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I have written a function for amortising a loan given the loan characteristics and for providing me with the present value of the payment cash flows: 

mortgage<-function(P,I,L,amort,periodicity,n,residual,discount){


if(periodicity<-TRUE){i<-I/n}else{i<-I}
N<-n*L 
payment<- -((residual-(P*((1+i)^N)))/((((1+i)^N)-1)/i))
pay<<-payment
if(amort==T){
Pt<-P
currP<-NULL
while(Pt>=residual){
H<-Pt*i 
C<- payment-H
Q<-Pt-C
Pt<-Q
currP<-c(currP,Pt)}
monthP<-c(P,currP[1:(length(currP)-1)])-currP

adFmonth<<-data.frame(
  Amortization=c(P,currP[1:(length(currP)-1)]),
  Payment<-monthP+c((pay-monthP)[1:(length(monthP)-1)],0),
  Principal=monthP,
  Interest=c((pay-monthP)[1:(length(monthP)-1)],0),
  Time=sort(rep(1:ceiling(L*n/n),n))[1:length(monthP)]
)
adFmonth<<-adFmonth
}

AssetValue<-pv(discount/n,N,residual,pay)

return(AssetValue)
}

Given a set of (different) loans and a matrix containing the value of each function argument per loan, I would now like to create a loop, as that the function loops through each column of the matrix and gives me the output of the function.

Can anyone help me with that endeavour?

Thank you very much.

David Raths
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  • You've written a function that is difficult to program with because it uses global assignment with `<<-` instead of just returning the results. You should rewrite your function to not use `<<-` and just return everything - use a `list()` if you need to return multiple objects. Then the `for` loop should be trivial, it would be best if you showed an attempt. `results = list(); for(i in 1:ncol(your_matrix)) { results[[i]] = mortgage(your_matrix[1, i], your_matrix[2, i], ...)}` – Gregor Thomas Nov 09 '17 at 16:11
  • Your answer seems appealing, however if I do as stated, I get the following error message: Error in I/n : non-numeric argument to binary operator This makes no sense as both are defined as a number in my formula. Is it possible that when I import a CSV file with the same values from Excel, that the values are no longer treated as numerical? If yes, how can I change that? – David Raths Nov 09 '17 at 16:46
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    You can make a reproducible example. [See here for a complete guide](https://stackoverflow.com/q/5963269/903061). I can't help more based on the information you have provided.Importing a CSV, you almost certainly have a `data.frame`, not a `matrix`. Its entirely possible that your data is poorly imported and non-numeric - you can find plenty of questions about that. Look at `str(your_data)` to see the classes of everything. – Gregor Thomas Nov 09 '17 at 17:00

1 Answers1

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I would definitely advise you to take @Gregor's advice in the comment on your first post - avoid the global assignment within a function. That said, here is how I would approach your problem:

Assumption: For your data you are using a dataframe where each row maps to a unique loan and each column entry has a parameter (named A, B, and C in my examples) for the corresponding row.

My simple example should be easily extendable to your problem:

df <- data.frame(A=c(1,0,3),B=c(3,3,3),C=c(4,3,0)) # Replace with your data

# Replace with your function
myFn <- function(A,B,C) {
  x <- (A+B)/C
  return(x)
  }

# Appends 'answer' to your dataframe with the answer correctly mapped to each loan
df$answer <- with(df,myFn(A=A,B=B,C=C))

If you wish to have multiple 'answers' returned, change the code to be like:

df <- data.frame(A=c(1,0,3),B=c(3,3,3),C=c(4,3,0)) # Replace with your data

# Replace with your function
myFn <- function(A,B,C) {
  x <- (A+B)
  y <- C
  z <- x/y
  return(data.frame(A=A,B=B,C=C,numerator=x,denominator=y,answer=z))
  }

# Creates a new dataframe from the original with each 'multivariate' answer correctly mapped to each loan
df2 <- with(df,myFn(A=A,B=B,C=C))

# Examine your output:
head(df2)
str(df2)

There are probably more elegant solutions, but this should serve your purposes.

Nate
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