3

I'm newish to R and stack overflow, plus I'm not experienced at coding, and I'm hoping for some assistance. I have a dataframe where I'd like to do the same action on multiple variables. I wrote a function for the actions I'd like to take, but I'm not sure how to change the column names so the function acts on each variable separately.

#Fake Data

#index for a list of traits, and the current food type for each pet

shelterpets <- base::data.frame(
    ID                  = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10"),
    index_agility       = round(runif(10, min=-0.4, max=0.4), digits = 2),
    index_boldness      = round(runif(10, min=-0.4, max=0.4), digits = 2),
    index_curiousity    = round(runif(10, min=-0.4, max=0.4), digits = 2),
    index_dexterity     = round(runif(10, min=-0.4, max=0.4), digits = 2),
    index_empathy       = round(runif(10, min=-0.4, max=0.4), digits = 2),
    food_type           = c("diet_food", "diet_food", "regular_food", "diet_food", "regular_food", "regular_food", "regular_food", "diet_food", "diet_food", "regular_food")
                                )


 
# function to look at index for each trait, current food type, and suggest changes to food type
function(petfood) {
 
# variable to capture predicted food type: diet_food, regular_food
shelterpets$food10_trait  <- NA

 
#pet previously on diet_food and above 0.10 then confirm diet_food, else predict regular_food
shelterpets$food10_trait  <- ifelse(shelterpets$food_type == "diet_food",
                                        ifelse(shelterpets$index_trait >= 0.10, "diet_food",  "regular_food"),
                                    shelterpets$food10_trait)

 
#pet previously on regular_food and below -0.10 then confirm regular_food, else predict diet_food
shelterpets$food10_trait  <- ifelse(shelterpets$food_type == "regular_food",
                                        ifelse(shelterpets$index_trait <=  -0.10, "regular_food",  "diet_food" ),
                                    shelterpets$food10_trait)

 
#typecast
shelterpets$food10_trait  <- as.factor(shelterpets$food10_trait)

 
#update trait so replace "trait" with "agility", then "boldness", etc.
       }

And what I want it to look like is

 ID index_agility index_boldness index_curiousity index_dexterity index_empathy    food_type food10_agility food10_boldness
1  1          0.26          -0.28             0.17            0.17          0.28    diet_food      diet_food    regular_food
2  2          0.17          -0.12            -0.25            0.06          0.06    diet_food      diet_food    regular_food
3  3          0.24           0.14            -0.13            0.25          0.28 regular_food      diet_food       diet_food
4  4         -0.07           0.30            -0.32            0.06          0.23    diet_food   regular_food       diet_food
5  5          0.33           0.00             0.13            0.23         -0.18 regular_food      diet_food       diet_food
6  6          0.17          -0.20             0.01            0.25          0.17 regular_food      diet_food    regular_food

  food10_curiousity food10_dexterity food10_empathy
1         diet_food        diet_food      diet_food
2      regular_food     regular_food   regular_food
3      regular_food        diet_food      diet_food
4      regular_food     regular_food      diet_food
5         diet_food        diet_food   regular_food
6         diet_food        diet_food      diet_food

I made this to start


#get names in array to hopefully pass to the function, so drop ID and food_type
pet <- as.matrix(colnames(shelterpets))
pet <- pet[-c(1,7),,drop=F]

and I saw these questions, but I didn't quite follow how they worked enough to adapt them:

  1. Pass a data.frame column name to a function
  2. How to iteratively pass arguments into an R function

Thanks for any pointers you can give.

plover
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    Ideally. your data should be in long format with a single `index` and single indicator column for *agility*, *boldness*, etc. Then, no loop is needed! Just two `ifelse` calls to calculate `food_trait` column. – Parfait Apr 13 '21 at 23:22
  • Thank you for that idea! Is this something that would work when the datasets get very large (the real one has over 100k observations), or best with smaller ones? – plover Apr 15 '21 at 01:09
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    For processing and storing, long data even at millions of rows is usually better than wide data. – Parfait Apr 15 '21 at 15:19

3 Answers3

3

Your attempt is almost there, but you can loop over each trait and assign a new column to your data frame with the result of the function. I made some minor changes:

ifelse(shelterpets$index_trait

to

ifelse(shelterpets[, paste0('index_', trait)]

the input can be each trait as a character string, and the return value can just be as.factor(...)

# function to look at index for each trait, current food type, and suggest changes to food type
f <- function(trait, data = shelterpets) {
  
  # variable to capture predicted food type: diet_food, regular_food
  data$food10_trait  <- NA
  
  
  #pet previously on diet_food and above 0.10 then confirm diet_food, else predict regular_food
  data$food10_trait  <- ifelse(data$food_type == "diet_food",
                                      ifelse(data[, paste0('index_', trait)] >= 0.10, "diet_food",  "regular_food"),
                               data$food10_trait)
  
  
  #pet previously on regular_food and below -0.10 then confirm regular_food, else predict diet_food
  data$food10_trait  <- ifelse(data$food_type == "regular_food",
                                      ifelse(data[, paste0('index_', trait)] <=  -0.10, "regular_food",  "diet_food" ),
                               data$food10_trait)
  
  
  #typecast
  as.factor(data$food10_trait)
  
  
  #update trait so replace "trait" with "agility", then "boldness", etc.
}

## test
f('agility')
# [1] diet_food    diet_food    regular_food regular_food diet_food    regular_food regular_food diet_food    regular_food diet_food   
# Levels: diet_food regular_food

Apply over each trait

traits <- gsub('.*_', '', grep('index', names(shelterpets), value = TRUE))
shelterpets[, paste0('food10_', traits)] <- lapply(traits, f)

#    ID index_agility index_boldness index_curiousity index_dexterity index_empathy    food_type food10_agility food10_boldness food10_curiousity food10_dexterity food10_empathy
# 1   1          0.06          -0.34            -0.25            0.28          0.22    diet_food   regular_food    regular_food      regular_food        diet_food      diet_food
# 2   2          0.37          -0.01            -0.13            0.22          0.35    diet_food      diet_food    regular_food      regular_food        diet_food      diet_food
# 3   3          0.33          -0.07            -0.03            0.20          0.22 regular_food      diet_food       diet_food         diet_food        diet_food      diet_food
# 4   4          0.07          -0.23            -0.14           -0.29          0.05    diet_food   regular_food    regular_food      regular_food     regular_food   regular_food
# 5   5          0.23           0.06             0.09            0.24         -0.17 regular_food      diet_food       diet_food         diet_food        diet_food   regular_food
# 6   6         -0.27          -0.19            -0.23            0.37         -0.35 regular_food   regular_food    regular_food      regular_food        diet_food   regular_food
# 7   7          0.17           0.30            -0.14           -0.14         -0.11 regular_food      diet_food       diet_food      regular_food     regular_food   regular_food
# 8   8         -0.22           0.13             0.21           -0.06          0.08    diet_food   regular_food       diet_food         diet_food     regular_food   regular_food
# 9   9         -0.25           0.21            -0.02            0.09         -0.29    diet_food   regular_food       diet_food      regular_food     regular_food   regular_food
# 10 10         -0.35           0.39            -0.34            0.20          0.13 regular_food   regular_food       diet_food      regular_food        diet_food      diet_food
rawr
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  • Thank you! I didn't know the paste0 idea or command and am just learning about grep and sub. I really appreciate the clarity in your explanation. – plover Apr 15 '21 at 01:06
  • And I didn't know about the '.*_' part but just found another question here that was also useful, thanks again! https://stackoverflow.com/questions/11776287/remove-pattern-from-string-with-gsub – plover Apr 15 '21 at 01:17
2

You could simply write a function like below:

my_function<- function(x, y){
  ifelse(y == "diet_food",
       ifelse(x >= 0.10,  "diet_food", "regular_food"),
       ifelse(x <= -0.10, "regular_food",  "diet_food"))
}

data.frame(lapply(df[2:6], my_function, y=df[,7]))
  index_agility index_boldness index_curiousity index_dexterity index_empathy
1     diet_food   regular_food        diet_food       diet_food     diet_food
2     diet_food   regular_food     regular_food    regular_food  regular_food
3     diet_food      diet_food     regular_food       diet_food     diet_food
4  regular_food      diet_food     regular_food    regular_food     diet_food
5     diet_food      diet_food        diet_food       diet_food  regular_food
6     diet_food   regular_food        diet_food       diet_food     diet_food

Then you could use cbind to bind the results to the original df. You can also use sapply instead of lapply

Onyambu
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2

As mentioned, consider using long data usually preferred format for analytical operations including merging, plotting, modeling, etc. rather than wide usually better format for presentation and reporting. While data does result with more rows, you avoid looping for vectorized operations. Consider also within to avoid the repeating data frame name in new column calculation.

index_cols <- names(shelterpets)[grep("index", names(shelterpets))]

shelterpets_long <- stats::reshape(
    shelterpets, varying=index_cols, times=index_cols,
    v.names="value", timevar="indicator", ids=NULL, 
    idvar=c("ID", "food_type"), direction="long",
    new.row.names = 1:1E5
)

shelterpets_long <- base::within(shelterpets_long, {
  # pet previously on diet_food and above 0.10 then confirm diet_food, else predict regular_food
  food10_trait <- ifelse(food_type == "diet_food", 
                         ifelse(value >= 0.10, "diet_food", "regular_food"),
                         NA)
  
  # pet previously on regular_food and below -0.10 then confirm regular_food, else predict diet_food
  food10_trait  <- ifelse(food_type == "regular_food",
                          ifelse(value <=  -0.10, "regular_food",  "diet_food" ),
                          food10_trait)
  # typecast
  food10_trait  <- as.factor(food10_trait)
})

head(shelterpets_long)
#   ID    food_type     indicator value food10_trait
# 1  1    diet_food index_agility -0.29 regular_food
# 2  2    diet_food index_agility -0.39 regular_food
# 3  3 regular_food index_agility  0.23    diet_food
# 4  4    diet_food index_agility -0.36 regular_food
# 5  5 regular_food index_agility -0.34 regular_food
# 6  6 regular_food index_agility -0.01    diet_food

For possibly faster reshape to long format:

shelterpets_long2 <- base::data.frame(
    base::expand.grid(ID=unique(shelterpets$ID), indicator=index_cols, 
                      stringsAsFactors = FALSE),
    food_type = shelterpets$food_type,
    index = base::matrix(data.matrix(shelterpets[index_cols]), 
                         ncol=1, byrow=TRUE)
)

all.equal(shelterpets_long[c("ID", "food_type", "indicator", "value")],
          shelterpets_long2[c("ID", "food_type", "indicator", "value")])
# [1] TRUE

And if ever you need to reshape to wide, use this faster approach from slight modification of @Moody_Mudskipper's answer. As needed, merge to original, shelterpet, by ID variable:

### FASTER RESHAPE WIDE
### (https://stackoverflow.com/a/55973705/1422451)
matrix_spread <- function(df1, id, key, value){
  unique_ids <- unique(df1[[key]])
  mat <- matrix( df1[[value]], ncol=length(unique_ids), byrow = FALSE)
  df2 <- data.frame(unique(df1[[id]]), mat)
  names(df2) <- c(id, paste0(value,"_",unique_ids))
  df2
}

shelterpets_wide <- matrix_spread(
    shelterpets_long, 
    id = "ID",
    key = "indicator",
    value = "food10_trait"
)

shelterpets_wide
#    ID food10_trait_index_agility food10_trait_index_boldness food10_trait_index_curiousity food10_trait_index_dexterity food10_trait_index_empathy
# 1   1               regular_food                regular_food                     diet_food                 regular_food               regular_food
# 2   2               regular_food                regular_food                  regular_food                 regular_food               regular_food
# 3   3                  diet_food                   diet_food                     diet_food                    diet_food                  diet_food
# 4   4               regular_food                regular_food                  regular_food                 regular_food               regular_food
# 5   5               regular_food                   diet_food                     diet_food                 regular_food                  diet_food
# 6   6                  diet_food                regular_food                     diet_food                 regular_food               regular_food
# 7   7               regular_food                   diet_food                     diet_food                    diet_food               regular_food
# 8   8               regular_food                   diet_food                     diet_food                 regular_food                  diet_food
# 9   9                  diet_food                regular_food                  regular_food                 regular_food               regular_food
# 10 10               regular_food                regular_food                  regular_food                 regular_food               regular_food
Parfait
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  • This is super helpful, thank you. Quick question: what does the 1E5 do in the command "new.row.names = 1:1E5"? Also, when I tried this and went from long to wide, the foodtrait values didn't match, but I'm not sure why. – plover Apr 15 '21 at 19:48
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    `1:1E5` is a short hand for `1:100000` where 5 helps to know how many zeros. This range should be adjusted to accommodate actual final data. Re mismatch, try `byrow = FALSE` in `matrix_spread` function. See edit with adjusted results. Be sure `set.seed(###)` at beginning to reproduce random data. – Parfait Apr 15 '21 at 20:16