im quite new to R but wanted to use the packages "nls" and "nlstools" since it has nice tools for analysis and evaluation.
the code I use is:
conB1_2015 = read.csv("C:\\Path_to_File\\conB1_2015.csv")
conB1_2015 = na.omit(conB1_2015)
tRef <- mean(conB1_2015$Mean_Soil_Temp_V2..C., na.rm=TRUE)
rRef <- conB1_2015$Lin_Flux..mymol.m.2.s.1.[which.min(abs(conB1_2015$Mean_Soil_Temp_V2..C.-tRef))]
rMax <- max(conB1_2015$Lin_Flux..mymol.m.2.s.1., na.rm=TRUE)
half <- rMax/2
half_SM <- conB1_2015$Soil_Moist_V3[which.min(abs(conB1_2015$Lin_Flux..mymol.m.2.s.1.-half))]
form <- as.formula(Lin_Flux..mymol.m.2.s.1. ~ (rRef)*a*exp(b*Mean_Soil_Temp_V2..C.)*Soil_Moist_V3/(half_SM)+Soil_Moist_V3)
preview(form, data = conB1_2015, start = c(a = -1.98, b = -0.05), variable = 1)
The Problem is, that i get this Error running this code:
Error in data.frame(value, row.names = rn, check.names = FALSE) :
row names supplied are of the wrong length
When i change the variables in form <- as.formula(Lin_Flux..mymol.m.2.s.1. ~ (rRef)*a*exp(b*Mean_Soil_Temp_V2..C.)*Soil_Moist_V3/(half_SM)+Soil_Moist_V3)
to form <- as.formula(Lin_Flux..mymol.m.2.s.1.~(rRef<-4.41)*a*exp(b*Mean_Soil_Temp_V2..C.)*Soil_Moist_V3/(half_SM<-7.19)+Soil_Moist_V3)
the function works fine.
I wanted to automate the script to run over several csv's to test different models on different data. Is it really not possible to pass variables into the preview function or am I missing something? There can't be a problem with headers or the data table since it's working fine in the second example.
Related
I am trying to use the lqmm package in R and receiving the error Error in f(arg, ...) : NA/NaN/Inf in foreign function call (arg 1). I can successfully use it for a version of my data in which a variable called cluster_name is averaged over.
I've tried to verify that there are no NaNs or infinite values in my dataset this way:
na_data = mydata
new_DF <- na_data[rowSums(is.na(mydata)) > 0,] # yields a dataframe with no observations
is.na(na_data) <- sapply(na_data, is.infinite)
new_DF <- na_data[rowSums(is.na(mydata)) > 0,] # still a dataframe with no observations
There are no variables in my dataframe that are type char -- every such variable has been converted to a factor.
When I run my model
m1 = lqmm(std_brain ~ std_beh*type*taught, random = ~1, group=subject, data = begin_data, tau=.5, na.action=na.exclude)
on the first 12,528 lines of my dataset, the model works fine. Line 12,529 looks totally normal.
Similarly, if I run tail(mydata, 11943) I get a dataframe that runs without error, but tail(mydata, 11944) gives me a dataframe that generates the error. I can also run a subset from 9990:21825 without error, but extending the dataframe on either side generates the error. The whole dataframe is 29450 observations, and thus this middle slice contains the supposedly problematic observations. I tried making a smaller version of my dataset that contained just the borders of problems, and some observations around them, and I can see that 3/4 cases involve the same subject (7645), but I don't know what to make of that. I don't see how to make this reproducible without providing the whole dataframe (in case you were wondering, the small dataset doesn't cause any error). So here is the csv file I used.
Here is the function that gets the dataframe ready for analysis:
prep_data_set <- function(data_file, brain_var = 'beta', beh_var = 'accuracy') {
data = read.csv(data_file)
data$subject <- factor(data$subject)
data$type <- factor(data$type)
data$type <- relevel(data$type, ref = "S")
data$taught <- factor(data$taught)
data <- subset(data, data$run_num < 13)
data$run = factor(data$run_num)
brain_mean <- mean(data[[brain_var]])
brain_sd <- sd(data[[brain_var]])
beh_mean <- mean(data[[beh_var]])
beh_sd <- sd(data[[beh_var]])
data <- subset(data, data$cluster_name != "")
data$cluster_name <- factor(data$cluster_name)
data$mean_centered_brain <- data[[brain_var]]
data$std_brain <- data$mean_centered_brain/brain_sd
data$mean_centered_beh <- data[[beh_var]]
data$std_beh <- data$mean_centered_beh/beh_sd
return(data)
}
I run
mydata = prep_data_set(file.path(resdir, 'robust0005', 'pos_rel_con__all_clusters.csv'))
m1 = lqmm(std_brain ~ std_beh*type*taught, random = ~1, group=subject, data = mydata, tau=.5, na.action=na.exclude)
to generate the error.
By comparison
regular_model = lmer(std_brain ~ type*taught*std_beh + (1|subject/run) +
(1|subject:cluster_name), data = mydata)
runs fine.
I hope there is something interesting and generalizable in this question; I know it's kind of annoying to post to Stack Overflow with some idiosyncratic problem in a ~30000 line dataset.
I am trying to write a function in R, which contains a function from another package. The code works perfectly outside a function.
I am guessing, it might have got to do something with the package I am using (survey).
A self-contained code example:
#activating the package
library(survey)
#getting the dataset into R
tm <- read.spss("tm.sav", to.data.frame = T, max.value.labels = 5)
# creating svydesign object (it basically contains the weights to adjust the variables (~persgew: also a column variable contained in the tm-dataset))
tm_w <- svydesign(ids=~0, weights = ~persgew, data = tm)
#getting overview of the welle-variable
#this variable is part of the tm-dataset. it is needed to execute the following steps
table(tm$welle)
# data manipulation as in: taking the v12d_gr-variable as well as the welle-variable and the svydesign-object to create a longitudinal variable which is transformed into a data frame that can be passed to ggplot
t <- svytable(~v12d_gr+welle, tm_w)
tt <- round(prop.table(t,2)*100, digits=0)
v12d <- tt[2,]
v12d <- as.data.frame(v12d)
this is the code outside the function, working perfectly. since I have to transform quite a few variables in the exact same way, I aim to create a function to save up some time.
The following function is supposed to take a variable that will be transformed as an argument (v12sd2_gr).
#making sure the survey-object is loaded
tm_w <- svydesign(ids=~0, weights = ~persgew, data = data)
#trying to write a function containing the code from above
ltd_zsw <- function(variable1){
t <- svytable(~variable1+welle, tm_w)
tt <- round(prop.table(t,2)*100, digits=0)
var_ltd_zsw <- tt[2,]
var_ltd_zsw <- as.data.frame(var_ltd_zsw)
return(var_ltd_zsw)
}
Calling the function:
#as v12d has been altered already, I am trying to transform another variable v12sd2_gr
v12sd2 <- ltd_zsw(v12sd2_gr)
Console output:
Error in model.frame.default(formula = weights ~ variable1 + welle, data = model.frame(design)) :
invalid type (closure) for variable 'variable1'
Called from: model.frame.default(formula = weights ~ variable1 + welle, data = model.frame(design))
How do I fix it? And what does it mean to dynamically build a formula and reformulating?
PS: I hope it is the appropriate way to answer to the feedback in the comments.
Update: I think I was able to trace the problem back to the argument I am passing (variable1) and I am guessing it has got something to do with the fact, that I try to call a formula within the function. But when I try to call the svytable with as.formula(svytable(~variable1+welle, tm_w))it still doesn't work.
What to do?
I have found a solution to the problem.
Here is the tested and working function:
ltd_test <- function (var, x, string1="con", string2="pro") {
print (table (var))
x$w12d_gr <- ifelse(as.numeric(var)>2,1,0)
x$w12d_gr <- factor(x$w12d_gr, levels = c(0,1), labels = c(string1,string2))
print (table (x$w12d_gr))
x_w <- svydesign(ids=~0, weights = ~persgew, data = x)
t <- svytable(~w12d_gr+welle, x_w)
tt <- round(prop.table(t,2)*100, digits=0)
w12d <- tt[2,]
w12d <- as.data.frame(w12d)
}
The problem appeared to be caused by the svydesgin()-fun. In its output it produces an object which is then used by the formula for svytable()-fun. Thats why it is imperative to first create the x_w-object with svydesgin() and then use the svytable()-fun to create the t-object.
Within the code snippet I posted originally in the question the tm_w-object has been created and stored globally.
Thanks for the help to everyone. I hope this is gonna be of use to someone one day!
When I run the t-test for a numeric and a dichotomous variable there in no problem and I can see the results. The problem is when I run the ggttest of the same t-test. There is an error and says that one of my variable is not found. I do not why that happens. The aml dataset I used is from package boot. Below you can see the code:
https://i.stack.imgur.com/7kuaA.png
library(gginference)
time_group.test16537 = t.test(formula = time~group,
data = aml,
alternative = "two.sided",
paired = FALSE,
var.equal = FALSE,
conf.level = 0.95)
time_group.test16537
ggttest(time_group.test16537,
colaccept="lightsteelblue1",
colreject="gray84",
colstat="navyblue")
The problem comes with these lines of code in ggttest:
datnames <- strsplit(t$data.name, splitter)
len1 <- length(eval(parse(text = datnames[[1]][1])))
len2 <- length(eval(parse(text = datnames[[1]][2])))
It tries to find the len of group and time, but it doesn't see that it came from a data.frame. Pretty bad bug...
For your situation, supposedly you have less than 30 in each group and it plots a t-distribution, so do:
library(gginference)
library(boot)
gginference:::normt(t.test(time~group,data=aml),
colaccept = "lightsteelblue1",colreject = "grey84",
colstat = "navyblue")
t.test doesn't store your data in the output so there is no way that you could extract the data from the list of the output of t.test.
The only way to use formula is:
library(gginference)
t_test <- t.test(questionnaire$pulse ~ questionnaire$gender)
ggttest(t_test)
Original answer here: How to extract the dataset from an "htest" object when using formula in r
I have a mixed type data set, one continuous variable, and eight categorical variables, so I wanted to try kamila clustering. It gives me an error when I use one continuous variable, but when I use two continuous variables it is working.
library(kamila)
data <- read.csv("mixed.csv",header=FALSE,sep=";")
conInd <- 9
conVars <- data[,conInd]
conVars <- data.frame(scale(conVars))
catVarsFac <- data[,c(1,2,3,4,5,6,7,8)]
catVarsFac[] <- lapply(catVarsFac, factor)
kamRes <- kamila(conVars, catVarsFac, numClust=5, numInit=10,calcNumClust = "ps",numPredStrCvRun = 10, predStrThresh = 0.5)
Error in kamila(conVar = conVar[testInd, ], catFactor =
catFactor[testInd, : Input datasets must be dataframes
I think the problem is that the function assumes that you have at least two of both data types (i.e. >= 2 continuous variables, and >= 2 categorical variables). It looks like you supplied a single column index (conInd = 9, just column 9), so you have only one continuous variable in your data. Try adding another continuous variable to your continuous data.
I had the same problem (with categoricals) and this approach fixed it for me.
I think the ultimate source of the error in the program is at around line 170 of the source code. Here's the relevant snippet...
numObs <- nrow(conVar)
numInTest <- floor(numObs/2)
for (cvRun in 1:numPredStrCvRun) {
for (ithNcInd in 1:length(numClust)) {
testInd <- sample(numObs, size = numInTest, replace = FALSE)
testClust <- kamila(conVar = conVar[testInd,],
catFactor = catFactor[testInd, ],
numClust = numClust[ithNcInd],
numInit = numInit, conWeights = conWeights,
catWeights = catWeights, maxIter = maxIter,
conInitMethod = conInitMethod, catBw = catBw,
verbose = FALSE)
When the code partitions your data into a training set, it's selecting rows from a one-column data.frame, but that returns a vector by default in that case. So you end up with "not a data.frame" even though you did supply a data.frame. That's where the error comes from.
If you can't dig up another variable to add to your data, you could edit the code such that the calls to kamila in the cvRun for loop wrap the data.frame() function around any subsetted conVar or catFactor, e.g.
testClust <- kamila(conVar = data.frame(conVar[testInd,]),
catFactor = data.frame(catFactor[testInd,], ... )
and just save that as your own version of the function called say, my_kamila, which you could use instead.
Hope this helps.
I'm fairly new to using the caret library and it's causing me some problems. Any
help/advice would be appreciated. My situations are as follows:
I'm trying to run a general linear model on some data and, when I run it
through the confusionMatrix, I get 'the data and reference factors must have
the same number of levels'. I know what this error means (I've run into it before), but I've double and triple checked my data manipulation and it all looks correct (I'm using the right variables in the right places), so I'm not sure why the two values in the confusionMatrix are disagreeing. I've run almost the exact same code for a different variable and it works fine.
I went through every variable and everything was balanced until I got to the
confusionMatrix predict. I discovered this by doing the following:
a <- table(testing2$hold1yes0no)
a[1]+a[2]
1543
b <- table(predict(modelFit,trainTR2))
dim(b)
[1] 1538
Those two values shouldn't disagree. Where are the missing 5 rows?
My code is below:
set.seed(2382)
inTrain2 <- createDataPartition(y=HOLD$hold1yes0no, p = 0.6, list = FALSE)
training2 <- HOLD[inTrain2,]
testing2 <- HOLD[-inTrain2,]
preProc2 <- preProcess(training2[-c(1,2,3,4,5,6,7,8,9)], method="BoxCox")
trainPC2 <- predict(preProc2, training2[-c(1,2,3,4,5,6,7,8,9)])
trainTR2 <- predict(preProc2, testing2[-c(1,2,3,4,5,6,7,8,9)])
modelFit <- train(training2$hold1yes0no ~ ., method ="glm", data = trainPC2)
confusionMatrix(testing2$hold1yes0no, predict(modelFit,trainTR2))
I'm not sure as I don't know your data structure, but I wonder if this is due to the way you set up your modelFit, using the formula method. In this case, you are specifying y = training2$hold1yes0no and x = everything else. Perhaps you should try:
modelFit <- train(trainPC2, training2$hold1yes0no, method="glm")
Which specifies y = training2$hold1yes0no and x = trainPC2.