I was just trying some code with data.table (library(data.table)) and noticed behaviour I found odd. Why does the first code here put rownames into the rn variable, while the second snippet doesn't? I am curious as to why this is happening. I would have thought the copy() and assign are done before the setDT() so it shouldn't have to be performed in two separate steps.
Keeps rownames:
dtcars <- copy(mtcars)
setDT(dtcars, keep.rownames=TRUE)
Does not keep rownames:
setDT(dtcars <- copy(mtcars), keep.rownames=TRUE)
I even tried with the assignment expressed as a function to make sure that was run first with
setDT(`<-`(dtcars, copy(mtcars)), keep.rownames=TRUE)
Compare this to the following where x is assigned before the call to mean() — I would expect my second snippet to behave like this but if the following behaved like my second snippet it would return NA.
mean(x <- c(rnorm(10, 0, 1), NA), na.rm=TRUE)
As you know, the <- operator is parsed to a function call (to the `<-` function) in R. What you don't seem to be aware of is that each function has a return value (even if it is just NULL and invisible).
The `<-` function has the side effect of assigning a value to a symbol. Its (invisible) return value is the value that is assigned. That's why you can do stuff like y <- x <- 1.
In your second and third example, setDT gets passed the return value of `<-`, which is a data.frame but not bound to a symbol. However, you want to pass it the symbol dtcars, which is what happens in your first example.
Since the return value is a shallow copy of the assigned data.frame, the assigned data.frame is also turned into a data.table. However, I don't understand C sufficiently to understand (from the source code) why adding the rn column to the return value doesn't also add it to the assigned data.table. I suggest submitting an issue.
Related
I noticed that if I called setNames() in ifelse() the returned object does not preserved the names from setNames().
x <- 1:10
#no names kept
ifelse(x <5, setNames(x+1,letters[1:4]), setNames(x^3, letters[5:10]))
#names kept
setNames(ifelse(x <5, x+1,x^3), letters[1:10])
After looking at the code I realize that the second way is more concise but still would be interested to know why the names are not preserved when setNames() is called in ifelse(). ifelse() documentation warns of :
The mode of the result may depend on the value of test (see the examples), and the class attribute (see oldClass) of the result is taken from test and may be inappropriate for the values selected from yes and no.
Is the named list being stripped related to this warning?
It's not really specific to setNames. ifelse simply doesn't preserve names for the TRUE/FALSE parameter. It would get confusing if your TRUE and FALSE values had different names so it just doesn't bother. However, according to the Value session of the help page
A vector of the same length and attributes (including dimensions and "class") as test
Since names are stored as attributes, names are only preserved from the the test parameter. Observe these simple examples
ifelse(TRUE, c(a=1), c(x=4))
# [1] 1
ifelse(c(g=TRUE), c(a=1), c(x=4))
# g
# 1
So in your examples you need to move the names to the test condition
ifelse(setNames(x <5,letters[1:10]), x+1, x^3)
I have a dataframe with 10 columns in it. I want to retrieve the columns 1:4.
However I am storing the value 1:4 in a variable. Let's call it a. When I try to use df[,c(a)] it just gives me the numbers back. However, if I give df[,c(1:4)] it gives me the columns. What is the difference? If I use the function identical on these two: identical(c(a),c(1:4)), it gives me TRUE. What is that I am doing wrong?
I was able to figure this out. It was giving me this error :
Error in [.data.table(tr, 1, a) :
j (the 2nd argument inside [...]) is a single symbol but column name 'a' is not found. Perhaps you intended DT[,..a] or DT[,a,with=FALSE]. This difference to data.frame is deliberate and explained in FAQ 1.1.
So I had to use with=FALSE. This solved it.
Thanks!
I have a couple of data frames that I want to be ordered by its last column respectively, I've been trying since a while but nothing succeeds, the main idea is to create a function to avoid doing this over and over for each data frame, the function I'm building is this:
order_dataF = function(x){
tCol = colnames(x[length(x)])
print(tCol)
#x <- x[with(x, order(-tCol),)]
#x <- x[with(x, order(-(paste(tCol))),)]
#x[do.call( order, x[,match(tCol,names(x))]),]
#x <- x[order(x$tCol),]
}
All the lines that have a comment on it are the ones I tested none of this are working as expected, I know this is because order needs the column name instead the variable I'm giving.
tCol always always bring to me the last column name, when I run this function this is the result:
[1] "TotalSearches"
Error in -(paste(tCol)) : invalid argument to unary operator
Calls: main ... [.data.frame -> with -> with.default -> eval -> eval -> order
Execution halted
I'm printing tCol to see if this is really containing the last column name, in this case, indeed it does have exactly what I need.
Perhaps this is a silly question/problem and it's too easy to solve but I cannot move forward as this is slowing me down, I'm frustrated.
Also I'm seeing this looks like duplicated but is not, nobody is being asking the right question (perhaps not even me) but the idea is "Order my the content of a string variable which is obtained from the data frame column names"
Generally, don't try to use with (or other "nonstandard" evaluation functions like subset) inside functions.
order_by_last_col = function(df) {
df[order(df[, ncol(df)]), ]
}
# test
order_by_last_col(mtcars)
If using column names stored as character strings, you must use [, not $, because $ is also a non-standard evaluation shortcut, and it never evaluates the code that comes after $, it just looks for a column with that exact name. If you'd rather use names than indices (like above), do it this way with [:
order_by_last_col = function(df) {
last_col_name = tail(names(df), 1)
df[order(df[, last_col_name]), ]
}
Edit: Just a few more experiments to see why your initial attempts didn't work. they don't need to be in a function to not work, they just never work.
col = "wt"
mtcars$col # NULL
with(mtcars, head(col)) # "wt"
mtcars[, match(col, names(mtcars))] # this does work but is unnecessarily long
mtcars[, col] # works, easy
mtcars[[col]] # also works
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 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