I'm trying to convert a data.frame in R to mpfr format by multiplying by an mpfr unit constant. This works, as demonstrated in the code below, when applied to a column (result variable 'mpfr_col'), but for both approaches shown for working with a data.frame, it does not. The relevant errors for each attempt are listed in comment.
library(Rmpfr)
prec <- 256
m1 <- mpfr(1,prec)
col_build <- 1:10
test_df <- data.frame(col_build, col_build, col_build)
mpfr_col <- m1*(col_build)
mpfr_df <- m1*test_df # (list) object cannot be coerced to type 'double'
for(colnum in 1:length(colnames(test_df))){
test_df[,colnum] <- m1*test_df[,colnum] # attempt to replicate an object of type 'S4'
}
Answer:
Use [[colnum]] to access the columns instead of [,colnum]:
for(colnum in length(colnames(test_df))){
test_df[[colnum]] <- m1*test_df[[colnum]]
}
(Note: the print method of data.frame will fail, but the 'mpfr-izing' work. You can print it either by printing the columns individually or using as_tibble(test_df).
Explanation
The original fails because the [,colnum] assignment doesn't coerce the argument, I think. Using [[ returns an element (aka a column) of the list (aka the data.frame).
See this bit of Hadley Wickham's Advanced R book:
[ selects sub-lists. It always returns a list; if you use it with a
single positive integer, it returns a list of length one. [[ selects
an element within a list. $ is a convenient shorthand: x$y is
equivalent to x[["y"]].
And the help from Extract.data.frame {base}:
When [ and [[ are used to add or replace a whole column, no coercion
takes place but value will be replicated (by calling the generic
function rep) to the right length if an exact number of repeats can be
used.
Related
Using the default "iris" DataFrame in R, how come when creating a new column "NewCol"
iris[,'NewCol'] = as.POSIXlt(Sys.Date()) # throws Warning
BUT
iris$NewCol = as.POSIXlt(Sys.Date()) # is correct
This issue doesn't exist when assigning Primitive types like chr, int, float, ....
First, notice as #sindri_baldur pointed, as.POSIXlt returns a list.
From R help ($<-.data.frame):
There is no data.frame method for $, so x$name uses the default method which treats x as a list (with partial matching of column names if the match is unique, see Extract). The replacement method (for $) checks value for the correct number of rows, and replicates it if necessary.
So, if You try iris[, "NewCol"] <- as.POSIClt(Sys.Date()) You get warning that You're trying assign a list object to a vector. So only the first element of the list is used.
Again, from R help:
"For [ the replacement value can be a list: each element of the list is used to replace (part of) one column, recycling the list as necessary".
And in Your case, only one column is specified meaning only the first element of the as.POSIXlt's result (list) will be used. And You are warned of that.
Using $ syntax the iris data.frame is treated as a list and then the result of as.POSIXlt - a list again - is appended to it. Finally, the result is data.frame, but take a look at the type of the NewCol - it's a list.
iris[, "NewCol"] <- as.POSIXlt(Sys.Date()) # warning
iris$NewCol2 <- as.POSIXlt(Sys.Date())
typeof(iris$NewCol) # double
typeof(iris$NewCol2) # list
Suggestion: maybe You wanted to use as.POSIXct()?
For a data frame (data) which has one columns as sulfate,
What is a difference between data[["sulfate"]] and data[[colnames(data)=="sulfate"]]?
data["sulfate'] and data[colnames(data)=="sulfate"] yields same valued result and have data frame class but data[["sulfate"]] results into a numeric vector in my case but data[[colnames(data)=="sulfate"]] turns out to be an error. Why?
First - here are some ways to achieve what you are trying to achieve:
data$sulfate
getElement(data, "sulfate")
Next a short explanation why data[[colnames(data)=="sulfate"]] does not work.
1) The expression within [[ is colnames(data)=="sulfate" which is a logical vector.
2) Function [[ accepts a single element (because it's used to select a single element) or a numeric vector in which case it is used to select elements of a nested list. For example:
a <- list(list(2,3), list(3,4))
> a[[c(2,1)]]
[1] 3
The help page help(`[[`) will have more information about how it works.
3) The data.frame object in R is a list, you can confirm this by doing is.list(data). So the function [[ works the same way.
Now what happens when you pass it a vector instead of a single number - it gets turned into a numeric representation of 0s and 1s. For example inspect as.numeric(colnames(data)=="sulfate")).
Then the subsetting [[ encounters 0 entries and when you try to subset using a 0 it throws an error that you are attempting to select less than one element.
data[[0]]
Notice that the error is the same as when doing data[[colnames(data)=="sulfate"]]
I have two lists of lists. humanSplit and ratSplit. humanSplit has element of the form::
> humanSplit[1]
$Fetal_Brain_408_AGTCAA_L001_R1_report.txt
humanGene humanReplicate alignment RNAtype
66 DGKI Fetal_Brain_408_AGTCAA_L001_R1_report.txt 6 reg
68 ARFGEF2 Fetal_Brain_408_AGTCAA_L001_R1_report.txt 5 reg
If you type humanSplit[[1]], it gives the data without name $Fetal_Brain_408_AGTCAA_L001_R1_report.txt
RatSplit is also essentially similar to humanSplit with difference in column order. I want to apply fisher's test to every possible pairing of replicates from humanSplit and ratSplit. Now I defined the following empty vector which I will use to store the informations of my fisher's test
humanReplicate <- vector(mode = 'character', length = 0)
ratReplicate <- vector(mode = 'character', length = 0)
pvalue <- vector(mode = 'numeric', length = 0)
For fisher's test between two replicates of humanSplit and ratSplit, I define the following function. In the function I use `geneList' which is a data.frame made by reading a file and has form:
> head(geneList)
human rat
1 5S_rRNA 5S_rRNA
2 5S_rRNA 5S_rRNA
Now here is the main function, where I use a function getGenetype which I already defined in other part of the code. Also x and y are integers :
fishertest <-function(x,y) {
ratReplicateName <- names(ratSplit[x])
humanReplicateName <- names(humanSplit[y])
## merging above two based on the one-to-one gene mapping as in geneList
## defined above.
mergedHumanData <-merge(geneList,humanSplit[[y]], by.x = "human", by.y = "humanGene")
mergedRatData <- merge(geneList, ratSplit[[x]], by.x = "rat", by.y = "ratGene")
## [here i do other manipulation with using already defined function
## getGenetype that is defined outside of this function and make things
## necessary to define following contingency table]
contingencyTable <- matrix(c(HnRn,HnRy,HyRn,HyRy), nrow = 2)
fisherTest <- fisher.test(contingencyTable)
humanReplicate <- c(humanReplicate,humanReplicateName )
ratReplicate <- c(ratReplicate,ratReplicateName )
pvalue <- c(pvalue , fisherTest$p)
}
After doing all this I do the make matrix eg to use in apply. Here I am basically trying to do something similar to double for loop and then using fisher
eg <- expand.grid(i = 1:length(ratSplit),j = 1:length(humanSplit))
junk = apply(eg, 1, fishertest(eg$i,eg$j))
Now the problem is, when I try to run, it gives the following error when it tries to use function fishertest in apply
Error in humanSplit[[y]] : recursive indexing failed at level 3
Rstudio points out problem in following line:
mergedHumanData <-merge(geneList,humanSplit[[y]], by.x = "human", by.y = "humanGene")
Ultimately, I want to do the following:
result <- data.frame(humanReplicate,ratReplicate, pvalue ,alternative, Conf.int1, Conf.int2, oddratio)
I am struggling with these questions:
In defining fishertest function, how should I pass ratSplit and humanSplit and already defined function getGenetype?
And how I should use apply here?
Any help would be much appreciated.
Up front: read ?apply. Additionally, the first three hits on google when searching for "R apply tutorial" are helpful snippets: one, two, and three.
Errors in fishertest()
The error message itself has nothing to do with apply. The reason it got as far as it did is because the arguments you provided actually resolved. Try to do eg$i by itself, and you'll see that it is returning a vector: the corresponding column in the eg data.frame. You are passing this vector as an index in the i argument. The primary reason your function erred out is because double-bracket indexing ([[) only works with singles, not vectors of length greater than 1. This is a great example of where production/deployed functions would need type-checking to ensure that each argument is a numeric of length 1; often not required for quick code but would have caught this mistake. Had it not been for the [[ limit, your function may have returned incorrect results. (I've been bitten by that many times!)
BTW: your code is also incorrect in its scoped access to pvalue, et al. If you make your function return just the numbers you need and the aggregate it outside of the function, your life will simplify. (pvalue <- c(pvalue, ...) will find pvalue assigned outside the function but will not update it as you want. You are defeating one purpose of writing this into a function. When thinking about writing this function, try to answer only this question: "how do I compare a single rat record with a single human record?" Only after that works correctly and simply without having to overwrite variables in the parent environment should you try to answer the question "how do I apply this function to all pairs and aggregate it?" Try very hard to have your function not change anything outside of its own environment.
Errors in apply()
Had your function worked properly despite these errors, you would have received the following error from apply:
apply(eg, 1, fishertest(eg$i, eg$j))
## Error in match.fun(FUN) :
## 'fishertest(eg$i, eg$j)' is not a function, character or symbol
When you call apply in this sense, it it parsing the third argument and, in this example, evaluates it. Since it is simply a call to fishertest(eg$i, eg$j) which is intended to return a data.frame row (inferred from your previous question), it resolves to such, and apply then sees something akin to:
apply(eg, 1, data.frame(...))
Now that you see that apply is being handed a data.frame and not a function.
The third argument (FUN) needs to be a function itself that takes as its first argument a vector containing the elements of the row (1) or column (2) of the matrix/data.frame. As an example, consider the following contrived example:
eg <- data.frame(aa = 1:5, bb = 11:15)
apply(eg, 1, mean)
## [1] 6 7 8 9 10
# similar to your use, will not work; this error comes from mean not getting
# any arguments, your error above is because
apply(eg, 1, mean())
## Error in mean.default() : argument "x" is missing, with no default
Realize that mean is a function itself, not the return value from a function (there is more to it, but this definition works). Because we're iterating over the rows of eg (because of the 1), the first iteration takes the first row and calls mean(c(1, 11)), which returns 6. The equivalent of your code here is mean()(c(1, 11)) will fail for a couple of reasons: (1) because mean requires an argument and is not getting, and (2) regardless, it does not return a function itself (in a "functional programming" paradigm, easy in R but uncommon for most programmers).
In the example here, mean will accept a single argument which is typically a vector of numerics. In your case, your function fishertest requires two arguments (templated by my previous answer to your question), which does not work. You have two options here:
Change your fishertest function to accept a single vector as an argument and parse the index numbers from it. Bothing of the following options do this:
fishertest <- function(v) {
x <- v[1]
y <- v[2]
ratReplicateName <- names(ratSplit[x])
## ...
}
or
fishertest <- function(x, y) {
if (missing(y)) {
y <- x[2]
x <- x[1]
}
ratReplicateName <- names(ratSplit[x])
## ...
}
The second version allows you to continue using the manual form of fishertest(1, 57) while also allowing you to do apply(eg, 1, fishertest) verbatim. Very readable, IMHO. (Better error checking and reporting can be used here, I'm just providing a MWE.)
Write an anonymous function to take the vector and split it up appropriately. This anonymous function could look something like function(ii) fishertest(ii[1], ii[2]). This is typically how it is done for functions that either do not transform as easily as in #1 above, or for functions you cannot or do not want to modify. You can either assign this intermediary function to a variable (which makes it no longer anonymous, figure that) and pass that intermediary to apply, or just pass it directly to apply, ala:
.func <- function(ii) fishertest(ii[1], ii[2])
apply(eg, 1, .func)
## equivalently
apply(eg, 1, function(ii) fishertest(ii[1], ii[2]))
There are two reasons why many people opt to name the function: (1) if the function is used multiple times, better to define once and reuse; (2) it makes the apply line easier to read than if it contained a complex multi-line function definition.
As a side note, there are some gotchas with using apply and family that, if you don't understand, will be confusing. Not the least of which is that when your function returns vectors, the matrix returned from apply will need to be transposed (with t()), after which you'll still need to rbind or otherwise aggregrate.
This is one area where using ddply may provide a more readable solution. There are several tutorials showing it off. For a quick intro, read this; for a more in depth discussion on the bigger picture in which ddply plays a part, read Hadley's Split, Apply, Combine Strategy for Data Analysis paper from JSS.
I had a few questions about subsetting a named list in R using the [] operator:
For example, consider the list formals <- list(x = DOUBLE, y = DOUBLE, z = NULL). In this example, DOUBLE is treated as a symbol in R.
1) How should I retrieve all elements that are not equal to NULL. I tried formals[formals != NULL] but this only returns an object of type listwith no members.
2) How should I retrieve elements whose names satisfy for a condition. For example, how would I get all elements whose names are not z? I could use names(formals) but this is cumbersome and I was hoping for a quick solution using [].
Another option for the first question:
Filter(Negate(is.null), formals)
For the second case, you'll have to use names. Here's one way:
formals[names(formals) != 'z']
formals is actually a function in R. It's best to avoid names of functions when naming your variables.
This will work for your first question:
formals[!unlist(lapply(formals, is.null))]
I don't think you can avoid using names for the second question.
I have:
z = data.frame(x1=a, x2=b, x3=c, etc)
I am trying to do:
for (i in 1:10)
{
paste(c('N'),i,sep="") -> paste(c('z$x'),i,sep="")
}
Problems:
paste(c('z$x'),i,sep="") yields "z$x1", "z$x1" instead of calling the actual values. I need the expression to be evaluated. I tried as.numeric, eval. Neither seemed to work.
paste(c('N'),i,sep="") yields "N1", "N2". I need the expression to be merely used as name. If I try to assign it a value such as paste(c('N'),5,sep="") -> 5, ie "N5" -> 5 instead of N5 -> 5, I get target of assignment expands to non-language object.
This task is pretty trivial since I can simply do:
N1 = x1...
N2 = x2...
etc, but I want to learn something new
I'd suggest using something like for( i in 1:10 ) z[,i] <- N[,i]...
BUT, since you said you want to learn something new, you can play around with parse and substitute.
NOTE: these little tools are funny, but experienced users (not me) avoid them.
This is called "computing on the language". It's very interesting, and it helps understanding the way R works. Let me try to give an intro:
The basic language construct is a constant, like a numeric or character vector. It is trivial because it is not different from its "unevaluated" version, but it is one of the building blocks for more complicated expressions.
The (officially) basic language object is the symbol, also known as a name. It's nothing but a pointer to another object, i.e., a token that identifies another object which may or may not exist. For instance, if you run x <- 10, then x is a symbol that refers to the value 10. In other words, evaluating the symbol x yields the numeric vector 10. Evaluating a non-existant symbol yields an error.
A symbol looks like a character string, but it is not. You can turn a string into a symbol with as.symbol("x").
The next language object is the call. This is a recursive object, implemented as a list whose elements are either constants, symbols, or another calls. The first element must not be a constant, because it must evaluate to the real function that will be called. The other elements are the arguments to this function.
If the first argument does not evaluate to an existing function, R will throw either Error: attempt to apply non-function or Error: could not find function "x" (if the first argument is a symbol that is undefined or points to something other than a function).
Example: the code line f(x, y+z, 2) will be parsed as a list of 4 elements, the first being f (as a symbol), the second being x (another symbol), the third another call, and the fourth a numeric constant. The third element y+z, is just a function with two arguments, so it parses as a list of three names: '+', y and z.
Finally, there is also the expression object, that is a list of calls/symbols/constants, that are meant to be evaluated one by one.
You'll find lots of information here:
https://github.com/hadley/devtools/wiki/Computing-on-the-language
OK, now let's get back to your question :-)
What you have tried does not work because the output of paste is a character string, and the assignment function expects as its first argument something that evaluates to a symbol, to be either created or modified. Alternativelly, the first argument can also evaluate to a call associated with a replacement function. These are a little trickier, but they are handled by the assignment function itself, not by the parser.
The error message you see, target of assignment expands to non-language object, is triggered by the assignment function, precisely because your target evaluates to a string.
We can fix that building up a call that has the symbols you want in the right places. The most "brute force" method is to put everything inside a string and use parse:
parse(text=paste('N',i," -> ",'z$x',i,sep=""))
Another way to get there is to use substitute:
substitute(x -> y, list(x=as.symbol(paste("N",i,sep="")), y=substitute(z$w, list(w=paste("x",i,sep="")))))
the inner substitute creates the calls z$x1, z$x2 etc. The outer substitute puts this call as the taget of the assignment, and the symbols N1, N2 etc as the values.
parse results in an expression, and substitute in a call. Both can be passed to eval to get the same result.
Just one final note: I repeat that all this is intended as a didactic example, to help understanding the inner workings of the language, but it is far from good programming practice to use parse and substitute, except when there is really no alternative.
A data.frame is a named list. It usually good practice, and idiomatically R-ish not to have lots of objects in the global environment, but to have related (or similar) objects in lists and to use lapply etc.
You could use list2env to multiassign the named elements of your list (the columns in your data.frame) to the global environment
DD <- data.frame(x = 1:3, y = letters[1:3], z = 3:1)
list2env(DD, envir = parent.frame())
## <environment: R_GlobalEnv>
## ta da, x, y and z now exist within the global environment
x
## [1] 1 2 3
y
## [1] a b c
## Levels: a b c
z
## [1] 3 2 1
I am not exactly sure what you are trying to accomplish. But here is a guess:
### Create a data.frame using the alphabet
data <- data.frame(x = 'a', y = 'b', z = 'c')
### Create a numerical index corresponding to the letter position in the alphabet
index <- which(tolower(letters[1:26]) == data[1, ])
### Use an 'lapply' to apply a function to every element in 'index'; creates a list
val <- lapply(index, function(x) {
paste('N', x, sep = '')
})
### Assign names to our list
names(val) <- names(data)
### Observe the result
val$x