在数据帧的每一行上运行函数，将结果存储在新列 R

Question

I would like to run the function sum_differences on every row of a dataframe and store the result in a new column "di_Flex". a is column Company ID and b is column max_di_Flex . Can anyone help me out with the for loop? Thank you!

sum_differences <- function(a,b) {
  a <- unique(a)
  new_list <- c()

  for (i in a) {
    for (j in a) {
      if(i != j) {
        new_list <- c(new_list, abs(i-j))
      }
    }
  }
  outcome <- round((sum(new_list) / length(a)), 2)
  percent <- outcome/b
  return(percent)
}

data <- data.frame(structure(list(`Company ID` = c(0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 
5, 5, 6, 6, 6, 6, 6, 6, 6, 6), Flexibility_Thinking = c(7, 2, 
5, 1, 6, 3, 6, 6, 7, 5, 7, 7, 4, 7, 5, 2, 3, 3, 3, 3), max_di_Flex = c(12.8, 
12.8, 12.8, 12.8, 12.8, 0, 0, 0, 0, 8, 8, 8, 16, 16, 16, 16, 
16, 16, 16, 16))))

Answer 1

Try:

data$di_Flex <- apply(data, 1, function(x) sum_differences(x[1], x[3]))

Or:

data$di_Flex <- apply(as.data.frame(data), 1, function(x) sum_differences(x[1], x[3]))

Answer 2

Tidyverse solution using map2_dbl (when you have a function that takes 2 inputs and returns a number):

df <- as.data.frame(data)
library(tidyverse)
df %>%
    mutate(diFlex = map2_dbl(Company.ID, max_di_Flex, sum_differences))
#>    Company.ID Flexibility_Thinking max_di_Flex diFlex
#> 1           0                    7        12.8      0
#> 2           0                    2        12.8      0
#> 3           0                    5        12.8      0
#> 4           0                    1        12.8      0
#> 5           0                    6        12.8      0
#> 6           1                    3         0.0    NaN
#> 7           2                    6         0.0    NaN
#> 8           3                    6         0.0    NaN
#> 9           4                    7         0.0    NaN
#> 10          5                    5         8.0      0
#> 11          5                    7         8.0      0
#> 12          5                    7         8.0      0
#> 13          6                    4        16.0      0
#> 14          6                    7        16.0      0
#> 15          6                    5        16.0      0
#> 16          6                    2        16.0      0
#> 17          6                    3        16.0      0
#> 18          6                    3        16.0      0
#> 19          6                    3        16.0      0
#> 20          6                    3        16.0      0

Edit

From the comments I saw on the other answer, I realized the function is made to take vectors instead of individual values. That means you don't need a loop, and should just call the function directly on the vectors:

df %>%
    mutate(diFlex = sum_differences(df$Company.ID, df$max_di_Flex))
#>    Company.ID Flexibility_Thinking max_di_Flex diFlex
#> 1           0                    7        12.8   1.25
#> 2           0                    2        12.8   1.25
#> 3           0                    5        12.8   1.25
#> 4           0                    1        12.8   1.25
#> 5           0                    6        12.8   1.25
#> 6           1                    3         0.0    Inf
#> 7           2                    6         0.0    Inf
#> 8           3                    6         0.0    Inf
#> 9           4                    7         0.0    Inf
#> 10          5                    5         8.0   2.00
#> 11          5                    7         8.0   2.00
#> 12          5                    7         8.0   2.00
#> 13          6                    4        16.0   1.00
#> 14          6                    7        16.0   1.00
#> 15          6                    5        16.0   1.00
#> 16          6                    2        16.0   1.00
#> 17          6                    3        16.0   1.00
#> 18          6                    3        16.0   1.00
#> 19          6                    3        16.0   1.00
#> 20          6                    3        16.0   1.00