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Consider the following toy data frame of my seed study: 

site <- c(LETTERS[1:12])          
site1 <- rep(site,each=80)

fate <- c('germinated', 'viable', 'dead')
fate1 <- rep(fate,each=320)

number <- c(41:1000)

df <- data.frame(site1,fate1,number)

> str(df)
'data.frame':   960 obs. of  3 variables:
 $ site1 : Factor w/ 12 levels "A","B","C","D",..: 1 1 1 1 1 1 1 1 1 1 ...
 $ fate1 : Factor w/ 3 levels "dead","germinated",..: 2 2 2 2 2 2 2 2 2 2 ...
 $ number: int  41 42 43 44 45 46 47 48 49 50 ...

I want R to go through all observations which are "dead" and assign "0" to every single one of them. Similarly, I want to assign "1" to all "viable" observations and "2" to all "germinated" observations.

My final data frame would be a single column, somewhat like this:

> year16
  [1] 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0
 [38] 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1

All suggestions are highly welcome

Muneer
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3 Answers3

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As zx8754 mentioned, you can have a look at the properties of a factor. 

year16 <- as.numeric(factor(df$fate1, levels = c("dead", "viable", "germinated")))-1

Here first I reorder the levels of df$fate1, so dead is assigned to 1, viable to 2 and germinated to 3. You want to start the sequence at 0, so I have to substract 1 after turning the factor in a numeric variable.

kath
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  • This idea is clever, and I like it +1, but I wonder about the performance implication of rehashing every entry in the factor. – Tim Biegeleisen May 10 '18 at 10:30
  • Thank you @Kath. It works the same way Tim Biegeleisen suggested. Kindly read my comment under Tim's answer for one more confusion. – Muneer May 10 '18 at 10:40
  • @TimBiegeleisen Yes you're right! I actually like your solution much better but I wanted to provide an base R approach. – kath May 10 '18 at 10:40
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    @TimBiegeleisen I've tested on a 15MM data set and the verse (your) solution is slower by factor of 6 – David Arenburg May 10 '18 at 10:57
  • @ Kath. There some problem: It gives me 0, 1 and 2 twelve times each one, which I do not want. However, I want R to repeat 0, 1 or 2 the number of times it occurs in column "number". Is it possible? – Muneer May 10 '18 at 11:04
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    @Muneer does `rep(year16, df$number)` give the result you're looking for?? – kath May 10 '18 at 11:15
  • @Kath. Yes, it has solved my problem. Exactly the same way I wanted. Thank you so much. – Muneer May 10 '18 at 11:21
2

Using case_when from the dplyr library:

df$year16 <-
case_when(
    levels(df$fate1)[df$fate1] == "dead" ~ 0,
    levels(df$fate1)[df$fate1] == "viable" ~ 1,
    levels(df$fate1)[df$fate1] == "germinated" ~ 2,
    TRUE ~ -1
)

Note: The solutions given by @David and @kath are much more graceful than this, but what I gave above would still work even if we had non numerical replacements.

Tim Biegeleisen
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  • @ Tim Biegeleisen, Thank you very much. It solved my problem. But, there is one more issue: e.g. if in site A there are 40 "germinated" seeds, it just assigns 1 to all 40. However, I would like to repeat 1 fourty times. Is it possible? – Muneer May 10 '18 at 10:37
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    What is the difference between assigning 1 40 times and repeating 1 40 times? – Tim Biegeleisen May 10 '18 at 10:43
  • Literally, there is no difference. It might have some effect once I run model. – Muneer May 10 '18 at 10:47
  • The problem is both solutions give me 0, 1 and 2 twelve times each one, which I do not want. – Muneer May 10 '18 at 11:00
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Base R solution:

assignnum <- function(x) {

  if (x == 'viable') {
    z <- 1
} else if (x == 'dead') {
  z <- 0
} else if (x == 'germinated') {
  z <- 2  
}
  return(z)
}

df['result'] <- sapply(df$fate1, assignnum)
Ollie Perkins
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