I am going to rank a large dataset based on multiple different values for several times, and I would like to use a loop to do this for me. I have created a combination named
parameter <- c("market_value_LOCAL", "ep", "book_price",
"TOTAL_RATING_SCORE", "ENVIRONMENT", "SOCIAL", "GOVERNANCE")
Q1_R1000_parameter <- Q1_R1000[order(Q1_R1000$parameter[X]),]
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
`$`(df , V1)
`$`(df , "V1")
`$`(df , paste0("V1") )
...for instance will never work, nor will anything else that must first be evaluated in the second argument. You may only pass a string which is never evaluated.
[[ if you want to extract only a single column as a vector).
var <- "mpg" #Doesn't work mtcars$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