Dynamically select data frame columns using $ and a vector of column names
I wish to order a data frame based on different columns, one at a turn. I have a character vector with the relevant column names on which the order should be based:
parameter <- c("market_value_LOCAL", "ep", "book_price", "sales_price", "dividend_yield",
"beta", "TOTAL_RATING_SCORE", "ENVIRONMENT", "SOCIAL", "GOVERNANCE")
I wish to loop over the names in parameter and dynamically select the column to be used to order my data:
Q1_R1000_parameter <- Q1_R1000[order(Q1_R1000$parameter[X]), ]
where X is 1:10 (because I have 10 items in parameter ).
To make my example reproducible, consider the data set mtcars and some variable names stored in a character vector cols . When I try to select a variable from mtcars using a dynamic subset of cols , in a similar way as above ( Q1_R1000$parameter[X] ), the column is not selected:
cols <- c("cyl", "am")
mtcars$cols[1]
# NULL
You can't do that kind of subsetting with $ . In the source code ( R/src/main/subset.c ) it states:
/*The $ subset operator.
We need to be sure to only evaluate the first argument.
The second will be a symbol that needs to be matched, not evaluated.
*/
Second argument? What?! You have to realise that $ , like everything else in R, (including for instance ( , + , ^ etc) is a function, that takes arguments and is evaluated. df$V1 could be rewritten as
`$`(df , V1)
or indeed
`$`(df , "V1")
But...
`$`(df , paste0("V1") )
...for instance will never work, nor will anything else that must first be evaluated in the second argument. You may only pass a string which is never evaluated.
Instead use [ (or [[ if you want to extract only a single column as a vector).
For example,
var <- "mpg"
#Doesn't work
mtcars$var
#These both work, but note that what they return is different
# the first is a vector, the second is a data.frame
mtcars[[var]]
mtcars[var]
You can perform the ordering without loops, using do.call to construct the call to order . Here is a reproducible example below:
# set seed for reproducibility
set.seed(123)
df <- data.frame( col1 = sample(5,10,repl=T) , col2 = sample(5,10,repl=T) , col3 = sample(5,10,repl=T) )
# We want to sort by 'col3' then by 'col1'
sort_list <- c("col3","col1")
# Use 'do.call' to call order. Seccond argument in do.call is a list of arguments
# to pass to the first argument, in this case 'order'.
# Since a data.frame is really a list, we just subset the data.frame
# according to the columns we want to sort in, in that order
df[ do.call( order , df[ , match( sort_list , names(df) ) ] ) , ]
col1 col2 col3
10 3 5 1
9 3 2 2
7 3 2 3
8 5 1 3
6 1 5 4
3 3 4 4
2 4 3 4
5 5 1 4
1 2 5 5
4 5 3 5
If I understand correctly, you have a vector containing variable names and would like loop through each name and sort your data frame by them. If so, this example should illustrate a solution for you. The primary issue in yours (the full example isn't complete so I"m not sure what else you may be missing) is that it should be order(Q1_R1000[,parameter[X]]) instead of order(Q1_R1000$parameter[X]) , since parameter is an external object that contains a variable name opposed to a direct column of your data frame (which when the $ would be appropriate).
set.seed(1)
dat <- data.frame(var1=round(rnorm(10)),
var2=round(rnorm(10)),
var3=round(rnorm(10)))
param <- paste0("var",1:3)
dat
# var1 var2 var3
#1 -1 2 1
#2 0 0 1
#3 -1 -1 0
#4 2 -2 -2
#5 0 1 1
#6 -1 0 0
#7 0 0 0
#8 1 1 -1
#9 1 1 0
#10 0 1 0
for(p in rev(param)){
dat <- dat[order(dat[,p]),]
}
dat
# var1 var2 var3
#3 -1 -1 0
#6 -1 0 0
#1 -1 2 1
#7 0 0 0
#2 0 0 1
#10 0 1 0
#5 0 1 1
#8 1 1 -1
#9 1 1 0
#4 2 -2 -2
Using dplyr provides an easy syntax for sorting the data frames
library(dplyr)
mtcars %>% arrange(gear, desc(mpg))
It might be useful to use the NSE version to allow dynamically building the sort list
sort_list <- c("gear", "desc(mpg)")
mtcars %>% arrange_(.dots = sort_list)
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