If then command yielding string as output in r

Question

I am trying to calculate after-tax income for each household in a data frame like this:

     id  hhinc  
1     1  53880  
2     2  49501  
3     3  37525  
4     4  28791   
5     5  91049    
6     6 133000   
7     7  12299        
8     8  23000   
9     9  58100   
10   10   9764

where hhinc is household income.

I then created the following function to calculate the taxes paid by each household:


taxpaid = function(hhinc) { 
  if (hhinc > 0 & hhinc <= 19999) {tax = 0} 
  else if (hhinc > 20000 & hhinc <= 49999) {tax = (hhinc - 20000)*.15} 
  else if (hhinc > 50000 & hhinc <= 199999) {tax = 4499.85 + ((hhinc - 50000)*.25)} 
  else if (hhinc > 200000 & hhinc <= 999999) {tax <- 37499.75 + ((hhinc - 200000)*.39)} 
  else if (hhinc > 1000000) {tax <- 311999.61 + ((hhinc - 1000000)*.85)}
  return(tax)
}

Since this function only works for a scalar input, I vectorized the function:

taxpaid_vec = Vectorize(taxpaid, vectorize.args = "hhinc")

However, when I use this function to calculate the taxes paid, I receive non-numeric outputs. I therefore cannot subtract the taxes paid from each household's income to determine the after-tax income. I would like to know how to fix my code so that get a numeric output for taxes paid.

Ronak Shah · Accepted Answer · 2019-12-20 04:51:33Z

3

Replace if/else to ifelse to make your function vectorized.

taxpaid = function(hhinc) { 
   ifelse(hhinc > 0 & hhinc <= 19999, 0,
    ifelse(hhinc > 20000 & hhinc <= 49999, (hhinc - 20000)*.15, 
     ifelse(hhinc > 50000 & hhinc <= 199999, 4499.85 + ((hhinc - 50000)*.25),
      ifelse(hhinc > 200000 & hhinc <= 999999, 37499.75 + ((hhinc - 200000)*.39), 
       ifelse(hhinc > 1000000, 311999.61 + ((hhinc - 1000000)*.85), NA)))))
}

Apply the function

df$tax_income <- taxpaid(df$hhinc)
df

#   id  hhinc tax_income
#1   1  53880    5469.85
#2   2  49501    4425.15
#3   3  37525    2628.75
#4   4  28791    1318.65
#5   5  91049   14762.10
#6   6 133000   25249.85
#7   7  12299       0.00
#8   8  23000     450.00
#9   9  58100    6524.85
#10 10   9764       0.00

You might also look in to ?dplyr::case_when for handling such nested conditions.

answered Dec 20, 2019 at 4:51

Ronak Shah

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