I'll start with what I've done already. I'm looking for a way to solve equation f by changing the b and s parameters for each row, Q and n are constants. I know that apply() works for this type of problem, but that doesn't seem to work for me. The variable that I want to find doesn't give a unique solution.
Q = 0.203
n = 0.014
f <- function(y) (Q - (1/n)*(y*b)*((y*b)/(2*y+b))^(2/3)*sqrt(s))
With these parameters, let's say for b = 0.5 and s = 0.01 using uniroot() I get the following. Which is the result I want.
uniroot(f, lower = 0.000001, upper = 1000000)$root
[1] 0.2328931
(those lower and upper values seemed to work out well for me)
Now what I need is to solve this function for a large dataset.
set.seed(123)
tibble::tibble(b = runif(n = 1000, min = 0.1, max = 1.5),
s = runif(n = 1000, min = 0.001, max = 5)) %>%
dplyr::mutate(yn = uniroot(f, lower = 0.000001, upper = 1000000)$root) %>%
head(5)
And this is my desired output.
b s yn
1 0.503 1.37 0.0434
2 1.20 2.97 0.0194
3 0.673 0.802 0.0421
4 1.34 4.27 0.0163
5 1.42 4.24 0.0157
Actually you are very close to the goal. Here is a base R option using Vectorize + do.call which may help you
f <- function(b, s) {
fn <- function(y) (Q - (1 / n) * (y * b) * ((y * b) / (2 * y + b))^(2 / 3) * sqrt(s))
uniroot(fn, lower = 0.000001, upper = 1000000)$root
}
df$yn <- do.call(Vectorize(f), df)
such that
> df
# A tibble: 1,000 x 3
b s yn
<dbl> <dbl> <dbl>
1 0.503 1.37 0.0435
2 1.20 2.97 0.0194
3 0.673 0.802 0.0422
4 1.34 4.27 0.0163
5 1.42 4.24 0.0157
6 0.164 2.39 0.0912
7 0.839 3.87 0.0224
8 1.35 1.48 0.0223
9 0.872 0.329 0.0468
10 0.739 2.20 0.0289
# ... with 990 more rows
Data
set.seed(123)
df <- tibble::tibble(b = runif(n = 1000, min = 0.1, max = 1.5),
s = runif(n = 1000, min = 0.001, max = 5))