I can't figure out why this isn't working. I have a data set with 5 columns, n rows. I just want to apply a function to each row and have the result returned in an n by 1 vector.
Just to test out how everything works, i made this simple function:
f1 <- function(uniqueid,Perspvalue,expvalue,stddevi,stddevc) {
uniqueid+ Perspvalue- expvalue+ stddevi+stddevc
}
and here's the first few rows of my data set:
> data
uniqueid Perspvalue expvalue stddevi stddevc
1 1 2.404421e+03 3337239.00 8.266566e+03 3.324624e+03
2 2 1.345307e+03 3276559.87 7.068823e+03 2.648072e+03
3 3 1.345307e+03 3276559.87 7.068823e+03 2.648072e+03
Note that it's a data frame (i think), and not a matrix. I loaded in the data from a csv using read.csv.
So i try this: apply(data,1,f1)
But my result is this: Error in uniqueid + Perspvalue : 'Perspvalue' is missing
I expected a number instead of an error.
You'll need to use mapply for this, or even more convienient mdply from the plyr package.
Some example code:
spam_function = function(a, b) {
return(a*b)
}
require(plyr)
input_args = data.frame(a = runif(1000), b = runif(1000))
result = mdply(input_args, spam_function)
> head(result)
a b V1
1 0.46902575 0.6865863 0.32202668
2 0.56837805 0.2400993 0.13646717
3 0.07185661 0.2334754 0.01677675
4 0.15589191 0.6636891 0.10346377
5 0.98317092 0.8895609 0.87459042
6 0.46070479 0.4301685 0.19818071
If you just want the vector of results:
result_vector = result$V1
Or alternatively, a base R solution using mapply:
result_mapply = mapply(spam_function, a = input_args$a, b = input_args$b)
> head(result_mapply)
[1] 0.2757767 0.1268879 0.5851026 0.7904186
[5] 0.2186079 0.1091692
Related
I want to create a function in R that will create a numerical column based on a character/categorical column. In order to do this I need to get the distinct values in the categorical column. I can do this outside a function well, but would like to make a reusable function to do it. The issue I've run into is that the same distinct() formula that works outside the function doesn't behave the same way within the formula. I've created a demo below:
# test of call to db to numericize
DF <- data.frame("a" = c("a","b","c","a","b","c"),
"b" = paste(0:5, ".1", sep = ""),
"c" = letters[1:6],
stringsAsFactors = FALSE)
catnum <- function(db, inputcolname) {
x <- distinct(db,inputcolname);
print(x);
return(x);
}
y <- distinct(DF,a)
y
catnum(DF,'a')
While y gives the correct distinct one column answer (one column with (a,b,c) in it), x within the function is the entire dataframe. I have tried with and without the ' ', as in catnum(DF,a) but the results are the same.
Could someone tell me what is happening or suggest some code that would work?
One solution is to use distinct_ function inside function. The distinct expect column name and it doesn't work with column names in a variable.
For example distinct(DF, "a") will not work. The actual syntax is: distinct(DF, a). Notice the missing quotes. When distinct is called from function then column name was provided as variable name (i.e inputcolname) which was evaluated. Hence unexpected result. But distinct_ works on variable name for columns.
library(dplyr)
catnum <- function(db, inputcolname) {
x <- distinct_(db,inputcolname);
#print(x);
return(x);
}
#With modified function results were as expected.
catnum(DF,'a')
# a
# 1 a
# 2 b
# 3 c
Not sure what you are trying to do and where distinct function is coming from. Are you looking for this?
catnum<-function(DF,var){
length(unique(DF[[var]]))
}
catnum(DF,'a')
You're inputs are not the same, and so you get different results. If you give distinct the same arguments you give catnum, you will get the same result:
isTRUE(all.equal(distinct(DF, a),
catnum(DF, "a")))
## [1] FALSE
isTRUE(all.equal(distinct(DF, "a"),
catnum(DF, "a")))
##[1] TRUE
Unfortunately, this does not work:
catnum(DF, a)
## a b c
## 1 a 0.1 a
## 2 b 1.1 b
## 3 c 2.1 c
## 4 a 3.1 d
## 5 b 4.1 e
## 6 c 5.1 f
The reason, as explained in
vignette("programming")
is that you must jump through several annoying hoops if you want to write functions that use functions from dplyr. The solution (as you will learn in the vignette) is as follows:
catnum <- function(db, inputcolname) {
inputcolname <- enquo(inputcolname)
distinct(db, !!inputcolname)
}
catnum(DF, a)
## a
## 1 a
## 2 b
## 3 c
Or you could conclude that this is all too confusing and do something like
catnum <- function(db, inputcolname) {
unique(db[, inputcolname, drop = FALSE])
}
catnum(DF, "a")
## a
## 1 a
## 2 b
## 3 c
instead.
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've figured out if I use as.character(df[x,y]) or as.<whatever>df[x,y] I can get/coerce what I need, every time from my data frames
What I cant seem to find/figure out is why. Details below.
When I access df[1,1] (or anything in column 1) I get
df[1,1]
[1] a
Levels: a b c
but when I access 1,3 it works fine
> df[1,3]
[1] 10
but then when I use as.character() it works.
> as.character(df[1,1])
[1] "a"
The data frame was built using this line
df = data.frame(names = c("a","b","c"), size = c(1,2,3),num = c(10,20,30) )
> df
names size num
1 a 1 10
2 b 2 20
3 c 3 30
But in this data frame
imp2met = read.csv('tomet.csv', header = TRUE, sep=",",dec='.')
> imp2met
unit mult ret
1 (yd) 0.9100 (m)
2 (in) 2.5200 (cm)
3 .....
I get these results for 1,3
> imp2met[1,3]
[1] (m)
Levels: (c) (cm) (cm^2) ....
>
> as.character(imp2met[1,3])
[1] "(m)"
So why the "random" results? Why do I need as.<whatever>() but only some of the time?
data.frame default is to convert character vectors to factors. You can change this with the argument stringsAsFactors=FALSE
Also, when you subset a dataframe using [, you can add the drop=FALSE argument to simplify the results in some cases.
I'd like to perform different aggregations in a loop to be applied to different row subsets of my data, but it seems tricky to achieve (if possible at all):
t <- data.frame(agg=c(list("field1"=field1, "field2"=field2), ...),
fun=c(mean, ...))
f <- function(x) {
for (i in 1:nrow(t) {
y <- aggregate(x, by=t$agg[i], FUN=t$fun[i])
# do something with y
}
}
One problem is that the field list agg triggers an error when trying to build the data frame ("object 'field1' not found"), and the other problem is that R does not like to assign a function value to fun ("cannot coerce class ""function"" to a data.frame").
Appendix:
A concrete example for my data (just to match the definitions above) could be:
> d <- data.frame(field1=round(rnorm(5, 10, 1)),field2=letters[round(rnorm(5, 10, 1))], field3=1:5)
> d
field1 field2 field3
1 11 j 1
2 11 i 2
3 10 j 3
4 12 i 4
5 11 j 5
> with(d, aggregate(d$field3,by=list(field1, field2),FUN=mean))
Group.1 Group.2 x
1 11 i 2
2 12 i 4
3 10 j 3
4 11 j 3
Playing tricks with the variable names in the data frame, I still get this:
> with(d,t <- data.frame(agg=c(list("field1"=field1, "field2"=field2)),fun=c(mean)))
Error in as.data.frame.default(x[[i]], optional = TRUE) :
cannot coerce class ""function"" to a data.frame
The problems were several, mostly caused by R making exceptions to general processing:
First a vector cannot be nested, but only lists can. Still all the elements are required to have the same type.
Second, data.frame does some magic treatment when constructing the variables (causing the inability to assign closures), so it cannot be used.
Finally I had to refer to variables to aggregate by name
So the definition looks like this (where , ... means "add more similar items"):
t <- list(agg=list(c("field1", "field2"), ...),
fun=list(mean, ...))
f <- function(x) {
for (i in 1:length(t$agg)) {
agg <- t$agg[[i]]
aggList <- lapply(agg, FUN=function(e) x[[e]])
names(aggList) <- agg
y <- aggregate(x, by=aggList, FUN=t$fun[[i]])
# do something with y
}
}
Note: In the actual solution I added another list holding the names of the columns to select for the aggregated data frame to avoid warnings about mean returning NA.
I am struggling to make my apply() work: I have two dataframes:
from <- c(1,2,3)
to <- c(2,3,4)
df1 <- data.frame(from, to)
long <-c(9,9.2,9.4,9.6)
lat <- c(45,45.2,45.4,45.6)
id <- c(1,2,3,4)
df2 <- data.frame(long, lat, id)
Now I want something like this:
myFunction <- function(arg){
>>> How do I access arg$from and arg$to? <<<<
}
apply(df1,1,myFunction)
In myFunction I need to make some calculations and return a value for each from-to pair. I don't understand how to access parts of the arg, since arg[0] gives me numeric(0) and arg$from just crashes.
The problem is that apply(...) requires a matrix or array as the first argument. If you pass a dataframe, it will coerce that to a matrix. Matrices are 1 indexed, so the upper left element is [1,1], not [0,0]. Also, matrix columns cannot be referenced using the $ notation.
So,
f <- function(x) {
from <- x[1]
to <- x[2]
# do stuff with from and to...
}
apply(df,1,f)
would work.
One other thing to watch out for is that if your dataframe has (other) columns that have character strings, the conversion will make everything character (including the numbers!). This is because, by definition, all elements of a matrix must have the same data type. Your example does not have that problem, though.
Try mapply(). It's a multivariate version of sapply(). For example:
> myFunction <- function(arg1, arg2){
+ return(sum(arg1, arg2))
+ }
>
> mapply(myFunction, df1$from, df1$to)
[1] 3 5 7
You can also use it to make a new variable in your data frame.
> df1$newvar <- mapply(myFunction, df1$from, df1$to)
> df1
from to newvar
1 1 2 3
2 2 3 5
3 3 4 7