CODEWITHSUNDEEP
rif-statementlapplysapply
stackbeats
stackbeats

Reputation: 173

R sapply/lapply with multiple ifelse statements

I'm trying to simplify the following multiple ifelse code using sapply or lapply (still can't distinguish them).

My goal is to allocate points based on placement like shown below.

df$Point <- ifelse(df$Placement_v2 <= 1, 10,
ifelse(df$Placement_v2 <= 10, 9, 
ifelse(df$Placement_v2 <= 25, 8,
ifelse(df$Placement_v2 <= 50, 7, 1) )))

This code works okay, but I want to make a dataframe and simply my code above using sapply or lapply (or anyother function).

I've tried this code but is not working as expected. Only the rows with placement 1 get 10 points and other rows end up with 1.

<2nd code>

df$Point <- sapply(df2$Placement, function(x) ifelse(df$Placement_v2 <= x, df2$Point[df2$Placement == x], 1 ) )

How can I solve this problem?

Upvotes: 1

Views: 1795

Answers (2)

d.b
d.b

Reputation: 32558

You could create a dataframe with values and replacements. Then you can use cut to lookup the appropriate value

dict = data.frame(replacement = c(10, 9, 8, 7, 1, 1),
    values = c(0, 1, 10, 25, 50, 1e5))

#DATA
set.seed(42)
placement = sample(1:100, 15)

cbind(placement,
    new_placement = dict$replacement[as.integer(cut(placement, breaks = dict$values))])
#     placement new_placement
# [1,]        92             1
# [2,]        93             1
# [3,]        29             7
# [4,]        81             1
# [5,]        62             1
# [6,]        50             7
# [7,]        70             1
# [8,]        13             8
# [9,]        61             1
#[10,]        65             1
#[11,]        42             7
#[12,]        91             1
#[13,]        83             1
#[14,]        23             8
#[15,]        40             7

Upvotes: 0

Eric Watt
Eric Watt

Reputation: 3240

A few ways to go about this. I'll use data.table.

library(data.table)

set.seed(123)
df <- data.table(Placement_v2 = runif(200, -10, 100))

First option, move the evaluation out to a function, and then lapply the function to your Placement_v2 column. This has the benefit of being much cleaner than your nested ifelse statements.

funky <- function(x) {

  if (x <= 1) {
    val <- 10
  } else if (x <= 10){
    val <- 9
  } else if (x <= 25){
    val <- 8
  } else if (x <= 50){
    val <- 7
  } else {
    val <- 1
  }

  return(val)

}

df[, Point := unlist(lapply(Placement_v2, funky))]

Result:

     Placement_v2 Point
  1:    21.633527     8
  2:    76.713565     1
  3:    34.987461     7
  4:    87.131914     1
  5:    93.451401     1
 ---                   
196:    41.318597     7
197:    34.751585     7
198:    62.515336     1
199:     6.758128     9
200:    53.015376     1

I would instead approach this by subsetting the data, and assigning by each subset. You could do this by specifying each subset df[Placement_v2 <= 1], df[Placement_v2 >= 1 & Placement_v2 <= 10], etc. But, if you do it in the correct order, you can avoid the double equality evaluation.

df[, Point := 1]
df[Placement_v2 <= 50, Point := 7]
df[Placement_v2 <= 25, Point := 8]
df[Placement_v2 <= 10, Point := 9]
df[Placement_v2 <=  1, Point := 10]

Which gives the same result:

     Placement_v2 Point
  1:    21.633527     8
  2:    76.713565     1
  3:    34.987461     7
  4:    87.131914     1
  5:    93.451401     1
 ---                   
196:    41.318597     7
197:    34.751585     7
198:    62.515336     1
199:     6.758128     9
200:    53.015376     1

Upvotes: 1

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