Basically in SAS I could just do an if statement without an else. For example:
if species='setosa' then species='regular';
there is no need for else.
How to do it in R? This is my script below which does not work:
attach(iris)
iris2 <- iris
iris2$Species <- ifelse(iris2$Species=='setosa',iris2$Species <- 'regular',iris2$Species <- iris2$Species)
table(iris2$Species)
A couple options. The best is to just do the replacement, this is nice and clean:
iris2$Species[iris2$Species == 'setosa'] <- 'regular'
ifelse returns a vector, so the way to use it in cases like this is to replace the column with a new one created by ifelse. Don't do assignment inside ifelse!
iris2$Species <- ifelse(iris2$Species=='setosa', 'regular', iris2$Species)
But there's rarely need to use ifelse if the else is "stay the same" - the direct replacement of the subset (the first line of code in this answer) is better.
New factor levels
Okay, so the code posted above doesn't actually work - this is because iris$Species is a factor (categorical) variable, and 'regular' isn't one of the categories. The easiest way to deal with this is to coerce the variable to character before editing:
iris2$Species <- as.character(iris2$Species)
iris2$Species[iris2$Species == 'setosa'] <- 'regular'
Other methods work as well, (editing the factor levels directly or re-factoring and specifying new labels), but that's not the focus of your question so I'll consider it out of scope for the answer.
Also, as I said in the comments, don't use attach. If you're not careful with it you can end up with your columns out of sync creating annoying bugs. (In the code you post, you're not using it anyway - the rest runs just as well if you delete the attach line.)
I would recommend looking at the base R documentation for help with this. You can find the documentation of if, else, and ifelse here. For use of if and else, refer to ?Control.
Regular control flow in code is done with the basic if and else statements, as in most languages. ifelse() is used for vectorized operations--it will return the same shape as your vector based on the test. Regular if and else expressions do not necessarily have those properties.
Related
I'm quite new to R and I've been learning with the available resources on the internet.
I came across this issue where I have a vector (a) with vars "1", "2", and "3". I want to use the count function to generate a new df with the categories for each of those variables and its frequencies.
The function I want to use in a loop is this
b <- count(mydata, var1)
However, when I use this loop below;
for (i in (a)) {
'j' <- count(mydata[, i])
print (j)
}
The loop happens but the frequencies which gets saved on j is only of the categorical variable "var 3".
Can someone assist me on this code please?
TIA!
In R there are generally better ways than to use loops to process data. In your particular case, the “straightforward” way fails, because the idea of the “tidyverse” is to have the data in tidy format (I highly recommend you read this article; it’s somewhat long but its explanation is really fundamental for any kind of data processing, even beyond the tidyverse). But (from the perspective of your code) your data is spread across multiple columns (wide format) rather than being in a single column (long form).
The other issue is that count (like many other tidyverse functions) expect an unevaluated column name. It does not accept the column name via a variable. akrun’s answer shows how you can work around this (using tidy evaluation and the bang-bang operator) but that’s a workaround that’s not necessary here.
The usual solution, instead of using a loop, would first require you to bring your data into long form, using pivot_longer.
After that, you can perform a single count on your data:
result <- mydata %>%
pivot_longer(all_of(a), names_to = 'Var', values_to = 'Value') %>%
count(Var, Value)
Some comments regarding your current approach:
Be wary of cryptic variable names: what are i, j and a? Use concise but descriptive variable names. There are some conventions where i and j are used but, if so, they almost exclusively refer to index variables in a loop over vector indices. Using them differently is therefore quite misleading.
There’s generally no need to put parentheses around a variable name in R (except when that name is the sole argument to a function call). That is, instead of for (i in (a)) it’s conventional to write for (i in a).
Don’t put quotes around your variable names! R happens to accept the code 'j' <- … but since quotes normally signify string literals, its use here is incredibly misleading, and additionally doesn’t serve a purpose.
I'm creating some routines in R to ease model creation and to distinguish several groups based on several parameters (ex: original watches VS fakes ones using watches common attributes).
During the proccess, I keep track of the potential excluded lines in a vector (empty at first), and I get ride of them at the end using:
model$var <- raw_data[-line_excluded,]
The problem is that if line_excluded is c() (ndlr no line exlcuded), model$var is an empty dataframe then in that case I want all the lines of the dataframe.
The only solution I have think about is the us of
if (!is.null(line_excluded)){
model$var <- raw_data[-line_excluded,]}
But that's not really pretty, and I have several tracking variables as line_excluded which need that.
Thanks for the help
You can make it in another way using setdiff(), which can deal with empty line_excluded i.e.,
model$var <- raw_data[setdiff(seq(nrow(raw_data)),line_excluded),]
You can also try:
model$var <- raw_data[!(1:nrow(raw_data) %in% line_excluded),]
This is similar to what #THomasIsCoding suggested, you look for the row numbers that are not in your line_excluded..
I am trying to make a function in R that calculates the mean of nitrate, sulfate and ID. My original dataframe have 4 columns (date,nitrate, sulfulfate,ID). So I designed the next code
prueba<-read.csv("C:/Users/User/Desktop/coursera/001.csv",header=T)
columnmean<-function(y, removeNA=TRUE){ #y will be a matrix
whichnumeric<-sapply(y, is.numeric)#which columns are numeric
onlynumeric<-y[ , whichnumeric] #selecting just the numeric columns
nc<-ncol(onlynumeric) #lenght of onlynumeric
means<-numeric(nc)#empty vector for the means
for(i in 1:nc){
means[i]<-mean(onlynumeric[,i], na.rm = TRUE)
}
}
columnmean(prueba)
When I run my data without using the function(), but I use row by row with my data it will give me the mean values. Nevertheless if I try to use the function so it will make all the steps by itself, it wont mark me error but it also won't compute any value, as in my environment the dataframe 'prueba' and the columnmean function
what am I doing wrong?
A reproducible example would be nice (although not absolutely necessary in this case).
You need a final line return(means) at the end of your function. (Some old-school R users maintain that means alone is OK - R automatically returns the value of the last expression evaluated within the function whether return() is specified or not - but I feel that using return() explicitly is better practice.)
colMeans(y[sapply(y, is.numeric)], na.rm=TRUE)
is a slightly more compact way to achieve your goal (although there's nothing wrong with being a little more verbose if it makes your code easier for you to read and understand).
The result of an R function is the value of the last expression. Your last expression is:
for(i in 1:nc){
means[i]<-mean(onlynumeric[,i], na.rm = TRUE)
}
It may seem strange that the value of that expression is NULL, but that's the way it is with for-loops in R. The means vector does get changed sequentially, which means that BenBolker's advice to use return(.) is correct (as his advice almost always is.) . For-loops in R are a notable exception to the functional programming paradigm. They provide a mechanism for looping (as do the various *apply functions) but the commands inside the loop exert their effects in the calling environment via side effects (unlike the apply functions).
I've been trying to learn more about environments in R. Through reading, it seemed that I should be able to use functions like with() and transform() to modify variables in a data.frame as if I was operating within that object's environment. So, I thought the following might work:
X <- expand.grid(
Cond=c("baseline","perceptual","semantic"),
Age=c("child","adult"),
Gender=c("male","female")
)
Z <- transform(X,
contrasts(Cond) <- cbind(c(1,0,-1)/2, c(1,-2,1))/4,
contrasts(Age) <- cbind(c(-1,1)/2),
contrasts(Gender) <- cbind(c(-1,1)/2)
)
str(Z)
contrasts(Z$Cond)
But it does not. I was hoping someone could explain why. Of course, I understand that contrasts(X$Cond) <- ... would work, but I'm curious about why this does not.
In fact, this does not work either [EDIT: false, this does work. I tried this quickly before posting originally and did something wrong]:
attach(X)
contrasts(Cond) <- cbind(c(1,0,-1)/2, c(1,-2,1))/4
contrasts(Age) <- cbind(c(-1,1)/2)
contrasts(Gender) <- cbind(c(-1,1)/2)
detach(X)
I apologize if this is a "RTFM" sort of thing... it's not that I haven't looked. I just don't understand. Thank you!
[EDIT: Thank you joran---within() instead of with() or transform() does the trick! The following syntax worked.]
Z <- within(X, {
contrasts(Cond) <- ...
contrasts(Age) <- ...
contrasts(Gender) <- ...
}
)
transform is definitely the wrong tool, I think. And you don't want with, you probably want within, in order to return the entire object:
X <- within(X,{contrasts(Cond) <- cbind(c(1,0,-1)/2, c(1,-2,1))/4
contrasts(Age) <- cbind(c(-1,1)/2)
contrasts(Gender) <- cbind(c(-1,1)/2)})
The only tricky part here is to remember the curly braces to enclose multiple lines in a single expression.
Your last example, using attach, works just fine for me.
transform is only set up to evaluate expressions of the form tag = value, and because of the way it evaluates those expressions, it isn't really set up to modify attributes of a column. It is more intended for direct modifications to the columns themselves. (Scaling, taking the log, etc.)
The difference between with and within is nicely summed up by the Value section of ?within:
Value For with, the value of the evaluated expr. For within, the modified object.
So with only returns the result of the expression. within is for modifying an object and returning the whole thing.
While I agree with #Jornan that within is the best strategy here, I will point out it is possible to use transform you just need to do so in a different way
Z <- transform(X,
Cond = `contrasts<-`(Cond, value=cbind(c(1,0,-1)/2, c(1,-2,1))/4),
Age = `contrasts<-`(Age, value=cbind(c(-1,1)/2)),
Gender= `contrasts<-`(Gender, value=cbind(c(-1,1)/2))
)
Here we are explicitly calling the magic function that is used when you run contrasts(a)=b. This actually returns a value that can be used with the a=b format that transform expects. And of course it leaves X unchanged.
The within solution looks much cleaner of course.
I'm trying to subset a dataframe within a function using a mixture of fixed variables and some variables which are created within the function (I only know the variable names, but cannot vectorise them beforehand). Here is a simplified example:
a<-c(1,2,3,4)
b<-c(2,2,3,5)
c<-c(1,1,2,2)
D<-data.frame(a,b,c)
subbing<-function(Data,GroupVar,condition){
g=Data$c+3
h=Data$c+1
NewD<-data.frame(a,b,g,h)
subset(NewD,select=c(a,b,GroupVar),GroupVar%in%condition)
}
Keep in mind that in my application I cannot compute g and h outside of the function. Sometimes I'll want to make a selection according to the values of h (as above) and other times I'll want to use g. There's also the possibility I may want to use both, but even just being able to subset using 1 would be great.
subbing(D,GroupVar=h,condition=5)
This returns an error saying that the object h cannot be found. I've tried to amend subset using as.formula and all sorts of things but I've failed every single time.
Besides the ease of the function there is a further reason why I'd like to use subset.
In the function I'm actually working on I use subset twice. The first time it's the simple subset function. It's just been pointed out below that another blog explored how it's probably best to use the good old data[colnames()=="g",]. Thanks for the suggestion, I'll have a go.
There is however another issue. I also use subset (or rather a variation) in my function because I'm dealing with several complex design surveys (see package survey), so subset.survey.design allows you to get the right variance estimation for subgroups. If I selected my group using [] I would get the wrong s.e. for my parameters, so I guess this is quite an important issue.
Thank you
It's happening right as the function is trying to define GroupVar in the beginning. R is looking for the object h by itself (not within the dataframe).
The best thing to do is refer to the column names in quotes in the subset function. But of course, then you'd have to sidestep the condition part:
subbing <- function(Data, GroupVar, condition) {
....
DF <- subset(Data, select=c("a","b", GroupVar))
DF <- DF[DF[,3] %in% condition,]
}
That will do the trick, although it can be annoying to have one data frame indexing inside another.