CODEWITHSUNDEEP
rdplyr
dianaiii
dianaiii

Reputation: 89

R: use the newly generated data in the previous row

I group the observations in a dataframe by ID. I want to keep the value of the first row in each group as it is, while fill in the remaining rows in the same group with lagged values + 2 and raise it to the power of 2. I want to update the lagged values as I proceed.

Please take the following dataset as an example:

ID <- c("1","1","1","2","2","3","3","3")
val <- c(1:8)
df <- data.frame(ID,val)

  ID val
1  1   1
2  1   2
3  1   3
4  2   4
5  2   5
6  3   6
7  3   7
8  3   8

I am expecting to see:

  ID val
1  1   1
2  1   (1 + 2)^2 = 9
3  1   (9 + 2)^2 = 121
4  2   4
5  2   (4 + 2)^2 = 36
6  3   6
7  3   (6 + 2)^2 = 64
8  3   (64 + 2)^2 = 4356

What I have tried:

df <- df %>% group_by(ID) %>% mutate(val = ifelse(row_number()!=1,(lag(val)+2)^2, val))

But what I seemed to get is:

  ID val
1  1   1
2  1   (1 + 2)^2 = 9
3  1   (2 + 2)^2 = 16
4  2   4
5  2   (4 + 2)^2 = 36
6  3   6
7  3   (6 + 2)^2 = 64
8  3   (7 + 2)^2 = 81

I guess R does not update the value in the previous row before it proceeds to the next row, so it was using the old lagged value. Is there a way to fix this? Also, I have to apply similar but more complicated calculations (a lot of exponents) on a large dataset, so if there is a quick way, it will be perfect!

Thank you!

Upvotes: 0

Views: 272

Answers (4)

akrun
akrun

Reputation: 886938

Using base R

df$val <- with(df,  ave(val, ID, FUN = function(u)
              Reduce(function(x, y) x+ 2, u, accumulate = TRUE)))

Or using seq

library(dplyr)
df %>%
    group_by(ID) %>%
    mutate(val =  seq(first(val), length.out = n(), by = 2))

Upvotes: 1

ThomasIsCoding
ThomasIsCoding

Reputation: 101034

Below is one option using Reduce in data.table

> setDT(df)[, val := Reduce(function(x, y) (x + 2)^2, val, accumulate = TRUE), ID][]
   ID  val
1:  1    1
2:  1    9
3:  1  121
4:  2    4
5:  2   36
6:  3    6
7:  3   64
8:  3 4356

Upvotes: 1

AnilGoyal
AnilGoyal

Reputation: 26218

If you are pssing two arguments use accumulate2 instead. ..1 will represent previously iterated value, ..2 will represent first argument and ..3 will represent second argument.

Note ..2 is not used directly here but you are using it for invisble iteration only


df %>% 
  group_by(ID) %>% 
  mutate(val = accumulate2(val, pw, ~(..1+ 2)^..3)) %>%
  ungroup

Upvotes: 3

Ronak Shah
Ronak Shah

Reputation: 388797

You can use accumulate to perform such recursive calculation.

library(dplyr)
library(purrr)

df %>% 
  group_by(ID) %>% 
  mutate(val = accumulate(val, ~.x + 2)) %>%
  ungroup

# A tibble: 8 x 2
#  ID      val
#  <chr> <dbl>
#1 1         1
#2 1         3
#3 1         5
#4 2         4
#5 2         6
#6 3         6
#7 3         8
#8 3        10

Upvotes: 2

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