Solution for generating experimental treatment combinations

1 Solution with R output

1.1 Task 1

The number of possible treatment combinations is the product of the number of levels of each variable:

4 (N) * 3 (P) * 2 (irrigation) = 24 possible combinations

1.2 Task 2

library(tidyr)
library(dplyr)

# No. of rows: 4 * 3 * 2 = 24 rows

# Define levels as vectors
n_levels <- c(100,200,300,400)
p_levels <- c(50,100,150)
irri_levels <- c("yes","no")

1.2.1 One potential approach using nested rep-calls

dat_exp1 <- tibble(nitrogen = rep(n_levels, each = 2*3),
                   phosphorus = rep( rep(p_levels, each = 2), times = 4),
                   irrigation = rep(irri_levels, times = 4 * 3))
dat_exp1

# A tibble: 24 × 3
   nitrogen phosphorus irrigation
      <dbl>      <dbl> <chr>     
 1      100         50 yes       
 2      100         50 no        
 3      100        100 yes       
 4      100        100 no        
 5      100        150 yes       
 6      100        150 no        
 7      200         50 yes       
 8      200         50 no        
 9      200        100 yes       
10      200        100 no        
# ℹ 14 more rows

1.2.2 Another approach with one table for each irrigation levels, which are combined

dat_irri <- tibble(nitrogen = rep(n_levels, each = 3),
                   phosphorus = rep(p_levels, times = 4),
                   irrigation = "yes")

dat_no_irri <- dat_irri %>% mutate(irrigation = "no")

dat_exp2 <- bind_rows(dat_irri, dat_no_irri)
dat_exp2

# A tibble: 24 × 3
   nitrogen phosphorus irrigation
      <dbl>      <dbl> <chr>     
 1      100         50 yes       
 2      100        100 yes       
 3      100        150 yes       
 4      200         50 yes       
 5      200        100 yes       
 6      200        150 yes       
 7      300         50 yes       
 8      300        100 yes       
 9      300        150 yes       
10      400         50 yes       
# ℹ 14 more rows

1.3 Extra:

With a single call to tidyr::expand_grid

dat_exp3 <- expand_grid(n_levels, p_levels, irri_levels)
dat_exp3

# A tibble: 24 × 3
   n_levels p_levels irri_levels
      <dbl>    <dbl> <chr>      
 1      100       50 yes        
 2      100       50 no         
 3      100      100 yes        
 4      100      100 no         
 5      100      150 yes        
 6      100      150 no         
 7      200       50 yes        
 8      200       50 no         
 9      200      100 yes        
10      200      100 no         
# ℹ 14 more rows

2 Solution as one script without output

library(tidyr)
library(dplyr)

# No. of rows: 4 * 3 * 2 = 24 rows

# Define levels as vectors
n_levels <- c(100,200,300,400)
p_levels <- c(50,100,150)
irri_levels <- c("yes","no")

##One potential approach using nested rep-calls
dat_exp1 <- tibble(nitrogen = rep(n_levels, each = 2*3),
                  phosphorus = rep(rep(p_levels, each = 2), times = 4),
                  irrigation = rep(irri_levels, times = 4*3))
dat_exp1

# Another approach with one table for each irrigation levels, which are combined
dat_irri <- tibble(nitrogen = rep(n_levels, each = 3),
                   phosphorus = rep(p_levels, times = 4),
                   irrigation = "yes")

dat_no_irri <- dat_irri %>% mutate(irrigation = "no")
dat_exp2 <- bind_rows(dat_irri, dat_no_irri)
dat_exp2

# Extra: With a single call to tidyr::expand_grid

dat_exp3 <- expand_grid(n_levels, p_levels, irri_levels)
dat_exp3