roc11111111 - 22 days ago 4x

R Question

I have an expression

`qbinom(0.05, n, .47) - 1`

and I want to create a loop which iterates this expression over n for n = (20,200). For each iteration of this loop, this function will produce a number. I want to take the maximum of the 180 numbers it will produce. So, something like.

`for (n in 20:200) {`

max(qbinom(0.05, n, .47)-1)

But I'm not sure how exactly to do this.

Thanks!

Answer

First, I will show you how to do this with a loop.

```
n <- 20:200
MAX = -Inf ## initialize maximum
for (i in 1:length(n)) {
x <- qbinom(0.05, n[i], 0.47) - 1
if (x > MAX) MAX <- x
}
MAX
# [1] 81
```

Note, I am not keeping a record of all 181 values generated. Each value is treated as a temporary value and will be overwritten in the next iteration. In the end, we only have a single value `MAX`

.

If you want to at the same time retain all the records, we need first initialize a vector to hold them.

```
n <- 20:200
MAX = -Inf ## initialize maximum
x <- numeric(length(n)) ## vector to hold record
for (i in 1:length(n)) {
x[i] <- qbinom(0.05, n[i], 0.47) - 1
if (x[i] > MAX) MAX <- x[i]
}
## check the first few values of `x`
head(x)
# [1] 5 5 6 6 6 7
MAX
# [1] 81
```

Now I am showing the **vectorization** solution.

```
max(qbinom(0.05, 20:200, 0.47) - 1)
# [1] 81
```

R functions related to probability distributions are vectorized **in the same fashion**. For those related to binomial distributions, you can read `?rbinom`

for details.

Note, the vectorization is achieved with recycling rule. For example, by specifying:

```
qbinom(0.05, 1:4, 0.47)
```

R will first do recycling:

```
p: 0.05 0.05 0.05 0.05
mean: 1 2 3 4
sd: 0.47 0.47 0.47 0.47
```

then evaluate

```
qbinom(p[i], mean[i], sd[i])
```

**via a C-level loop**.

**Follow-up**

How would I be able to know which of the 20:200 corresponds to the maximum using the vectorization solution?

We can use

```
x <- qbinom(0.05, 20:200, 0.47) - 1
i <- which.max(x)
# [1] 179
```

Note, `i`

is the position in vector `20:200`

. To get the `n`

you want, you need:

```
(20:200)[i]
# 198
```

The maximum is

```
x[i]
# [1] 81
```

Source (Stackoverflow)

Comments