#Related to Q1
P.= @. 1e-2*ones(I,J)*1; #Option 1.1, Don't work
P.= 1e-2.*ones(I,J).*1; #Option1.2, works
#Related to Q2
P.= @. 1e-2*ones(I,J)*1; #Option 2.1, Don't work
P.= @. 1e-2*$ones(I,J)*1; #Option2.2, works
Q-1 What is the exact difference between options 1.1 and 1.2 ? Why does one work and other don’t ?
Q-2 What does “$” exactly do so that option 2.2 gives the relevant output?
@. adds a dot to every function call in the expression. In the case of 1.1, that includes the function
ones, which isn’t what you want. We can write out what the
@. is producing by hand like this:
P .= 1e-2 .* ones.(I, J) .* 1 # doesn't work
This fails because it’s trying to assign the entire output of
I x J matrix) to each element of
P. Or, other words, it’s trying to do:
for i in eachindex(P)
P[i] = 1e-2 * ones(I, J) * 1
which doesn’t work.
$ syntax within
@. means “don’t add a dot to this particular function call”, so:
@. 1e-2 * $ones(I, J) * 1
is equivalent to:
1e-2 .* ones(I, J) .* 1
i.e. the same as Option 1.2.