I’m working on a dataset that will solve simple equation systems of 2 equations with 2 unknowns. However, as there are a hundreds of such systems, I will need to find some algorithm, perhaps a loop, that will perform the task for n rows in a data frame.
Consider the following data frame:
df <- as.data.frame(rbind(c(1, 0.214527122, 166486.7605, 1, -0.003217907, 0),
c(1, 0.227828903, 160088.1385, 1, -0.003417434, 0),
c(1, 0.214324606, 154445.4132, 1, -0.003214869, 0),
c(1, 0.218698883, 147921.278, 1, -0.003280483, 0),
c(1, 0.201444364, 151268.6711, 1, -0.003021665, 0)))
colnames(df) = c("y1", "x1", "b1", "y2", "x2", "b2")
Notice how each row in this data set contains two linear equations, for two unknowns, x and y.
> df
y1 x1 b1 y2 x2 b2
1 1 0.2145271 166486.8 1 -0.003217907 0
2 1 0.2278289 160088.1 1 -0.003417434 0
3 1 0.2143246 154445.4 1 -0.003214869 0
4 1 0.2186989 147921.3 1 -0.003280483 0
5 1 0.2014444 151268.7 1 -0.003021665 0
The two equations represent supply and demand curves in an economic model. I want to calculate the resulting equilibrium price and quantity for each row in the data frame.
I can, for example, solve the first row with the following code:
A <- rbind(c(df$y1[1], df$x1[1]),
c(df$y2[1], df$x2[1]))
B <- c(df$b1[1],df$b2[1])
> solve(A,B)
[1] 2460.396 764595.000
At equilibrium, price is 2460.4 and quantity is 764 595. I can then add this result to the data frame in two new columns:
df$Price[1] <- solve(A,B)[1]
df$Quantity[1] <- solve(A,B)[2]
> df
y1 x1 b1 y2 x2 b2 Price Quantity
1 1 0.2145271 166486.8 1 -0.003217907 0 2460.396 764595
2 1 0.2278289 160088.1 1 -0.003417434 0 NA NA
3 1 0.2143246 154445.4 1 -0.003214869 0 NA NA
4 1 0.2186989 147921.3 1 -0.003280483 0 NA NA
5 1 0.2014444 151268.7 1 -0.003021665 0 NA NA
That was the first row. But I haven’t figured out a good way of automating this to solve for each row in the data frame. I suppose some type of loop might be an option, but I’ve had no luck in solving it.
>Solution :
Rearrange the columns so that the first 4 are the matrix and then run the following apply :
df2 <- df[c("y1", "y2", "x1", "x2", "b1", "b2")]
t(apply(df2, 1, function(x) solve(matrix(x[1:4], 2), x[5:6])))
giving
[,1] [,2]
[1,] 2460.396 764595
[2,] 2365.835 692284
[3,] 2282.444 709965
[4,] 2186.029 666374
[5,] 2235.497 739823