CODEWITHSUNDEEP
rdataframe
BigD
BigD

Reputation: 581

ifelse didn't work in dataframe in R

I have a question about ifelse in data.frame in R. I checked several SO posts about it, and unfortunately none of these solutions fitted my case.

My case is, making a conditional calculation in a data frame, but it returns the condition has length > 1 and only the first element will be used even after I used ifelse function in R, which should work perfectly according to the SO posts I checked.

Here is my sample code:

library(scales)
head(temp[, 2:3])
  previous current
1        0      10
2       50      57
3       92     177
4       84     153
5       30      68
6      162     341
temp$change = ifelse(temp$previous > 0, rate(temp$previous, temp$current), temp$current)
rate = function(yest, tod){
  value = tod/yest
  if(value>1){
    return(paste("+", percent(value-1), sep = ""))
  }
  else{
    return(paste("-", percent(1-value), sep = ""))
  }
}

So if I run the ifelse one, I will get following result:

head(temp[, 2:4])
  previous current change
1        0      10     10
2       50      57  +NaN%
3       92     177  +NaN%
4       84     153  +NaN%
5       30      68  +NaN%
6      162     341  +NaN%

So my question is, how should I deal with it? I tried to assign 0 to the last column before I run ifelse, but it still failed.

Many thanks in advance!

Upvotes: 0

Views: 1637

Answers (4)

BigD
BigD

Reputation: 581

Adopting the advice I received from warm-hearted SO users, I vectorized some of my functions and it worked! Raise a glass to SO community!

Here is the solution:

temp$rate = ifelse(temp$previous > 0, ifelse(temp$current/temp$previous > 1, 
                                             temp$current/temp$previous - 1, 
                                             1 - temp$current/temp$previous), 
                   temp$current)

This will return rate with scientific notation. If "regular" notation is needed, here is an update:

temp$rate = format(temp$rate, scientific = F)

Upvotes: 0

kangaroo_cliff
kangaroo_cliff

Reputation: 6222

Try the following two segments, both should does what you wanted. May be it is the second one you are looking for.

library(scales)
set.seed(1)
temp <- data.frame(previous = rnorm(5), current = rnorm(5))
rate <- function(i) {
  yest <- temp$previous[i] 
  tod <- temp$current[i]
  if (yest <= 0)
    return(tod)
  value = tod/yest
 if (value>1) {
   return(paste("+", percent(value-1), sep = ""))
 } else {
   return(paste("-", percent(1-value), sep = ""))
 }
}

temp$change <- unlist(lapply(1:dim(temp)[1], rate))

Second:

ind <- which(temp$previous > 0)
temp$change <- temp$current
temp$change[ind] <- unlist(lapply(ind, 
                      function(i)  rate(temp$previous[i], temp$current[i])))

In the second segment, the function rate is same as you've coded it.

Upvotes: 1

kangaroo_cliff
kangaroo_cliff

Reputation: 6222

Only the first element in value is evaluated. So, the output of rate solely depend on the first row of temp.

Upvotes: 0

pachadotdev
pachadotdev

Reputation: 3775

Here's another way to do the same

# 1: load dplyr
#if needed install.packages("dplyr")
library(dplyr)

# 2: I recreate your data
your_dataframe = as_tibble(cbind(c(0,50,92,84,30,162),
                                 c(10,57,177,153,68,341))) %>% 
  rename(previous = V1, current = V2)

# 3: obtain the change using your conditions
your_dataframe %>% 
  mutate(change = ifelse(previous > 0,
                         ifelse(current/previous > 1,
                                paste0("+%", (current/previous-1)*100),
                                paste0("-%", (current/previous-1)*100)), 
                         current))

Result:

# A tibble: 6 x 3
  previous current             change
     <dbl>   <dbl>              <chr>
1        0      10                 10
2       50      57               +%14
3       92     177 +%92.3913043478261
4       84     153 +%82.1428571428571
5       30      68 +%126.666666666667
6      162     341 +%110.493827160494

Upvotes: 1

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