I programmed a function, which created (or at least tried to create) a data frame of numeric values. I need to retrieve these numeric values later on in the function. For that purpose, I explicitly assigned all values in the data frame a numeric class, using
as.numeric()
Later on in my function, when I extract the elements from the data frame, using
mydataframe[1,2]
I get an error "non-numeric argument to binary operator". I don't really understand what is non-numeric in my data frame.
If I ask for class and mode of the values in the data frame, they are both "numeric", storage mode is "double". Can anyone enlighten me? Where do I go wrong?
By the way, I can extract elements without error, if I use
as.numeric(mydataframe[1,2])
But I need to extract quite a lot of elements, so I prefer all elements of my data frame being numeric.
My code:
mydata <- by(data, data[,index], function(data) {
*myfunction including a for-loop, creating a vector of numbers (subvar1)*}
var1 <- as.numeric(sum(subvar1) / n)
var2 <- as.numeric(mean(data[,value]))
var3 <- nrow(data)
var3 <- as.numeric(var3)
list(var1=var1, var2=var2, var3=var3)})
mydataframe <- data.matrix(do.call(rbind, mydata))
Thanks in advance!
Related
I'm working, in RStudio, with data for patients that are either normal, have Crohn's disease, or ulcerative colitis. Now, the data is structured in such a way that patient information is in a separate data frame (called sampleInfo), and the data I want to use for analysis is in a different data frame (called expressionData). For my analysis, I would like to remove the patients that are 'normal' from the dataset and only keep those with Crohn's disease or ulcerative colitis.
So, what I did was first run the following command to make a new data frame from sampleInfo containing all the patients (aka rows) with the normal disease state, using the following command:
bad_patients <- sampleInfo[sampleInfo$characteristics_ch1.3 == "disease state: normal", ]
bad_patients has a column called geoaccession, which contains the patient ID, which also corresponds with the column names for the same patient in expressionData.
I save the names of these IDs using
patient_names <- bad_patients$geo_accession.
Now, I want to remove the columns with these names from expressionData. I looked at a lot of different StackOverflow posts, as well as posts on the R help forum, and found two main ways, both of which I have tried. The first is done with the following command:
newDataFrame <- expressionData[ , !names(expressionData) %in% patient_names]
Though this method does produce a new matrix called newDataFrame, attempting to view this matrix in RStudio gives the following error:
Error in View : 'names' attribute [1] must be the same length as the vector [0]
I also tried a second subset method with the following command:
newDataFrame <- subset(expressionData, -patient_names)
which raises the error: Error in -patient_names : invalid argument to unary operator
I also tried this subset method by explicity typing out the columns I wanted to remove as follows:
newDataFrame <- subset(expressionData, -c('ID090190', ...) (where ... corresponds to the rest of the IDs) and got the same exact error.
Can someone tell me what I'm doing wrong, or how to work around this?
Couple of solutions:
Subsetting based on names
newDataFrame <- expressionData[!(names(expressionData) %in% patient_names)]
One problem with your attempt was that you hadn't wrapped the whole expression evaluated by ! in parentheses. As it was, you were looking for !names(expressionData) in patient_names. ! here would coerce names(expressionData) into a logical and likely return a vector full of FALSEs
I've subset with only one dimension (x[this] rather than x[,this]). You can do this with the columns of data frames because a data frame is a list of its columns. This subsetting method preserves the data.frame class of the returned object, whereas the two-dimensional subset will just return a vector if you select only one column. (Tibbles will return a tibble with both methods, which is one big advantage of tibbles)
Tidyverse solution: use dplyr::select with dplyr::all_of
newDataFrame <- dplyr::select(expressionData, -dplyr::all_of(patientnames))
Edit: Make sure your data really is a data.frame
If you're getting this error Error in UseMethod("select_") : no applicable method for 'select_' applied to an object of class "c('matrix', 'array', 'double', 'numeric')", it's because your data is a matrix, rather than a data frame. You may have inadvertently coerced it in processing.
Use as.data.frame to return to a data frame object, which will be compabtible with the methods above. If you wish to keep your data as a matrix, use colnames:
expressionData[ , !(colnames(expressionData) %in% patient_names)] to subset the columns.
If expressionData is a matrix, you'll need to subset the columns with colnames, rather than names. The names of a data.frame are identical to its colnames (because a df is a list of its columns), but the names of a matrix are the names of every element in the matrix, because a matrix is just an array with dimensionality. You'll want to check colnames(expressionData) to make sure that there are colnames to subset.
You might want to try:
newDataFrame <- expressionData[ , !colnames(expressionData) %in% patient_numbers]
names(expressionData) is NULL, hence your error; you want the column names
in your example, your list of sample names was called patient_numbers, not patient_names
I have n data frames, each corresponding to data from a city.
There are 3 variables per data frame and currently they are all factor variables.
I want to transform all of them into numeric variables.
I have started by creating a vector with the names of all the data frames in order to use in a for loop.
cities <- as.vector(objects())
for ( i in cities){
i <- as.data.frame(lapply(i, function(x) as.numeric(levels(x))[x]))
}
Although the code runs and there I get no error code, I don't see any changes to my data frames as all three variables remain factor variables.
The strangest thing is that when doing them one by one (as below) it works:
df <- as.data.frame(lapply(df, function(x) as.numeric(levels(x))[x]))
What you're essentially trying to do is modify the type of the field if it is a factor (to a numeric type). One approach using purrr would be:
library(purrr)
map(cities, ~ modify_if(., is.factor, as.numeric))
Note that modify() in itself is like lapply() but it doesn't change the underlying data structure of the objects you are modifying (in this case, dataframes). modify_if() simply takes a predicate as an additional argument.
for anyone who's interested in my question, I worked out the answer:
for ( i in cities){
assign(i, as.data.frame(lapply(get(i), function(x) as.numeric(levels(x))[x])))
}
I have a data frame consisting of five character variables which represent specific bacteria. I then have thousands of observations of each variable that all begin with the letter K. eg
x <- c(K0001,K0001,K0003,K0006)
y <- c(K0001,K0001,K0002,K0003)
z <- c(K0001,K0002,K0007,K0008)
r <- c(K0001,K0001,K0001,K0001)
o <- c(K0003,K0009,K0009,K0009)
I need to identify unique observations in the first column that don't appear in any of the remaining four columns. I have tried the approach suggested here which I think would work if I could create individual vectors using select ...
How to tell what is in one vector and not another?
but when I try to create a vector for analysis using the code ...
x <- select(data$x)
I get the error
Error in UseMethod("select_") :
no applicable method for 'select_' applied to an object of class "character
I have tried to mutate the vectors using as.factor and as.numeric but neither of these approaches work as the first gives an equivalent error as above, and as.numeric returns NAs.
Thanks in advance
The reference that you cited recommended using setdiff. The only thing that you need to do to apply that solution is to convert the four columns into one, so that it can be treated as a set. You can do that with unlist
setdiff(data$x, unlist(data[,2:5]))
"K0006"
I've imported a data frame from a csv-file
dat3 <- read.csv(file.choose(),as.is = TRUE)
contains names and values. My problem is, that when I try to replace a value in the data frame, e.g.
dat3[3,6]<-12
then it just assumes, that "12" is a text string and not a value, thus preventing me from using that number to mathematical operations. I'd like to being able to replace some numbers in the data frame and using them for mathematical operations.
When I try adding 1 to dat3[3,6] I get: "Error in dat3[3, 6] + 1 : non-numeric argument to binary operator".
I've tried:
lapply(dat3[3,6], as.numeric)
dat3[3,6]<-as.numeric(12)
But it doesn't work. I have though no problems in using the already imported numbers in the data frame. This only happens for numbers which I replace.
Yes!
I've found the answer!
It is:
dat[, c(3:6)] <- sapply(dat[, c(3:6)], as.numeric)
to convert column to numbers.
Thank you all!
I am having trouble turning my data.frame into a matrix format. Because I wanted to change my data.frame with mostly factor variables into a numeric matrix, I used the following code
UN2010frame <- data.matrix(lapply(UN2010, as.numeric))
However when I checked the mode of the UN2010frame, it still showed up as a list. Because the code I want to run (Ordrating) does not accept data in a list format, I used UN2010matrix <- unlist(UN2010frame) to unlist my matrix. When I did this, my first row ( which was formerly a row with column names) turned into NAs. This was a problem for me because when I tried to run an ordinal IRT model using this data set, I got the following error message.
> Error in 1:nrow(Y) : argument of
> length 0
I think it is because all the values in my first row are now gone.
If you could help me on any front, It would be deeply appreciated.
Thank you very much!
Haillie
First, the correct use of data.matrix is :
data.matrix(UN2010)
as it converts automatically to numeric. The lapply in your code is the first source for the error you get. You put a list in the data.matrix function, not a dataframe. So it returns a list of matrices, and not a matrix.
Second, unlist returns a vector, not a matrix. So pretty sure you won't find a "first row with NA", as you have a vector. Which might explain part of your confusion.
You probably have a character column somewhere. Converting this to numeric gives NA. If you don't want this, then exclude them from the further analysis. One possibility is to use colwise() from the plyr package to convert only the factors:
colwise(as.numeric,is.factor)(UN2010)
Which returns a dataframe with only the factors. This can be easily converted by data.matrix() or as.matrix(). Alternatively you use the base solution :
id <- sapply(UN2010,is.character)
sapply(UN2010[!id],as.numeric)
which will return you a matrix with all non-character columns converted to numeric.If you really want to keep the dataframe with all original columns, you can do :
UN2010frame <- UN2010
UN2010frame[!id] <- lapply(UN2010[!id],as.numeric)
Toy example code :
UN2010 <- data.frame(
F1 = factor(rep(letters[1:3],10)),
F2 = factor(rep(letters[5:10],5)),
Char = rep(letters[11:16],each=5),
Num = 1:30,
stringsAsFactors=FALSE
)
Try as.data.frame instead of data.matrix.