My toy dataframe:
d <- data.frame(
value = sample(1:10),
class = sample(c("a","b"), 20, replace = TRUE)
)
I split my data frame up by values of 'class' and put them in a list where each list element is named after its class:
l <- dlply(d, .(class), function(x)return(x))
Then I want to lapply over each class and make a histogram. Note that I do NOT want a facet. I want as many individual files saved as classes. So I define a function doPlots that makes histograms, then ggsaves them (as a_hist.png and b_hist.png, in this example):
doPlots <- function(d, name){
g <- ggplot(data = d, aes(x=value)) +
geom_histogram(binwidth=1)
ggsave(filename=paste(name,"hist.png",sep="_"))
}
However, when I lapply:
lapply(l, FUN=doPlots, name=names(l))
I get Error: device must be NULL, a string or a function.
Thanks in advance.
Two problems with your code, one is that you are passing the entire vector of names to the function. Second, you have not added a plot to save to the ggsave function. You can use mapply to iterate over two or more lists.
doPlots <- function(d, name){
g <- ggplot(data = d, aes(x=value)) +
geom_histogram(binwidth=1)
ggsave(filename=paste(name, "hist.png", sep="_"), g)
}
mapply(doPlots, l, names(l))
Consider base R's by which slices a dataframe by factor levels into a list of elements. You can even pass your user-defined function into it all in one call:
dlist <- by(d, d$class, FUN=function(i) {
name <- max(as.character(i$class))
doPlots(i, name)
})
Related
I have written a function to change any one variable (i.e., column) in a data.frame to its unique levels and return the changed data.frame.
I wonder how to change multiple variables at once using my function and get one final data.frame with all the changes?
I have tried the following, but this gives multiple data.frames while only the last data.frame is the desired output:
data <- data.frame(sid = c(33,33, 41), pid = c('Bob', 'Bob', 'Jim'))
#== My function for ONE variable:
f <- function(data, what){
data[[what]] <- as.numeric(factor(data[[what]], levels = unique(data[[what]])))
return(data)
}
# Looping over `what`:
what <- c('sid', 'pid')
lapply(seq_along(what), function(i) f(data, what[i]))
In the function, we could change to return the data[[what]]
f <- function(data, what){
data[[what]] <- as.numeric(factor(data[[what]], levels = unique(data[[what]])))
data[[what]]
}
data[what] <- lapply(seq_along(what), function(i) f(data, what[i]))
Or do
data[what] <- lapply(what, function(x) f(data, x))
Or simply
data[what] <- lapply(what, f, data = data)
I have a dataset with numeric and factor variables. I want to do one page with numeric and other with factor var. First of all, i select factor var with his index.
My df is IRIS dataset.
df<-iris
df$y<-sample(0:1,nrow(iris),replace=TRUE)
fact<-colnames(df)[sapply(df,is.factor)]
index_fact<-which(names(df)%in%fact)
Then i calculate rest of it (numerics)
nm<-ncol(df)-length(fact)
Next step is create loop
i_F=1
i_N=1
list_plotN<- list()
list_plotF<- list()
for (i in 1:length(df)){
plot <- ggplot(df,aes(x=df[,i],color=y,fill=y))+xlab(names(df)[i])
if (is.factor(df[,i])){
p_factor<-plot+geom_bar()
list_plotF[[i_F]]<-p_factor
i_F=i_F+1
}else{
p_numeric <- plot+geom_histogram()
list_plotN[[i_N]]<-p_numeric
i_N=i_N+1
}
}
When i see list_plotF and list_plot_N,it didn't well. It always have same vars. i don't know what i'm doing wrong.
thanks!!!
I don't really follow your for loop code all that well. But from what I see it seems to be saving the last plot in every loop you make. I've reconstructed what I think you need using lapply. I generally prefer lapply to for loops whenever I can.
Lapply takes a list of values and a function and applies that function to every value. you can define your function separately like I have so everything looks cleaner. Then you just mention the function in the lapply command.
In our case the list is a list of columns from your dataframe df. The function it applies first creates our base plot. Then it does a quick check to see if the column it is looking at is a factor.. If it's a factor it creates a bar graph, else it creates a histogram.
histOrBar <- function(var) {
basePlot <- ggplot(df, aes_string(var))
if ( is.factor(df[[var]]) ) {
basePlot + geom_bar()
} else {
basePlot + geom_histogram()
}
}
loDFs <- lapply(colnames(df), histOrBar)
Consider passing column names with aes_string to better align x with df:
for (i in 1:length(df)){
plot <- ggplot(df, aes_string(x=names(df)[i], color="y", fill="y")) +
xlab(names(df)[i])
...
}
To demonstrate the problem using aes() and solution using aes_string() in OP's context, consider the following random data frame with columns of different data types: factor, char, int, num, bool, date.
Data
library(ggplot2)
set.seed(1152019)
alpha <- c(LETTERS, letters, c(0:9))
data_tools <- c("sas", "stata", "spss", "python", "r", "julia")
random_df <- data.frame(
group = sample(data_tools, 500, replace=TRUE),
int = as.numeric(sample(1:15, 500, replace=TRUE)),
num = rnorm(500),
char = replicate(500, paste(sample(LETTERS[1:2], 3, replace=TRUE), collapse="")),
bool = as.numeric(sample(c(TRUE, FALSE), 500, replace=TRUE)),
date = as.Date(sample(as.integer(as.Date('2019-01-01', origin='1970-01-01')):as.integer(Sys.Date()),
500, replace=TRUE), origin='1970-01-01')
)
Graph
fact <- colnames(random_df)[sapply(random_df,is.factor)]
index_fact <- which(names(random_df) %in% fact)
i_F=1
i_N=1
list_plotN <- list()
list_plotF <- list()
plot <- NULL
for (i in 1:length(random_df)){
# aes() VERSION
#plot <- ggplot(random_df, aes(x=random_df[,i], color=group, fill=group)) +
# xlab(names(random_df)[i])
# aes_string() VERSION
plot <- ggplot(random_df, aes_string(x=names(random_df)[i], color="group", fill="group")) +
xlab(names(random_df)[i])
if (is.factor(random_df[,i])){
p_factor <- plot + geom_bar()
list_plotF[[i_F]] <- p_factor
i_F=i_F+1
}else{
p_numeric <- plot + geom_histogram()
list_plotN[[i_N]] <- p_numeric
i_N=i_N+1
}
}
Problem (using aes() where graph outputs DO NOT change according to type)
Solution (using aes_string() where graphs DO change according to type)
The following toy code yields a density plot for each column of the y dataframe. However, sapply does not carry over the column name attributes. I'd like to name each new plot with the column name from which the data comes from. Any help is appreciated!
y <- data.frame(sample(1:50), sample(1:50), sample(1:50))
colnames(y) <- c("col1", "col2", "col3")
toy.func <- function(y) {
X11()
plot = plot(density(y), main = colnames(y))
return(plot)
}
result <- sapply(y, toy.func)
You are right and it makes sense: y is seen as a list and sapply goes over its element that are vectors, and we cannot assign a name to a vector. So, a somewhat minimal deviation from your approach that achieves what you want would be to use mapply:
toy.func <- function(y, name) {
X11()
plot = plot(density(y), main = name)
return(plot)
}
mapply(toy.func, y, colnames(y))
It applies toy.func by taking one element from y and one from colnames(y) in every step.
Another option would be to go over the column names at the same time providing the data frame
toy.func <- function(name, data) {
X11()
plot = plot(density(data[, name]), main = name)
return(plot)
}
sapply(colnames(y), toy.func, y)
Also note that your function can be simplified to, in this case,
toy.func <- function(name, data) {
X11()
plot(density(data[, name]), main = name)
}
The function pairs() produces p scatter plot of all the p x p pairs of variables.
x<-rnorm(100,0,1)
y<-rnorm(100,0,1)
z<-rnorm(100,1,1)
t<-rnorm(100,2,10)
dd<-cbind(x,y,z,t)
pairs(dd)
But I would like to be able to choose the ones for the 'lines' and columns of the grid. For instance plot these pairs (x,y) (x,z) (t,y) (t,z). Is there a function that accepts a formula like (x+t)~(z+y) for instance ?
You can either use the specific arguments of pairs or create a custom function that accepts a formula as input:
pairs(dd, horInd=c(1, 4), verInd=c(2,3))
Custom function:
my_pairs <- function(df, formula) {
form <- deparse(formula)
s <- strsplit(form, "~")
lhs <- trimws(unlist(strsplit(s[[1]][1], "\\+")))
rhs <- trimws(unlist(strsplit(s[[1]][2], "\\+")))
lhs.ind <- match(lhs, colnames(df))
rhs.ind <- match(rhs, colnames(df))
all_cmbs <- expand.grid(lhs.ind, rhs.ind)
rows <- all_cmbs[,1]
cols <- all_cmbs[,2]
par(mfrow=c(2,2))
for(i in 1:4) {
eval(substitute(plot(df[,rows[i]], df[,cols[i]],
xlab=colnames(df)[rows[i]],
ylab=colnames(df)[cols[i]])))
}
}
my_pairs(dd, x + t ~ y + z)
I'm not quite familiar with R function dealing with variables used.
Here's the problem:
I want to built a function, of which variables ... are column names of data frame used for table().
f <- function (data, ...){
T <- with(data, table(...) # ... variables input
return(T)
}
How can I deal with the code?
Thanks a lot for answering!
The order of evaluation doesn't quite work right with with() apparently. Here's an alternative that should work (using sample data from #DavidArenburg)
set.seed(1)
data1 <- data.frame(a = sample(5,5), b = sample(5,5))
f <- function (data, ...) {
xx <- lapply(substitute(...()), eval, data, parent.frame())
T <- do.call(table, xx)
return(T)
}
f(data = data1, a,b)
It is often far easier to avoid non-standard evaluation and use character strings to reference the columns within a data.frame.
set.seed(1)
data1 <- data.frame(a = sample(5,5), b = sample(5,5))
f <- function (data, ...) {
do.call(table,data[unlist(list(...))])
}
# the following calls to `f` return the same results
f(data = data1, 'a','b')
f(data = data1, c('a','b'))
a <- c('a','b')
f(data = data1, a)