CODEWITHSUNDEEP
r
crystal
crystal

Reputation: 37

Trying to make a prediction model in R

Collected several questions during the last days of programming. Trying to make a program that make a prognosis on how much money I get after working for 5 years. One problem is when I try to take the money in minus money out.

moneyover.year[i] =  moneyin.year[i] - moneyout.year[i]

Then it gives this error: "non-numeric argument to binary operator"

I would like to extract information from each year with the function "summary" but it doesn't really work... It just shows:

Length Class Mode [1,] 100 -none- numeric

Finally I want to predict how many holidays I will get. The graph works but I want to get the percentage sign on y-axis. Preferably a smooth curve with the histogram.

Here is the full code:

library(ggplot2)
library(mc2d)
library(scales)
moneyin.year= NULL # Is there a work around this?
moneyout.year= NULL # Is there a work around this?
moneyover.year = NULL # Is there a work around this?
n=100

for (i in 1:5 ) {
  moneyin.year[i] <- list(rpert(n, min=20000, mode=23000, max=30000, shape=30))
  moneyout.year[i] <- list(rpert(n, min=10000, mode=12500, max=19500, shape=20))
  moneyover.year[i] =  moneyin.year[i] - moneyout.year[i] # GIVES ERROR!
}
moneyin.year
moneyout.year
moneyover.year # GIVES ERROR!

#graph = moneyover.year[1]
graph = moneyin.year[1]
summary(graph) # Doesnt really work...
sd(graph) # GIVES ERROR!
var(graph) # GIVES ERROR!

p <- ggplot(data.frame(graph), aes(x = graph))  
p <- p + geom_bar(aes(y = (..count..)/sum(..count..)), color="black", binwidth = 500, fill = "steelblue") 
p <- p + scale_y_continuous(labels = percent)
p <- p + xlab("EUR") + ylab("Percent")
p <- p + theme_bw() 
print(p)


extraholidays = dpois(20:50,30)
barplot(extraholidays,names=20:50,xlab='Days',ylab='Percentage') 
# How to get "%" sign???
# How to get a smooth line?

Upvotes: 0

Views: 77

Answers (1)

Jthorpe
Jthorpe

Reputation: 10167

You're getting an error when you call moneyover.year[i] = blah blah blah because you haven't initialized moneyover.year and you cant take or assign to the subset of a vector that doesn't yet exist. (you initilized moneyover and not moneyover.year...).

Initilization is just a fact of programming, but you can use multiple assignment to initialize several variables at once, as in:

moneyin.year <- moneyout.year <- moneyover.year <- NULL

The second error has to do with R's type conversion. Specifically, in the first iteration of the for loop, this linemoneyin.year[i] <- list(blah blah blah) coerces moneyin.year to a list. This is a problem when this line is evaluated:

moneyover.year[i] =  moneyin.year[i] - moneyout.year[i]

because the single bracket operator ([) returns a sub-list, and not the first element of the list. This is a problem because the minus operator (-) is not defined for lists (hence, the somewhat non-intuitive error "non-numeric argument to binary operator"). Instead, you want to use the double-bracket operator ([[) which returns the value stored in the i'th element of the list. as in: 

moneyover.year[[i]] =  moneyin.year[[i]] - moneyout.year[[i]]

Since all three of the objects above are lists, it might save on confusion to initialize them as lists rather than as NULL values, as in:

moneyin.year <- moneyout.year <- moneyover.year <- list()

Upvotes: 1

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