I want to change the name of the output of my R function to reflect different strings that are inputted. Here is what I have tried:
kd = c("a","b","d","e","b")
test = function(kd){
return(list(assign(paste(kd,"burst",sep="_"),1:6)))
}
This is just a simple test function. I get the warning (which is just as bad an error for me):
Warning message:
In assign(paste(kd, "burst", sep = "_"), 1:6) :
only the first element is used as variable name
Ideally I would get ouput like a_burst = 1, b_burst = 2 and so on but am not getting close.
I would like split up a dataframe by contents of a vector and be able to name everything according to the name from that vector, similar to
How to split a data frame by rows, and then process the blocks?
but not quite. The naming is imperative.
Something like this, maybe?
kd = c("a","b","d","e","b")
test <- function(x){
l <- as.list(1:5)
names(l) <- paste(x,"burst",sep = "_")
l
}
test(kd)
You could use a vector instead of a list by way of setNames:
t1_6 <- setNames( 1:6, kd)
t1_6
a b d e b <NA>
1 2 3 4 5 6
> t1_6["a"]
a
1
Looking at the question again I wondered if you wnated to assign sequential names to a character vector:
> a1_5 <- setNames(kd, paste0("alpha", 1:5))
> a1_5
alpha1 alpha2 alpha3 alpha4 alpha5
"a" "b" "d" "e" "b"
Related
HEADLINE: Is there a way to get R to recognize data.frame column names contained within lists in the same way that it can recognize free-floating vectors?
SETUP: Say I have a vector named varA:
(varA <- 1:6)
# [1] 1 2 3 4 5 6
To get the length of varA, I could do:
length(varA)
#[1] 6
and if the variable was contained within a larger list, the variable and its length could still be found by doing:
list <- list(vars = "varA")
length(get(list$vars[1]))
#[1] 6
PROBLEM:
This is not the case when I substitute the vector for a dataframe column and I don't know how to work around this:
rows <- 1:6
cols <- c("colA")
(df <- data.frame(matrix(NA,
nrow = length(rows),
ncol = length(cols),
dimnames = list(rows, cols))))
# colA
# 1 NA
# 2 NA
# 3 NA
# 4 NA
# 5 NA
# 6 NA
list <- list(vars = "varA",
cols = "df$colA")
length(get(list$vars[1]))
#[1] 6
length(get(list$cols[1]))
#Error in get(list$cols[1]) : object 'df$colA' not found
Though this contrived example seems inane, because I could always use the simple length(variable) approach, I'm actually interested in writing data from hundreds of variables varying in lengths onto respective dataframe columns, and so keeping them in a list that I could iterate through would be very helpful. I've tried everything I could think of, but it may be the case that it's just not possible in R, especially given that I cannot find any posts with solutions to the issue.
You could try:
> length(eval(parse(text = list$cols[1])))
[1] 6
Or:
list <- list(vars = "varA",
cols = "colA")
length(df[, list$cols[1]])
[1] 6
Or with regex:
list <- list(vars = "varA",
cols = "df$colA")
length(df[, sub(".*\\$", "", list$cols[1])])
[1] 6
If you are truly working with a data frame d, then nrow(d) is the length of all of the variables in d. There should be no reason to use length in this case.
If you are actually working with a list x containing variables of potentially different lengths, then you should use the [[ operator to extract those variables by name (see ?Extract):
x <- list(a = 1:10, b = rnorm(20L))
l <- list(vars = "a")
length(d[[l$vars[1L]]]) # 10
If you insist on using get (you shouldn't), then you need to supply a second argument telling it where to look for the variable (see ?get):
length(get(l$vars[1L], x)) # 10
I haven't been able to find an answer to this, but I am guessing this is because I am not phrasing my question properly.
I want to combine two strings containing several comma-separated values into one string, alternating the inputs from each original string.
x <- '1,2'
y <- 'R,L'
# fictitious function
z <- combineSomehow(x,y)
z = '1R, 2L'
EDIT : Adding dataframe to better describe my issue. I would like to be able to accomplish the above, but within a mutate ideally.
df <- data.frame(
x = c('1','2','1,1','2','1'),
y = c('R','L','R,L','L','R'),
desired_result = c('1R','2L','1R,1L','2L','1R')
)
df:
x y desired_result
1 1 R 1R
2 2 L 2L
3 1,1 R,L 1R,1L
4 2 L 2L
5 1 R 1R
Final Edit/Answer: Based on #akrun's comment/response below and after removing the error originally in df, this ended up being the tidyverse answer:
mutate(desired_result = map2(.x=strsplit(x,','),.y=strsplit(y,','),
~ str_c(.x,.y, collapse=',')))
It can be done with strsplit and paste
combineSomehow <- function(x, y) {
do.call(paste0, c(strsplit(c(x,y),","), collapse=", "))
}
combineSomehow(x,y)
#[1] "1R, 2L"
Without modifying the function, we can Vectorize it to apply on multiple elements
df$desired_result2 <- Vectorize(combineSomehow)(df$x, df$y)
I am currently developing an application and I need to loop through the columns of the data frame. For instance, if the data frame has the columns
char_set <- data.frame(character(),character(),character(),character(),stringsAsFactors = FALSE)
names(char_set) <- c("a","b","c","d")
If the input is given as "a", then the column name "b" should be assigned to the variable, say promote.
It throws an error Error in[.data.frame(char_set, i + 1) : undefined columns selected. Is there any solution?
char_name <- "a"
char_set <- data.frame(character(),character(),character(),character(),stringsAsFactors = FALSE)
names(char_set) <- c("a","b","c","d")
for (i in 1:ncol(char_set)) {
promote <- ifelse(names(char_set) == char_name,char_set[i+1], "-")
print(promote)
}
Thanks in advance!!!
This is actually quite interesting. I would suggest doing something on those lines:
char_name <- "a"
char_set <- data.frame(
a = 1:2,
b = 3:4,
c = 5:6,
d = 8:9,
stringsAsFactors = FALSE
)
res_dta <- data.frame(matrix(nrow = 2, ncol = 3))
for (i in wrapr::seqi(1, NCOL(char_set) - 1)) {
print(i)
if (names(char_set)[i] == char_name) {
res_dta[i] <- char_set[i + 1]
} else {
res_dta[i] <- char_set[i]
}
}
Results
char_set
a b c d
1 1 3 5 8
2 2 4 6 9
res_dta
X1 X2 X3
1 3 3 5
2 4 4 6
There are few generic points:
When you are looping through columns be mindful not fall outside data frame dimensions; running i + 1 on i = 4 will give you column 5 which will return an error for data frame with four columns. You may then decide to run to one column less or break for a specific i value
Not sure if I got your request right, for column names a you want to take values of column b; then column b stays as it was?
Broadly speaking, I'm of a view that this names(char_set)[i] == char_name requires more thought but you have a start with this answer. Updating your post with desired results would help to design a solution.
The problem in your code is that you are looping from 1 to the number of columns of the char_set df, then you are calling the variable char_set[i+1].
This, when the i index takes the maximum value, the instruction char_set[i+1] returns an error because there is no element with that index.
You can try with this solution:
char_name<-"a"
promote<-ifelse((which(names(char_set)==char_name)+1)<ncol(char_set),names(char_set)[which(names(char_set)==char_name)+1],"-")
promote
> [1] "b"
char_name<-"d"
promote<-ifelse((which(names(char_set)==char_name)+1)<ncol(char_set),names(char_set)[which(names(char_set)==char_name)+1],"-")
promote
> [1] "-"
However. when the variable char_name takes the value a, the variable promote will take the value that the set char_set has at the position after the element named a, which matches char_name.
I suggest you to think about the case in which the variable char_name takes the value d and you don't have any values in the char_set after d.
I'm trying to call a vector "a" from a data frame "df" using a function. I know I could do this just fine with the following:
> df$a
[1] 1 2 3
But I'd like to use a function where both the data frame and vector names are input separately as arguments. This is the best that I've come up with:
show_vector <- function(data.set, column) {
data.set$column
}
But here's how it goes when I try it out:
> show_vector(df, a)
NULL
How could I change this function in order to successfully reference vector df$a where the names of both are input to a function as arguments?
It's actually possible to do this without passing the column name as a string (in other words, you can pass in the unquoted column name:
show_vector <- function(data.set, column) {
eval(substitute(column), envir = data.set)
}
Usage example:
df <- data.frame(a = 1:3, b = 4:6)
show_vector(df, b)
# 4 5 6
I've wondered about this kind of thing a lot in the past and haven't found an easy fix. The best I've come up with is this:
df <- data.frame(c(1, 2, 3), c(4, 5, 6))
colnames(df) <- c("A", "B")
test <- function(dataframe, columnName) {
return(dataframe[, match(columnName, colnames(dataframe))])
}
test(df, "A")
Your code would work if you only put the column name in quotes i.e. show_vector(df, "a")
Other multiple ways to do this:
Using base functionality
func <- function(df, cname){
return(df[, grep(cname, colnames(df))])
}
Or even
func <- function(df, cname){
return(df[, cname])
}
You can use substitute to capture the input vector name as it is then use `as.character to make it as a character.
show_vector <- function(data.set, column) {
data.set[,as.character(substitute(column))]
}
Now lets take a look:
(dat=data.frame(a=1:3,b=4:6,c=10:12))
a b c
1 1 4 10
2 2 5 11
3 3 6 12
show_vector(dat,a)
[1] 1 2 3
show_vector(dat,"a")
[1] 1 2 3
It works.
we can also write a simple one where we just input a character string:
show_vector1 <- function(data.set, column) {
data.set[,column]
}
show_vector1(dat,"a")
[1] 1 2 3
Although this will not work if the column name is not a character:
show_vector1(dat,a)
**Show Traceback
Rerun with Debug
Error in `[.data.frame`(data.set, , column) : undefined columns selected**
I have a data frame where the row names are words and I can call the first column of that row of data drame using something like
>df['rowB',1]
i know I can use paste to combine a variable and a string using paste to do something like
>paste("the value is ", df['rowB',1], "."]
and that will get me an output of the string with the value of the variable. what if rowname is a variable that equals 'rowB? I tried to do a first paste to put in the paste above, but the result of the first paste doesn't evaulate to the value, but rather is just a string that says
>rowname<-'rowB'
>type<-paste("relatype[\'", rowname, "\',1]", sep="")
'df['rowB',1]'
long story short, I want to input a value called 'rowname' as a parameter of a function and have it be evaluated for the value of rowname, so I can then put that value into a string within that same function.
I'm also open to a wholly different solution. any and all suggestions are welcome.
thanks
Not sure what the problem might be, not entirely clear from your description, but if rowname is a variable, you don't need anything special, because it will evaluate to it's value anyway. Let
mat <- matrix(1:10, nrow = 5)
rownames(mat) <- letters[1:5]
mat
## [,1] [,2]
##a 1 6
##b 2 7
##c 3 8
##d 4 9
##e 5 10
and rowname <- "b", then
rowname
##[1] "b"
so
mat[rowname, 1]
##b
##2
which is the same as mat["b", 1]. It only fails, if you use mat['rowname', 1].
If you want to put this in functions, you can do something like:
getElement <- function(mat, row.name, column.index) {
mat[row.name, column.index]
}
getElement(mat, "b", 1)
##b
##2
pasteSenstence <- function(mat, row.name, col.index) {
paste("The element of row", row.name, "and column", col.index, "is",
getElement(mat, row.name, col.index))
}
pasteSentence(mat, "b", 1)
##[1] "The element of row b and column 1 is 2"
which also works with rowname <- "b"
pasteSentence(mat, rowname, 1)
##[1] "The element of row b and column 1 is 2"
This should work:
paste("the value is ", get(df['rowname',1]), "."]
If you are not familiar, 'get' in r is similar to 'eval' in python.
x=c('a', 'c', 'b')
a=2
x[1]
'a'
get(x[1])
2
I'm afraid I don't understand the question; how is your function different from the following?
foo = function(rowname = "Species", d = t(iris)){
paste("I'm selecting", d[rowname, 1])
}
foo()
# [1] "I'm selecting setosa"