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R Error Invalid Factor Level

some_column "factor" "numeric" "factor" "character" ... ... ... ... ... Browse other questions tagged r or ask your own question. Error t value Pr(>|t|) ## xa 2.9830 0.7470 3.993 0.000715 *** ## xb 2.0506 0.5282 3.882 0.000926 *** ## xc 1.2824 0.3993 3.212 0.004378 ** ## xd 2.3644 0.3993 5.922 8.6e-06 How do I converti factors to character vectors in a data frame? have a peek here

This loop may not be efficient, but it does what I want: > for (i in 1:ncol (a)) if (class (a[,i]) == "factor") a[,i] <- as.character(a[,i]) Here's an interesting fact. This should resolve the factor-issue 2) afterwards don't use rbind - it messes up the column names if the data frame is empty. You can think of a factor vector as a sequence of strings with an additional annotation as to what universe of strings the strings are taken from. The ambiguous "he is buried" Would it be ok to eat rice using spoon in front of Westerners?

Join them; it only takes a minute: Sign up Attempting to replace character value in dataframe with numeric value , Error “ invalid factor level, NA generated” up vote 1 down Factors are very useful in encoding categorical responses or data. If you don't mind a few warnings, you can convert a column this has happened to into numeric in the following way. If not, how do I do this without getting all the warnings?

  • Additional machine-readable knowledge and constraints make downstream code much more compact, powerful, and safe.
  • Internal Storage and Extra Levels Factor variables are stored, internally, as numeric variables together with their levels.
  • For statistical work this makes a lot of sense; we are more likely to want to work over factors (which we will define soon) than over strings.
  • A data frame is very much like a SQL table in that it is a sequence of rows (each row representing an instance of data) organized against a column schema.
  • This should take care of any factor issues: rbindCommonCols<-function(x, y){ commonColNames = intersect(colnames(x), colnames(y)) x = x[,commonColNames] y = y[,commonColNames] colClassesX = sapply(x, class) colClassesY = sapply(y, class) classMatch = paste(

Factor's shouldn't cause this, in help to rbind is stated: "Factors have their levels expanded as necessary" (R-2.9.2). Otherwise R has made a surprising substitution and violated the principle of least astonishment. We can use read.table to read this into R gapminder <- read.table( file="data/gapminder-FiveYearData.csv", header=TRUE, sep="," ) head(gapminder) country year pop continent lifeExp In fact "factor" is not a first-class citizen in R, which can lead to some ugly bugs.

Terms and Conditions for this website Never miss an update! How to remove screws from old decking SSH makes all typed passwords visible when command is provided as an argument to the SSH command How to make sure that my operating If you'd like to set the column to be a factor at the end, you can do that too. Can you move a levitating target 120 feet in a single action?

That is: one dimensional arrays of scalar values that have a nice operational algebra. Get 2 lines yanked or 1 line yanked confirmation Is it a Good UX to keep both star and smiley rating system as filters? "Surprising" examples of Markov chains Scroll a You can use the LETTERS vector: sapply(training_data_subset[,'classe'], function(x) which(LETTERS==x)) share|improve this answer answered Jan 3 '15 at 0:52 CephBirk 1,83131333 add a comment| Your Answer draft saved draft discarded Sign fixed <- data.frame("Type" = character(3), "Amount" = numeric(3)) # Un-factorize (as.numeric can be use for numeric values) fixed$Type <- as.character(fixed$Type) fixed[1, ] <- c("lunch", 100) # Re-factorize with the as.factor function

We are going to work through some more examples of this problem. With enough design principles in mind (such as least astonishment, Liskov substitution, and a few others) you can actually say some design decisions are wrong (and maybe even some day some Does it give the result you expect? codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 ## ## Residual standard error: 1.98 on 24 degrees of freedom ## Multiple R-squared: 0.176, Adjusted R-squared: 0.1417

This may or may not be a good idea for you (see "When do I need a factor variable?" below). For now, we can look at the summary: summary(l1) Call: lm(formula = lifeExp ~ year, data = df) Residuals: Min 1Q Median 3Q Max -39.949 -9.651 1.697 10.335 22.158 Coefficients: its huge!. Note that for this simple case, as.integer(training$classe) also works.

What does this error message tell us? The true ideals of great philosophies always seem to get lost somewhere along the road.. I don't have emotions and sometimes that makes me very sad. Check This Out What we are trying to point out is: design is not always just a matter of taste.

I just found this little gem within the package data.tables; set() Look at the Benchmark results now when using set()! In both cases, the output object is stored in a list: str(dimnames(df)) List of 2 $ : chr [1:1704] "1" "2" "3" "4" ... $ : chr [1:6] "country" If you analyse rbind.data.frame code then you could see that the first argument initialized output types.

Customize ???

For example, "Sex" will usually take on only the values "M" or "F," whereas "Name" will generally have lots of possibilities. Let's look at the output: l1 Call: lm(formula = lifeExp ~ year, data = df) Coefficients: (Intercept) year -585.6522 0.3259 Not much there right? Why? Reply With Quote The Following 2 Users Say Thank You to TheEcologist For This Useful Post: Jake(10-03-2012), merik(10-07-2012) 10-07-201202:14 PM #9 merik View Profile View Forum Posts Posts 91 Thanks 22

To make sure our analysis is reproducible, we should put the code into a script file so we can come back to it later. Here's an example: Reordering the levels of a factor This question arises in some models. Customize ??? How to get the last monday of every month Numbers at the corners of concentric squares Cannot get promoted.

xMerged is now class integer # which is treated as numeric in lm, losing a lot of information model2 <- lm(y~0+xMerged,data=subset(d,train)) print(summary(model2)) ## ## Call: ## lm(formula = y ~ 0 This sort of fix would have worked if f had been a vector of characters or even a vector of integers, but for factors we get gibberish. When you use the matrix-style notation S-Plus will often factorize your character variables automatically. So it's better when types match. –Marek Oct 31 '09 at 0:20 add a comment| up vote 23 down vote An "easy" way is to simply not have your strings set

The actual values of the numeric variable are 1, 2, and so on. If you want to add multiple rows to a data.frame, you will need to separate the new columns in the list: df <- rbind(df, list(c("l",

Join them; it only takes a minute: Sign up Appending rows to a dataframe - the factor problem up vote 56 down vote favorite 19 I have a large dataframe (14552 For instance, factor-character discrepancy did not mess things up. –Farrel Oct 30 '09 at 19:16 You have right about factor-character, somewhere in code I found that levels for this First let's build a synthetic data set where y~f(x) where x is a factor or categorical variable.

 # build a synthetic data set set.seed(36236) n <- 50 d <- Eliminate cause not symptoms. 

Thanks for the help Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] More information about the R-help mailing list R › R help Search We can see that the object is a data.frame with 1,704 observations (rows), and 6 variables (columns). It is far more efficient memory wise to first create an "empty dataframe", the exact size of the end product, this is then allocated to the memory once. Then it works like a charm :-) convert.factors.to.strings.in.dataframe <- function(dataframe) { class.data <- sapply(dataframe, class) factor.vars <- class.data[class.data == "factor"] for (colname in names(factor.vars)) { dataframe[,colname] <- as.character(dataframe[,colname]) } return (dataframe)

Browse other questions tagged r or ask your own question. Typically we think of factor levels or categories taking values from a fixed set of strings. gsub(" ","",paste(label[1,2],"-",label[2,2])) doesn't give me that error message). In principle a factor is a value where the value is known to be taken from a known finite set of possible values called levels.

We can also modify this information: copy <- gapminder # lets create a copy so we don't mess up the original colnames(copy) <- c("a",

 fRevised <- ifelse(is.na(f),'a',f) print(fRevised) ## [1] "3" "1" "1" "a" "2" There are a few different ways you can change a factor.