CODEWITHSUNDEEP
r
Sutton Murray
Sutton Murray

Reputation: 35

If Else Statement with Four Criteria returning NA

I have this dataset:

Pars <-structure(list(Pars.PeopleSoftID = c(11, 18, 18, 22, 39, 181), 
    Pars..Central.1. = c(0, 1, 1, 1, 1, 1), Pars..Central.2. = c(0, 
    1, 1, 1, 1, 1), Pars.External = c(3, 2, 2, 1, 2, 5), Pars.Minimum = c(15, 
    2, 1.71, 1, 1, 4.39), Pars.Maximum = c(15, 2.76, 2, 1, 1, 
    5.48)), .Names = c("Pars.PeopleSoftID", "Pars..Central.1.", 
"Pars..Central.2.", "Pars.External", "Pars.Minimum", "Pars.Maximum"
), row.names = c(NA, 6L), class = "data.frame")

I'm trying to add a column that calculates a number based on what is in the columns Central 1, Central 2, and External and multiplies that number by the value in Pars$Minimum.

The logic is:

If Central 1 = 0, Central 2 = 0, External = 0 THEN <-- 0 * Pars$Minimum

If Central 1 = 1, Central 2 = 0, External = 0 THEN <-- 1 * Pars$Minimum

If Central 1 = 1, Central 2 = 1, External = 0 THEN <-- 0.7 * Pars$Minimum

If Central 1 = 1, Central 2 = 1, External > 0 THEN <-- 0.7 * Pars$Minimum

If Central 1 = 1, Central 2 = 0, External > 0 THEN <-- 0.8 * Pars$Minimum

I just tried this, but it seems to be just returning values of NA for everything:

 Pars$Cen1 <- with(Pars, if(`Central 1` == 0 , 0, ifelse(`Central 1` == 1 & `Central 2` == 0 & External == 0, 1, ifelse(`Central 1` == 1 & `Central 2` == 1 & External == 0, .7, ifelse(`Central 1` == 1 & `Central 2` == 1 & External > 0, .7, ifelse(`Central 1` == 1 & `Central 2` == 0 & External > 0, .8)

Upvotes: 1

Views: 718

Answers (4)

Christoph Wolk
Christoph Wolk

Reputation: 1758

Another possibility is to make a map of values and lookup the conditions pasted together there:

  valuemap = c("000" = 0, "001"= 0, "100"= 1, "110"= 0.7, "111" = 0.7, "101" = 0.8)

  Pars$Cen1 <- Pars$Pars.Minimum * 
    valuemap[paste0(Pars$Pars..Central.1.,Pars$Pars..Central.2., 
                    1 * (Pars$Pars.External > 0))]

Upvotes: 3

Adrian Martin
Adrian Martin

Reputation: 820

Your final ifelse statement has only two arguments, so if you ever get to trying to assign the value of that 'no' statement, you'll get an error.

This seemed to work for me:

Pars$Cen1 <- ifelse(Pars$Pars..Central.1. == 0, 0, 
                ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 0 & Pars$Pars.External == 0, 1, 
                       ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 1 & Pars$Pars.External == 0, .7, 
                              ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 1 & Pars$Pars.External > 0, .7, 
                                     ifelse(Pars$Pars..Central.1. == 1 & Pars$Pars..Central.2. == 0 & Pars$Pars.External > 0, .8, "Missing")))))

Upvotes: 1

Nate Thompson
Nate Thompson

Reputation: 635

I renamed some of your columns as well, wasnt sure why each column name needed to be called "Pars." if the table was already called Pars, but this works here to do what you want, should give you a format to use. 

Pars <-structure(list(Pars.PeopleSoftID = c(11, 18, 18, 22, 39, 181), 
    Pars..Central.1. = c(0, 1, 1, 1, 1, 1), Pars..Central.2. = c(0, 
    1, 1, 1, 1, 1), Pars.External = c(3, 2, 2, 1, 2, 5), Pars.Minimum = c(15, 
    2, 1.71, 1, 1, 4.39), Pars.Maximum = c(15, 2.76, 2, 1, 1, 
    5.48)), .Names = c("PeopleSoftID", "Central.1", 
"Central.2", "External", "Minimum", "Maximum"
), row.names = c(NA, 6L), class = "data.frame")

Pars$Cen1 <- with(Pars,ifelse(Central.1 == 0 & Central.2 == 0 & External == 0,0*Minimum,
                       ifelse(Central.1 == 1 & Central.2 == 0 & External == 0,1*Minimum, 
                       ifelse(Central.1 == 1 & Central.2 == 1 & External >= 0,0.7*Minimum,
                       ifelse(Central.1 == 1 & Central.2 == 0 & External > 0,.8*Minimum,0)))))

Upvotes: 3

pogibas
pogibas

Reputation: 28339

library(dplyr)
mutate(Pars, Cen1 = case_when(
    `Central 1` == 0 & `Central 2` == 0 & External == 0 ~ 0 * Minimum,
    `Central 1` == 1 & `Central 2` == 0 & External == 0 ~ 1 * Minimum,
    `Central 1` == 1 & `Central 2` == 1 & External == 0 ~ 0.7 * Minimum,
    `Central 1` == 1 & `Central 2` == 1 & External > 0 ~ 0.7 * Minimum,
    `Central 1` == 1 & `Central 2` == 0 & External > 0 ~ 0.8 * Minimum))

Upvotes: 1

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