA: ISLR package, Wage data
cv.err <- rep(NA, 10)
for (i in 2:10){
Wage$age.cut1 = cut(Wage$age, i)
fit = glm(wage~age.cut1, data = Wage)
cv.err[i] = cv.glm(Wage, fit, K = 10)$delta[2]
}
> cv.err
[1] NA 1733.646 1681.587 1636.521 1632.931 1623.112 1611.965 1600.903 1609.973
[10] 1607.234 # these are the expected results
cv.err <- rep(NA, 10)
for (i in 2:10){
age.cut1 = cut(Wage$age, i)
fit = glm(wage~age.cut1, data = Wage)
cv.err[i] = cv.glm(Wage, fit, K = 10)$delta[2]
}
> cv.err
[1] NA 1603.255 1608.617 1602.296 1606.265 1606.139 1602.448 1606.063 1605.100
[10] 1606.986
Yes, the difference of those two make a great difference in your loop logic. In first loop, age.cut1 is a column in Wage dataframe evidenced by the $ qualifier and is used in the glm formula. In second loop, age.cut1 is a standalone, separate named vector and is unused in the glm formula. Whenever a formula is used, the columns derive from the object referenced in data argument.
Unfamiliar with listed packages and data structures, most likely age.cut1 column does exist in Wage data frame prior to looping (since no error occurred in its reference in second loop's glm call). However, it is only updated in first loop with cut(Wage$age, i). Though a similar named object is assigned in second loop, the original column data remains untouched in glm.
Related
My file folder with the N datasets in the form of csv files is called sample_obs. The goal is to end up with two lists, one I have already figured out how to obtain, namely, a list of the names of each individual csv file that matches the format of their actual names in the folder all of them are in, not their file paths.
So, this is all the code I have written and the other list I need to create is a list of the factors/Independent Variables chosen by my Backward Elimination Stepwise Regression function only, no R-squared, Cp, AIC, BIC, or any other standard regression diagnostic tools; I don't want or need coefficient estimates either, just the regressors "chosen" for each dataset out of the 30 candidate regressors.
So far, in terms of my code that actually runs (except the last few lines):
# these 2 lines together create a simple character list of
# all the file names in the file folder of datasets you created
directory_path <- "~/DAEN_698/sample_obs"
file_list <- list.files(path = directory_path, full.names = TRUE, recursive = TRUE)
head(file_list, n = 2)
> head(file_list, n = 2)
[1] "C:/Users/Spencer/Documents/DAEN_698/sample_obs2/0-5-1-1.csv"
[2] "C:/Users/Spencer/Documents/DAEN_698/sample_obs2/0-5-1-2.csv"
# Create another list with the just the "n-n-n-n" part of the names of of each dataset
DS_name_list = stri_sub(file_list, 49, 55)
head(DS_name_list, n = 3)
> head(DS_name_list, n = 3)
[1] "0-5-1-1" "0-5-1-2" "0-5-1-3"
# This command reads all the data in each of the N csv files via their names
# stored in the 'file_list' list of characters.
csvs <- lapply(file_list, read.csv)
### Step 3: Run a Backward Elimination Stepwise Regression function on each of the N csvs.
# Assign the full models (meaning one with all 30 candidate regressors included in step 1)
# as the initial model that BE starts out with.
# This is crucial because if the initial model had less than the number of candidate factors # in the datasets, e.g. 25 (so, X1:X26), then it could miss 1 or more of the factors
# X26:X30 which ought to be 'chosen' in dataset j by Stepwise j.
full_model <- lapply(csvs, function(i) {
lm(formula = Y ~ ., data = i) })
Finally, this is the part where I get really tripped up. I have tried at least 6 different sets of arguments, different syntax, using different objects, etc. when running my BE Stepwise Regression on my N datasets, but I'll just include 2 of them below which take entirely different approaches but are both wrong:
# attempt 1
set.seed(50) # for reproducibility
BE_fits1 <- map(.x = full_model[-1], .f = function(i) { step(object = all_IVs_models2, direction = 'backward', scope = formula(full_model), trace = 0) })
# attempt 3
set.seed(50) # for reproducibility
BE_fits3 <- lapply(full_model, function(i) {
step(object = i[["coefficients"]], direction = 'backward',
scope = formula(full_model), trace = 0)
When I hit Ctrl+Enter on attempt 1, I get the following error message:
Error in x$terms %||% attr(x, "terms") %||% stop("no terms component nor attribute") :
no terms component nor attribute
And when I try to run my code for attempt #3, I get the following different error message:
Error in x$terms : $ operator is invalid for atomic vectors
I don't recognize either of these error messages.
p.s. If anyone looking over this question would like, I can re ask this question but including MUCH less minute details if you want me to.
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 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 have a dataframe dfab that contains 2 columns that I used as argument to generate a series of linear models as following:
models = list()
for (i in 1:10){
models[[i]] = lm(fc_ab10 ~ (poly(nUs_ab, i)), data = dfab)
}
dfab has 32 observations and I want to predict fc_ab10 for only 1 value.
I thought of doing so:
newdf = data.frame(newdf = nUs_ab)
newdf[] = 0
newdf[1,1] = 56
prediction = predict(models[[1]], newdata = newdf)
First I tried writing newdf as a dataframe with only one position, but since there are 32 in the dataset on which the model was built, I thought I had to provide at least 32 points as well. I don't think this is necessary though.
Every time I run that piece of code I am given the following error:
Error: variable 'poly(nUs_ab, i) was fitted with type “nmatrix.1” but type “numeric” was supplied.
In addition: Warning message:
In Z/rep(sqrt(norm2[-1L]), each = length(x)) :
longer object length is not a multiple of shorter object length
I thought all I need to use predict was a LM model, predictors (the number 56) given in a column-named dataframe. Obviously, I am mistaken.
How can I fix this issue?
Thanks.
newdf should be a data.frame with column name nUs_ab, otherwise R won't be able to know which column to operate upon (i.e., generate the prediction design matrix). So the following code should work
newdf = data.frame(nUs_ab = 56)
prediction = predict(models[[1]], newdata = newdf)
I need to reproduce this code using all of these variables.
composite <- read.csv("file.csv", header = T, stringsAsFactors = FALSE)
composite <- subset(composite, select = -Date)
model1 <- lm(indepvariable ~., data = composite, na.action = na.exclude)
composite is a data frame with 82 variables.
UPDATE:
What I have done is found a way to create an object that contains only the significantly correlated variables, to narrow the number of independent variables down.
I have a variable now: sigvars, which is the names of an object that sorted a correlation matrix and picked out only the variables with correlation coefficients >0.5 and <-0.5. Here is the code:
sortedcor <- sort(cor(composite)[,1])
regvar = NULL
k = 1
for(i in 1:length(sortedcor)){
if(sortedcor[i] > .5 | sortedcor[i] < -.5){
regvar[k] = i
k = k+1
}
}
regvar
sigvars <- names(sortedcor[regvar])
However, it is not working in my lm() function:
model1 <- lm(data.matrix(composite[1]) ~ sigvars, data = composite)
Error: Error in model.frame.default(formula = data.matrix(composite[1]) ~ sigvars, : variable lengths differ (found for 'sigvars')
Think about what sigvars is for a minute...?
After sigvars <- names(sortedcor[regvar]), sigvars is a character vector of column names. Say your data have 100 rows and 5 variables come out as significant using the method you've chosen (which doesn't sound overly defensible to be). The model formula you are using will result in composite[, 1] being a vector of length 100 (100 rows) and sigvars being a character vector of length 5.
Assuming you have the variables you want to include in the model, then you could do:
form <- reformulate(sigvars, response = names(composite)[1])
model1 <- lm(form, data = composite)
or
model1 <- lm(composite[,1] ~ ., data = composite[, sigvars])
In the latter case, do yourself a favour and write the name of the dependent variable into the formula instead of composite[,1].
Also, you don't seem to have appreciated the difference between [i] and [i,j] for data frames, hence you are doing data.matrix(composite[1]) which is taking the first component of composite, leaving it as a data frame, then converting that to a matrix via the data.matrix() function. All you really need is just the name of the dependent variable on the LHS of the formula.
The error is here:
model1 <- lm(data.matrix(composite[1]) ~ sigvars, data = composite)
The sigvars is names(data). The equation is usually of the form lm(var1 ~ var2+var3+var4), you however have it as lm(var1 ~ var2 var3 var4).
Hopefully that helps.