CODEWITHSUNDEEP
rapplylapply
Arol
Arol

Reputation: 31

divide a column of lists by numbers in another column

Hi there I have this data frame (test) and I want to divide every element in the list (AO) by a value in another column (DP):

df <- structure(list(DP = c("572", "594", "625", "594", "537", "513"
), AO2 = list(list(c(2, 2)), list(c(2, 2, 2)), list(c(4, 4)), 
    list(c(3, 2, 2, 2, 3)), list(c(2, 2)), list(c(2, 2)))), row.names = c(NA, 
-6L), class = c("data.table", 
"data.frame"))

df

I would like to create a new column in df where each value of the list is divided by the value in df$DP of the same row.

enter image description here

I have tried using mapply but it didn't work. Any idea?

test$AO2_DP <- mapply(FUN = `/`, list(as.numeric(unlist(test$AO2))), as.numeric(test$DP), SIMPLIFY = FALSE)

Upvotes: 3

Views: 572

Answers (6)

ThomasIsCoding
ThomasIsCoding

Reputation: 101733

You can try the code below

transform(
  df,
  AO2_DP = Map("/", unlist(AO2, recursive = FALSE), as.numeric(DP))
)

which gives

    DP       AO2                                                      AO2_DP
1: 572 <list[1]>                                     0.003496503,0.003496503
2: 594 <list[1]>                         0.003367003,0.003367003,0.003367003
3: 625 <list[1]>                                               0.0064,0.0064
4: 594 <list[1]> 0.005050505,0.003367003,0.003367003,0.003367003,0.005050505
5: 537 <list[1]>                                     0.003724395,0.003724395
6: 513 <list[1]>                                     0.003898635,0.003898635

Upvotes: 0

GKi
GKi

Reputation: 39667

As the column of lists can be nested you can use rapply to go to each element and make the division.

df$AO2_DP <- Map(function(x, y) rapply(x, function(z) z/y, how="replace"),
   df$AO2, as.numeric(df$DP))

df
#   DP           AO2                                                          AO2_DP
#1 572          2, 2                                        0.003496503, 0.003496503
#2 594       2, 2, 2                           0.003367003, 0.003367003, 0.003367003
#3 625          4, 4                                                  0.0064, 0.0064
#4 594 3, 2, 2, 2, 3 0.005050505, 0.003367003, 0.003367003, 0.003367003, 0.005050505
#5 537          2, 2                                        0.003724395, 0.003724395
#6 513          2, 2                                        0.003898635, 0.003898635

Upvotes: 1

Yuriy Saraykin
Yuriy Saraykin

Reputation: 8880

df %>%
  unnest(AO2) %>%
  mutate(DP = as.numeric(DP),
         res = map2(.x = AO2, .y = DP, .f = ~ .x / .y))

Upvotes: 0

Ronak Shah
Ronak Shah

Reputation: 389047

Base R option using Map -

df$result <- Map(`/`, unlist(df$AO2, recursive = FALSE),df$DP)
df

#    DP       AO2                                                      result
#1: 572 <list[1]>                                     0.003496503,0.003496503
#2: 594 <list[1]>                         0.003367003,0.003367003,0.003367003
#3: 625 <list[1]>                                               0.0064,0.0064
#4: 594 <list[1]> 0.005050505,0.003367003,0.003367003,0.003367003,0.005050505
#5: 537 <list[1]>                                     0.003724395,0.003724395
#6: 513 <list[1]>                                     0.003898635,0.003898635

If DP value is character in your real data turn it to numeric first by df$DP <- as.numeric(df$DP) before applying the answer.

Upvotes: 1

rifset
rifset

Reputation: 213

Here's dplyr solution using your original idea but modified

library(dplyr)

df %>% 
  rowwise() %>% 
  mutate(AO2_DP = list(mapply(FUN = "/", list(as.numeric(unlist(AO2))), as.numeric(DP), SIMPLIFY = FALSE))) %>% 
  as.data.frame()
#>    DP           AO2
#> 1 572          2, 2
#> 2 594       2, 2, 2
#> 3 625          4, 4
#> 4 594 3, 2, 2, 2, 3
#> 5 537          2, 2
#> 6 513          2, 2
#>                                                            AO2_DP
#> 1                                        0.003496503, 0.003496503
#> 2                           0.003367003, 0.003367003, 0.003367003
#> 3                                                  0.0064, 0.0064
#> 4 0.005050505, 0.003367003, 0.003367003, 0.003367003, 0.005050505
#> 5                                        0.003724395, 0.003724395
#> 6                                        0.003898635, 0.003898635

I added a quotation mark to the FUN of your mapply(), and then to perform the row-wise operation I use rowwise().

Upvotes: 0

Kra.P
Kra.P

Reputation: 15123

Maybe rowwise and sapply will help

test %>%
  rowwise() %>%
  mutate(DP = as.numeric(DP)) %>%
  mutate(AO2_DP = list(sapply(AO2, function(x) x/DP)))

Upvotes: 0

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