I have a several groups, let's say A,B,C and I want to cut another variable based on these groups, i.e. each group has specific breaks for the same variable.
If I had to calculate the groups mean, i´d use tapply like this:
tapply(mydata$var,mydata$group,mean)
Unfortunately I do not know how to fix this for cut with changing breaks=c(...) arguments for different groups.
tapply(mydata$var,mydata$group,cut)
Any suggestions? I´d like to do it with tapply but any other solution but a custom made function would be suitable too.
EDIT: some small example:
test <- data.frame(var=rnorm(100,0,1),
group=c(rep("A",30),
rep("B",20),
rep("C",50)))
# for group A:
cut(test$var,breaks=c(-4,0,4))
# for group B
cut(test$var,breaks=c(-4,1,4))
and so on...
I'm going to put my mind-reading hat on here and take a stab that you want something like this:
dat <- data.frame(x = runif(100),grp = rep(letters[1:3],length.out = 100))
mapply(cut,split(dat$x,dat$grp),list(c(-Inf,0.5,Inf),
c(-Inf,0.1,0.5,0.9,Inf),
c(-Inf,0.25,0.5,0.75,Inf)))
So this is simply splitting x by grp and applying cut to each piece using different breaks for each piece.
Actually R behaves quite clever here. I found a solution that does work the way I thought initially. Though it's not using the apply family. Somehow R creates integers here instead of factors – which is why in this solution, there is no problem with factor levels like Joran mentions.
dat <- data.frame(x = rnorm(100),grp = rep(letters[1:3],length.out = 100))
ifelse(dat$grp == "a",cut(dat$x,breaks=c(-Inf,0.1,0.2,Inf)),
ifelse(dat$grp == "b",cut(dat$x,breaks=c(-Inf,0.1,1,Inf)),
cut(dat$x,breaks=c(-Inf,0.9,2,Inf))) )
Related
I have a tibble called 'Volume' in which I store some data (10 columns - the first 2 columns are characters, 30 rows).
Now I want to calculate the relative Volume of every column that corresponds to Column 3 of my tibble.
My current solution looks like this:
rel.Volume_unmod = tibble(
"Volume_OD" = Volume[[3]] / Volume[[3]],
"Volume_Imp" = Volume[[4]] / Volume[[3]],
"Volume_OD_1" = Volume[[5]] / Volume[[3]],
"Volume_WS_1" = Volume[[6]] / Volume[[3]],
"Volume_OD_2" = Volume[[7]] / Volume[[3]],
"Volume_WS_2" = Volume[[8]] / Volume[[3]],
"Volume_OD_3" = Volume[[9]] / Volume[[3]],
"Volume_WS_3" = Volume[[10]] / Volume[[3]])
rel.Volume_unmod
I would like to keep the tibble structure and the labels. I am sure there is a better solution for this, but I am relative new to R so I it's not obvious to me. What I tried is something like this, but I can't actually run this:
rel.Volume = NULL
for(i in Volume[,3:10]){
rel.Volume[i] = tibble(Volume = Volume[[i]] / Volume[[3]])
}
Mockup Data
Since you did not provide some data, I've followed the description you provided to create some mockup data. Here:
set.seed(1)
Volume <- data.frame(ID = sample(letters, 30, TRUE),
GR = sample(LETTERS, 30, TRUE))
Volume[3:10] <- rnorm(30*8)
Solution with Dplyr
library(dplyr)
# rename columns [brute force]
cols <- c("Volume_OD","Volume_Imp","Volume_OD_1","Volume_WS_1","Volume_OD_2","Volume_WS_2","Volume_OD_3","Volume_WS_3")
colnames(Volume)[3:10] <- cols
# divide by Volumn_OD
rel.Volume_unmod <- Volume %>%
mutate(across(all_of(cols), ~ . / Volume_OD))
# result
rel.Volume_unmod
Explanation
I don't know the names of your columns. Probably, the names correspond to the names of the columns you intended to create in rel.Volume_unmod. Anyhow, to avoid any problem I renamed the columns (kinda brutally). You can do it with dplyr::rename if you wan to.
There are many ways to select the columns you want to mutate. mutate is a verb from dplyr that allows you to create new columns or perform operations or functions on columns.
across is an adverb from dplyr. Let's simplify by saying that it's a function that allows you to perform a function over multiple columns. In this case I want to perform a division by Volum_OD.
~ is a tidyverse way to create anonymous functions. ~ . / Volum_OD is equivalent to function(x) x / Volumn_OD
all_of is necessary because in this specific case I'm providing across with a vector of characters. Without it, it will work anyway, but you will receive a warning because it's ambiguous and it may work incorrectly in same cases.
More info
Check out this book to learn more about data manipulation with tidyverse (which dplyr is part of).
Solution with Base-R
rel.Volume_unmod <- Volume
# rename columns
cols <- c("Volume_OD","Volume_Imp","Volume_OD_1","Volume_WS_1","Volume_OD_2","Volume_WS_2","Volume_OD_3","Volume_WS_3")
colnames(rel.Volume_unmod)[3:10] <- cols
# divide by columns 3
rel.Volume_unmod[3:10] <- lapply(rel.Volume_unmod[3:10], `/`, rel.Volume_unmod[3])
rel.Volume_unmod
Explanation
lapply is a base R function that allows you to apply a function to every item of a list or a "listable" object.
in this case rel.Volume_unmod is a listable object: a dataframe is just a list of vectors with the same length. Therefore, lapply takes one column [= one item] a time and applies a function.
the function is /. You usually see / used like this: A / B, but actually / is a Primitive function. You could write the same thing in this way:
`/`(A, B) # same as A / B
lapply can be provided with additional parameters that are passed directly to the function that is being applied over the list (in this case /). Therefore, we are writing rel.Volume_unmod[3] as additional parameter.
lapply always returns a list. But, since we are assigning the result of lapply to a "fraction of a dataframe", we will just edit the columns of the dataframe and, as a result, we will have a dataframe instead of a list. Let me rephrase in a more technical way. When you are assigning rel.Volume_unmod[3:10] <- lapply(...), you are not simply assigning a list to rel.Volume_unmod[3:10]. You are technically using this assigning function: [<-. This is a function that allows to edit the items in a list/vector/dataframe. Specifically, [<- allows you to assign new items without modifying the attributes of the list/vector/dataframe. As I said before, a dataframe is just a list with specific attributes. Then when you use [<- you modify the columns, but you leave the attributes (the class data.frame in this case) untouched. That's why the magic works.
Whithout a minimal working example it's hard to guess what the Variable Volume actually refers to. Apart from that there seems to be a problem with your for-loop:
for(i in Volume[,3:10]){
Assuming Volume refers to a data.frame or tibble, this causes the actual column-vectors with indices between 3 and 10 to be assigned to i successively. You can verify this by putting print(i) inside the loop. But inside the loop it seems like you actually want to use i as a variable containing just the index of the current column as a number (not the column itself):
rel.Volume[i] = tibble(Volume = Volume[[i]] / Volume[[3]])
Also, two brackets are usually used with lists, not data.frames or tibbles. (You can, however, do so, because data.frames are special cases of lists.)
Last but not least, initialising the variable rel.Volume with NULL will result in an error, when trying to reassign to that variable, since you haven't told R, what rel.Volume should be.
Try this, if you like (thanks #Edo for example data):
set.seed(1)
Volume <- data.frame(ID = sample(letters, 30, TRUE),
GR = sample(LETTERS, 30, TRUE),
Vol1 = rnorm(30),
Vol2 = rnorm(30),
Vol3 = rnorm(30))
rel.Volume <- Volume[1:2] # Assuming you want to keep the IDs.
# Your data.frame will need to have the correct number of rows here already.
for (i in 3:ncol(Volume)){ # ncol gives the total number of columns in data.frame
rel.Volume[i] = Volume[i]/Volume[3]
}
A more R-like approach would be to avoid using a for-loop altogether, since R's strength is implicit vectorization. These expressions will produce the same result without a loop:
# OK, this one messes up variable names...
rel.V.2 <- data.frame(sapply(X = Volume[3:5], FUN = function(x) x/Volume[3]))
rel.V.3 <- data.frame(Map(`/`, Volume[3:5], Volume[3]))
Since you said you were new to R, frankly I would recommend avoiding the Tidyverse-packages while you are still learing the basics. From my experience, in the long run you're better off learning base-R first and adding the "sugar" when you're more familiar with the core language. You can still learn to use Tidyverse-functions later (but then, why would anybody? ;-) ).
I have a data matrix with approximately one hundred variables and I want to do box plots of these variables. Doing them one by one is possible, but tedious. The code I use for my box plots is:
boxplot(myVar ~ Group*Trt*Time,data=exp,col=c('red','blue'),frame.plot=T,las=2, ylab='Counts', at=c(1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19))
I started doing them one by one, but realized there must be better options. So, the boxplot call will take only one variable at at time (I may be wrong), so I am looking for a way to get it done in one go. A for loop? Next, I would like to print the name of the current variable (= the colName) on the plot in order to keep them apart.
Appreciate suggestions.
Thank you.
jd
Why not try the following:
data(something)
panel.bxp <- function(x, ...)
{
a <- par("a"); on.exit(par(a))
par(a = c(0, 2, a[3:4]))
boxplot(x, add=TRUE)
}
Then, to run the function, you can try something like the following:
pairs(something, diag.panel = panel.bxp, text.panel = function(...){})
EDIT: There is also a nice link to an article here on R-bloggers which you might want to have a look at.
Being very new to R, I've tried to follow my 'old' thinking - making a for-loop. Here is what I came up with. Probably very primitive, and therefore, I'd appreciate comments/suggestions. Anyway: the loop:
for (i in 1:ncol(final)) {
#print(i)
c <- colnames(final)[i]
#print(c)
b <- final[,i]
#b <- t(b)
#dim(b)
#print(b)
exp <- data.frame(Group,Trt,Time,b)
#dim(exp)
#print(exp)
boxplot(b ~ Group*Trt*Time,data=exp,col=c('red','blue'),frame.plot=T, las=2, ylab='Counts',main=c, at=c(1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19))
}
The loop runs through the data matrix 'final', (48rows x 67cols). Picks up the column header, c, which is used in the boxplot call as main title. Picks up the data column, b. Sets up the experiment using the Group, Trt, and Time factors established outside the loop, and calls the boxplot.
This seem to do what I want. Oddly, Rstudio does not allow more than 25 (approx) plots to be stored in the plots console, so I have to run this loop in a couple of rounds.
Anyway, sorry for answering my own question. Better solutions are greatly appreciated since my way is pretty amateourish, I suspect.
I am trying to use a custom function inside 'ddply' in order to create a new variable (NormViability) in my data frame, based on values of a pre-existing variable (CelltiterGLO).
The function is meant to create a rescaled (%) value of 'CelltiterGLO' based on the mean 'CelltiterGLO' values at a specific sub-level of the variable 'Concentration_nM' (0.01).
So if the mean of 'CelltiterGLO' at 'Concentration_nM'==0.01 is set as 100, I want to rescale all other values of 'CelltiterGLO' over the levels of other variables ('CTSC', 'Time_h' and 'ExpType').
The normalization function is the following:
normalize.fun = function(CelltiterGLO) {
idx = Concentration_nM==0.01
jnk = mean(CelltiterGLO[idx], na.rm = T)
out = 100*(CelltiterGLO/jnk)
return(out)
}
and this is the code I try to apply to my dataframe:
library("plyr")
df.bis=ddply(df,
.(CTSC, Time_h, ExpType),
transform,
NormViability = normalize.fun(CelltiterGLO))
The code runs, but when I try to double check (aggregate or tapply) if the mean of 'NormViability' equals '100' at 'Concentration_nM'==0.01, I do not get 100, but different numbers. The fact is that, if I try to subset my df by the two levels of the variable 'ExpType', the code returns the correct numbers on each separated subset. I tried to make 'ExpType' either character or factor but I got similar results. 'ExpType has two levels/values which are "Combinations" and "DoseResponse", respectively. I can't figure out why the code is not working on the entire df, I wonder if this is due to the fact that the two levels of 'ExpType' do not contain the same number of levels for all the other variables, e.g. one of the levels of 'Time_h' is missing for the level "Combinations" of 'ExpType'.
Thanks very much for your help and I apologize in advance if the answer is already present in Stackoverflow and I was not able to find it.
Michele
I (the OP) found out that the function was missing one variable in the arguments, that was used in the statements. Simply adding the variable Concentration_nM to the custom function solved the problem.
THANKS
m.
Related: Strings as variable references in R
Possibly related: Concatenate expressions to subset a dataframe
I've simplified the question per the comment request. Here goes with some example data.
dat <- data.frame(num=1:10,sq=(1:10)^2,cu=(1:10)^3)
set1 <- subset(dat,num>5)
set2 <- subset(dat,num<=5)
Now, I'd like to make a bubble plot from these. I have a more complicated data set with 3+ colors and complicated subsets, but I do something like this:
symbols(set1$sq,set1$cu,circles=set1$num,bg="red")
symbols(set2$sq,set2$cu,circles=set2$num,bg="blue",add=T)
I'd like to do a for loop like this:
colors <- c("red","blue")
sets <- c("set1","set2")
vars <- c("sq","cu","num")
for (i in 1:length(sets)) {
symbols(sets[[i]][,sq],sets[[i]][,cu],circles=sets[[i]][,num],
bg=colors[[i]],add=T)
}
I know you can have a variable evaluated to specify the column (like var="cu"; set1[,var]; I want to know how to get a variable to specify the data.frame itself (and another to evaluate the column).
Update: Ran across this post on r-bloggers which has this example:
x <- 42
eval(parse(text = "x"))
[1] 42
I'm able to do something like this now:
eval(parse(text=paste(set[[1]],"$",var1,sep="")))
In fiddling with this, I'm finding it interesting that the following are not equivalent:
vars <- data.frame("var1","var2")
eval(parse(text=paste(set[[1]],"$",var1,sep="")))
eval(parse(text=paste(set[[1]],"[,vars[[1]]]",sep="")))
I actually have to do this:
eval(parse(text=paste(set[[1]],"[,as.character(vars[[1]])]",sep="")))
Update2: The above works to output values... but not in trying to plot. I can't do:
for (i in 1:length(set)) {
symbols(eval(parse(text=paste(set[[i]],"$",var1,sep=""))),
eval(parse(text=paste(set[[i]],"$",var2,sep=""))),
circles=paste(set[[i]],".","circles",sep=""),
fg="white",bg=colors[[i]],add=T)
}
I get invalid symbol coordinates. I checked the class of set[[1]] and it's a factor. If I do is.numeric(as.numeric(set[[1]])) I get TRUE. Even if I add that above prior to the eval statement, I still get the error. Oddly, though, I can do this:
set.xvars <- as.numeric(eval(parse(text=paste(set[[i]],"$",var1,sep=""))))
set.yvars <- as.numeric(eval(parse(text=paste(set[[i]],"$",var2,sep=""))))
symbols(xvars,yvars,circles=data$var3)
Why different behavior when stored as a variable vs. executed within the symbol function?
You found one answer, i.e. eval(parse()) . You can also investigate do.call() which is often simpler to implement. Keep in mind the useful as.name() tool as well, for converting strings to variable names.
The basic answer to the question in the title is eval(as.symbol(variable_name_as_string)) as Josh O'Brien uses. e.g.
var.name = "x"
assign(var.name, 5)
eval(as.symbol(var.name)) # outputs 5
Or more simply:
get(var.name) # 5
Without any example data, it really is difficult to know exactly what you are wanting. For instance, I can't at all divine what your object set (or is it sets) looks like.
That said, does the following help at all?
set1 <- data.frame(x = 4:6, y = 6:4, z = c(1, 3, 5))
plot(1:10, type="n")
XX <- "set1"
with(eval(as.symbol(XX)), symbols(x, y, circles = z, add=TRUE))
EDIT:
Now that I see your real task, here is a one-liner that'll do everything you want without requiring any for() loops:
with(dat, symbols(sq, cu, circles = num,
bg = c("red", "blue")[(num>5) + 1]))
The one bit of code that may feel odd is the bit specifying the background color. Try out these two lines to see how it works:
c(TRUE, FALSE) + 1
# [1] 2 1
c("red", "blue")[c(F, F, T, T) + 1]
# [1] "red" "red" "blue" "blue"
If you want to use a string as a variable name, you can use assign:
var1="string_name"
assign(var1, c(5,4,5,6,7))
string_name
[1] 5 4 5 6 7
Subsetting the data and combining them back is unnecessary. So are loops since those operations are vectorized. From your previous edit, I'm guessing you are doing all of this to make bubble plots. If that is correct, perhaps the example below will help you. If this is way off, I can just delete the answer.
library(ggplot2)
# let's look at the included dataset named trees.
# ?trees for a description
data(trees)
ggplot(trees,aes(Height,Volume)) + geom_point(aes(size=Girth))
# Great, now how do we color the bubbles by groups?
# For this example, I'll divide Volume into three groups: lo, med, high
trees$set[trees$Volume<=22.7]="lo"
trees$set[trees$Volume>22.7 & trees$Volume<=45.4]="med"
trees$set[trees$Volume>45.4]="high"
ggplot(trees,aes(Height,Volume,colour=set)) + geom_point(aes(size=Girth))
# Instead of just circles scaled by Girth, let's also change the symbol
ggplot(trees,aes(Height,Volume,colour=set)) + geom_point(aes(size=Girth,pch=set))
# Now let's choose a specific symbol for each set. Full list of symbols at ?pch
trees$symbol[trees$Volume<=22.7]=1
trees$symbol[trees$Volume>22.7 & trees$Volume<=45.4]=2
trees$symbol[trees$Volume>45.4]=3
ggplot(trees,aes(Height,Volume,colour=set)) + geom_point(aes(size=Girth,pch=symbol))
What works best for me is using quote() and eval() together.
For example, let's print each column using a for loop:
Columns <- names(dat)
for (i in 1:ncol(dat)){
dat[, eval(quote(Columns[i]))] %>% print
}
I'm using the library poLCA. To use the main command of the library one has to create a formula as follows:
f <- cbind(V1,V2,V3)~1
After this a command is invoked:
poLCA(f,data0,...)
V1, V2, V3 are the names of variables in the dataset data0. I'm running a simulation and I need to change the formula several times. Sometimes it has 3 variables, sometimes 4, sometimes more.
If I try something like:
f <- cbind(get(names(data0)[1]),get(names(data0)[2]),get(names(data0)[3]))~1
it works fine. But then I have to know in advance how many variables I will use. I would like to define an arbitrary vector
vars0 <- c(1,5,17,21)
and then create the formula as follows
f<- cbind(get(names(data0)[var0]))
Unfortunaly I get an error. I suspect the answer may involve some form of apply but I still don't understand very well how this functions work. Thanks in advance for any help.
Using data from the examples in ?poLCA this (possibly hackish) idiom seems to work:
library(poLCA)
vec <- c(1,3,4)
M4 <- poLCA(do.call(cbind,values[,vec])~1,values,nclass = 1)
Edit
As Hadley points out in the comments, we're making this a bit more complicated than we need. In this case values is a data frame, not a matrix, so this:
M1 <- poLCA(values[,c(1,2,4)]~1,values,nclass = 1)
generates an error, but this:
M1 <- poLCA(as.matrix(values[,c(1,2,4)])~1,values,nclass = 1)
works fine. So you can just subset the columns as long as you wrap it in as.matrix.
#DWin mentioned building the formula with paste and as.formula. I thought I'd show you what that would look like using the election dataset.
library("poLCA")
data(election)
vec <- c(1,3,4)
f <- as.formula(paste("cbind(",paste(names(election)[vec],collapse=","),")~1",sep=""))