using rbind to create dataframe is not working - r

Question

I am trying to write a script to get some specific values for the equation 25a+20b=1600 with a in the range between 24:60 and b in 20:50
I need to get the pairs of a and b satisfying the equation.
My first problem was how to define a and b with a single digit decimal place (a=24.0,24.1,24.2...etc.) but I overcame that defining a<-c(240:600)/10, so my first question is: Is there any direct method to do that?
Now, I did a couple of nested loops and I am able to get each time the equation is satisfied in a vector, I want to use rbind() to attach this vector to a matrix or a dataframe but it is not working without any error or warning. it just takes the value of the first vector and that's it !
Here is my code, can someone help me define where the problem is?
solve_ms <- function() {
index<-1
sol<-data.frame()
temp<-vector("numeric")
a<-c(240:600)/10
b<-c(200:500)/10
for (i in 1:length(a)){
for (j in 1:length(b)) {
c <- 25*a[i]+20*b[j]
if(c == 1600) {
temp<-c(a[i], b[j])
if(index == 1) {
sol<-temp
index<-0
}
else rbind(sol,temp)
}
}
}
return(sol)
}
I found our where my code problem is, it is using rbind without assigning its return to a dataframe. I had to do {sol<-rbind(sol,temp)} and it will work.
I will check other suggestions as well.. thanks.

Answer 1

Try this instead:
#define a function
fun <- function(a,b) (25*a+20*b) == 1600
Since floating point precision could be an issue:
#alternative function
fun <- function(a,b,tol=.Machine$double.eps ^ 0.5) abs(25*a+20*b-1600) < tol
#create all possible combinations
paras <- expand.grid(a=c(240:600)/10, b=20:50)
paras[fun(paras$a,paras$b),]
a b
241 48.0 20
594 47.2 21
947 46.4 22
1300 45.6 23
1653 44.8 24
2006 44.0 25
2359 43.2 26
2712 42.4 27
3065 41.6 28
3418 40.8 29
3771 40.0 30
4124 39.2 31
4477 38.4 32
4830 37.6 33
5183 36.8 34
5536 36.0 35
5889 35.2 36
6242 34.4 37
6595 33.6 38
6948 32.8 39
7301 32.0 40
7654 31.2 41
8007 30.4 42
8360 29.6 43
8713 28.8 44
9066 28.0 45
9419 27.2 46
9772 26.4 47
10125 25.6 48
10478 24.8 49
10831 24.0 50

Answer 2

If the problem is really this simple i.e. solving for roots of 2 variable linear equation, you can always rearrange the equation to write b in terms of a i.e. b = (1600-25*a)/20 and get all the values of b for corresponding values of a and filter the combinations by b
e.g.
a = c(240:600)/10
b = 20:50
RESULTS <- data.frame(a, b = (1600 - 25 * a)/20)[((1600 - 25 * a)/20) %in% b, ]
RESULTS
## a b
## 1 24.0 50
## 9 24.8 49
## 17 25.6 48
## 25 26.4 47
## 33 27.2 46
## 41 28.0 45
## 49 28.8 44
## 57 29.6 43
## 65 30.4 42
## 73 31.2 41
## 81 32.0 40
## 97 33.6 38
## 105 34.4 37
## 121 36.0 35
## 137 37.6 33
## 145 38.4 32
## 161 40.0 30
## 177 41.6 28
## 185 42.4 27
## 193 43.2 26
## 201 44.0 25
## 209 44.8 24
## 217 45.6 23
## 225 46.4 22
## 233 47.2 21
## 241 48.0 20