CODEWITHSUNDEEP
rfunctiondataframe
Giuseppe Petri
Giuseppe Petri

Reputation: 640

Why is the function in R not running across all set of inputs? I am getting this error: replacement has 1 row, data has 5

I need to run a function for a set of inputs. There are 4 inputs (pd, cl, va, co) for 5 different situations (5 rows in the "inputs" dataframe). The function should run for each row, accross 100 values of P. And retain the response that minimizes the difference against a givin parameter (ratio in this case).

Below is a reproducible example.

The problem I have is that it seems to run well only for the first input row. Therefore, it gives me this error: Error in $<-.data.frame`(`*tmp*`, min.diff, value = list(response = 20.091674500638)) : replacement has 1 row, data has 5".

Going around this several hours but was not able to figure out the error.

Any hint?

Thanks a lot in advance

inputs <- data.frame(pd=c(38, 50, 50, 86, 38),
                     cl=c(15, 40, 30, 81, 15),
                     va=c(150, 145, 160, 141, 150),
                     co=c(3.0, 4.5, 4.0, 4.8, 1.5))

P <- c(seq(0,150,length.out=100))

eq1 <- function(a,b,c,x){
        (a*exp(-((((exp(b)*x))/1000))+c+b))/1000
}

resp.function <- function(pd, cl, va, co, P){
        a <- 10000 + 0.01*pd + (-0.1*cl)
        b <- 2
        c <- -1
        ratio <- co / (va/1000)
        min.diff <- as.data.frame(eq1(a,b,c,P)) %>% 
                        rename(response = "eq1(a, b, c, P)") %>% 
                        mutate(diff=abs(response - ratio)) %>% 
                        slice(which.min(diff)) %>% 
                        select(response)
        return(min.diff)
}

inputs$min.diff <- resp.function(inputs$pd, inputs$cl, inputs$va, inputs$co, P)

Upvotes: 1

Views: 204

Answers (1)

Ian Campbell
Ian Campbell

Reputation: 24838

The reason you are getting that error is because your function, as written, only returns 1 value. There are 5 columns of inputs, so that doesn't work.

A base R approach to this is to use the apply function. 

inputs$min.diff <- apply(inputs,MARGIN=1,function(x){resp.function(x["pd"], x["cl"], x["va"], x["co"], P)})

The second argument of apply indicates which direction to apply to. 1 indicates row wise, and 2 indicates column wise.

As you may know, your function resp.function does not actually use the arguments that you pass other than P. I don't believe there is enough information in your post for me to deduce the appropriate function, but if you clarify your intent, I may be able to help.

Upvotes: 2

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