CODEWITHSUNDEEP
r
Sankalp Mathur
Sankalp Mathur

Reputation: 119

Calculating category specific variable in R

I have big data with col_1 as the first category and col_2 as the second category. I am attaching a sample form(refer to the picture below). The data has the first four columns (col_1, col_2, ice, fd). I want to generate the variable "ice_new" for each of the categories of col_1 by taking the sum of the column fd as the denominator and the value of "ice" for different col_2 as the numerator and adding them up. I tried using the "aggregate" function in R, but it doesn't work. How do I execute this in R? Any help will be appreciated

col_1   col_2   ice   fd    ice_new

A       A1      0.3   0.1   (0.3/(0.1+0.4) + 0.2/(0.1+0.4)
A       A2      0.2   0.4   (0.3/(0.1+0.4) + 0.2/(0.1+0.4)
B       B1      1.2   1     1.2/(1+2+1.2) + 1.4/(1+2+1.2) + 0.6/ (1+2+1.2)
B       B2      1.4   2     1.2/(1+2+1.2) + 1.4/(1+2+1.2) + 0.6/ (1+2+1.2)
B       B3      0.6   1.2   1.2/(1+2+1.2) + 1.4/(1+2+1.2) + 0.6/ (1+2+1.2)

enter image description here

Upvotes: 0

Views: 302

Answers (3)

Shirin Yavari
Shirin Yavari

Reputation: 682

df1 <- data.frame("col_1" = c("A", "A", "B", "B", "B"), 
                        "col_2" = c("A1", "A2", "B1", "B2", "B3"), 
                        "ice" = c(.3,.2,1.2,1.4,.6), 
                        "fd" = c(.1,.4,1,2,1.2))
library(dplyr)
df2 <- df1 %>% 
         group_by(col_1) %>% 
           mutate(ice_new=sum(ice)/sum(fd))

df2
## A tibble: 5 x 5
## Groups:   col_1 [2]
#  col_1 Col_2   ice    fd ice_new
#  <fct> <fct> <dbl> <dbl>  <dbl>
#1 A     A1      0.3   0.1  1    
#2 A     A2      0.2   0.4  1    
#3 B     B1      1.2   1    0.762
#4 B     B2      1.4   2    0.762
#5 B     B3      0.6   1.2  0.762

Upvotes: 1

tmfmnk
tmfmnk

Reputation: 39858

One dplyr possibility could be:

df %>%
 group_by(col_1) %>%
 mutate(ice_new = sum(ice/sum(fd)))

  col_1 col_2   ice    fd ice_new
  <chr> <chr> <dbl> <dbl>   <dbl>
1 A     A1      0.3   0.1   1    
2 A     A2      0.2   0.4   1    
3 B     B1      1.2   1     0.762
4 B     B2      1.4   2     0.762
5 B     B3      0.6   1.2   0.762

Or the same with base R:

with(df, ave(ice/ave(fd, col_1, FUN = sum), col_1, FUN = sum))

Upvotes: 2

Rafael
Rafael

Reputation: 3196

You could also use summarise to get one value per group: 

library(dplyr)
df %>% 
  group_by(col_1) %>%
  summarise(ice_new = sum(ice / sum(fd)))

# A tibble: 2 x 2
  col1  ice_new
  <chr>   <dbl>
1 A       1    
2 B       0.762

Upvotes: 0

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