I want to use a set of 9 ordinal variables (scale of 0-3) to define a 10th binary variable. The binary variable is defined as:

Of the 9 items, 5 or more are >= 2.

I don’t know how to write code to calculate this other than to list all possible combinations of 5 questions >= 2, such as with the code below. Is there a simpler way?

```
library(tidyverse)
q1 <- c(0,0)
q2 <- c(2,3)
q3 <- c(3,3)
q4 <- c(1,1)
q5 <- c(0,2)
q6 <- c(1,1)
q7 <- c(1,2)
q8 <- c(2,2)
q9 <- c(3,0)
df <- data.frame(q1,q2,q3,q4,q5,q6,q7,q8,q9)
df <- df%>%
mutate(cutoff = ifelse((q1>=2 & q2>=2 & q3>=2 & q4>=2 & q5>=2) | q2>=2 #etc.
,1,0))
```

### >Solution :

You could try using `apply`

within your `ifelse`

like so:

```
df$q10 <- ifelse(apply(df, 1, function(x) sum(x >= 2)) >= 5, 1, 0)
```

Output:

```
# q1 q2 q3 q4 q5 q6 q7 q8 q9 q10
# 1 0 2 3 1 0 1 1 2 3 0
# 2 0 3 3 1 2 1 2 2 0 1
```

The `apply(df,1, function(x) sum(x >= 2))`

calculates the total number of values greater than or equal to 2 by row (observation), so the first row has 4, and the second row has 5.

```
apply(df,1, function(x) sum(x >= 2))
# [1] 4 5
```

Wrapping it in an `ifelse`

statement conditional on >= 5 converts it to the desired dichotomous variable.

Good luck!