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Johannes Nägele
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Oct 9, 2023
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using MacroTools | ||
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""" | ||
Some variable names are interpreted as operators. | ||
Thus, we decompose expressions like :(a in b) in individual Symbols. | ||
""" | ||
function remove_expr(x::Expr) | ||
if x.head == :call | ||
return [x.args[2], x.args[1], x.args[3]] | ||
else | ||
error("Can not handle $x") | ||
end | ||
end | ||
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""" | ||
We expect an array of variable names; some however form an Expr instead of being individual Symbols. | ||
""" | ||
function remove_expr(x::Array) | ||
# we need a deepcopy to avoid issues with recursive changes | ||
untangled = deepcopy(x) | ||
for i in eachindex(x)[end:begin] | ||
if typeof(untangled[i]) == Expr | ||
untangled = [ | ||
untangled[1:(i-1)]; | ||
remove_expr(untangled[i]); | ||
untangled[(i+1):end] | ||
] | ||
end | ||
end | ||
return untangled | ||
end | ||
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remove_blocks(e::Any) = e | ||
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""" | ||
Sometimes our equation input might have lines which are themselves blocks (with only a single line). | ||
This happens e.g. if we have two variables on the left hand side. | ||
In such a case we just want to convert this block into it's contents. | ||
""" | ||
function remove_blocks(expr::Expr) | ||
for (i, arg) in enumerate(expr.args) | ||
if typeof(arg) == Expr | ||
if arg.head == :block | ||
if (length(arg.args) == 1) | ||
expr.args[i] = remove_blocks(arg.args[1]) | ||
else | ||
error("Can not handle sub-blocks with multiple lines.") | ||
end | ||
else | ||
expr.args[i] = remove_blocks(arg) | ||
end | ||
end | ||
end | ||
return expr | ||
end | ||
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""" | ||
Find all symbols in an Expr aside from the math operators. | ||
""" | ||
function find_symbols(line::Expr) | ||
found::Set{Symbol} = Set([]) | ||
args = line.args | ||
for i in eachindex(args) | ||
if typeof(args[i]) == Symbol | ||
push!(found, args[i]) | ||
elseif typeof(args[i]) == Expr | ||
union!(found, find_symbols(args[i])) | ||
end | ||
end | ||
return setdiff(found, math_operators) | ||
end |