R: calculating new variables to a dataset by applying different formulas - r

Question

I have a dataframe "data" with the following structure:
structure(list(age = c(45, 4, 32, 45), sex = c(1, 0, 1, 0), height = c(165,
178, 145, 132), weight = c(65, 73, 60, 45)), row.names = c(NA,
-4L), class = c("tbl_df", "tbl", "data.frame"))
And I would like to add to this data.frame two new variables (var1, var2), which should be calculated with the two following formulas:
var1 = age*height + (4 if sex==1 OR 2 if sex==0)
var2 = height*weight + (1 if age>40 or 2 if age=<40)
I have a problem both in adding the two variables to the data frame, both in applying a function (I tried to build a function, but seems that can be applied only to a single value and not to all values from all rows).
Can anyone help me, please?

Answer 1

akrun's suggestion of using Boolean arithmetic is a good one but you could also do simply a Boolean version of your own expression substituting multiplication for the if statements.s (whit mild editing of the "=<" to "<=")
data <- structure(list(age = c(45, 4, 32, 45), sex = c(1, 0, 1, 0), height = c(165, 178, 145, 132), weight = c(65, 73, 60, 45)), row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"))
data <- within(data, {var1 = age*height + 4*(sex==1) + 2 *(sex==0);
var2 = height*weight + (age>40) + 2 *(age <= 40)})
#----
> data
age sex height weight var2 var1
1 45 1 165 65 10726 7429
2 4 0 178 73 12996 714
3 32 1 145 60 8702 4644
4 45 0 132 45 5941 5942
Since the two sets of conditions are each disjoint, the "non-qualifying" choice terms will each be 0.

Answer 2

the function ifelse() is vector based, so it will apply the conditions to each element in the vector.
df <- structure(list(age = c(45, 4, 32, 45), sex = c(1, 0, 1, 0), height = c(165,
178, 145, 132), weight = c(65, 73, 60, 45)), row.names = c(NA,
-4L), class = c("tbl_df", "tbl", "data.frame"))
df$var1 <- ifelse(df$sex == 1,(df$age * df$height) + 4,(df$age * df$height) + 2)
df$var2 <- ifelse(df$age > 40,(df$weight * df$height) + 1,(df$age * df$height) + 2)
final output
> df
# A tibble: 4 x 6
age sex height weight var1 var2
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 45 1 165 65 7429 10726
2 4 0 178 73 714 714
3 32 1 145 60 4644 4642
4 45 0 132 45 5942 5941

Answer 3

I rather the tool case_when() from dplyr package.
Your original data is:
data <-
structure(
list(age = c(45, 4, 32, 45),
sex = c(1, 0, 1, 0),
height = c(165, 178, 145, 132),
weight = c(65, 73, 60, 45)),
row.names = c(NA, -4L),
class = c("tbl_df", "tbl", "data.frame"))
The new variables are created by:
library(dplyr)
data ->
data %>% mutate(var1 = case_when(sex==1 ~ age*height + 4,
sex==0 ~ age*height + 2),
var2 = case_when(age>40 ~ height*weight + 1,
age<=40 ~ height*weight + 2)
)
The outcome is:
# A tibble: 4 x 6
age sex height weight var1 var2
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 45 1 165 65 7429 10726
2 4 0 178 73 714 12996
3 32 1 145 60 4644 8702
4 45 0 132 45 5942 5941

Answer 4

We convert the logical/binary to numeric index by adding 1 to it and use that to change the values to 2, 4, or just 1, 2 and use that in the calculation
library(dplyr)
data %>%
mutate(var1 = (age * height) + c(2, 4)[sex + 1],
var2 = (height * weight) + (age <= 40)+1)
# A tibble: 4 x 6
# age sex height weight var1 var2
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 45 1 165 65 7429 10726
#2 4 0 178 73 714 12996
#3 32 1 145 60 4644 8702
#4 45 0 132 45 5942 5941