CODEWITHSUNDEEP
dictionaryif-statementjulia
Jet.B.Pope
Jet.B.Pope

Reputation: 760

How to find if item is contained in Dict in Julia

I'm fairly new to Julia and am trying to figure out how to check if the given expression is contained in a Dict I've created.

function parse( expr::Array{Any} )
   if expr[1] == #check here if "expr[1]" is in "owl"
       return BinopNode(owl[expr[1]], parse( expr[2] ), parse( expr[3] ) )
   end
end

owl = Dict(:+ => +, :- => -, :* => *, :/ => /)

I've looked at Julia's documentation and other resources, but can't find any answer to this.

"owl" is the name of my dictionary that I'm trying to check. I want to run the return statement should expr[1] be either "+,-,* or /".

Upvotes: 3

Views: 2551

Answers (2)

Bogumił Kamiński
Bogumił Kamiński

Reputation: 69839

A standard approach to check if some dictionary contains some key would be:

:+ in keys(owl)

or

haskey(owl, :+)

Your solution depends on the fact that you are sure that 0 is not one of the values in the dictionary, which might not be true in general. However, if you want to use such an approach (it is useful when you do not want to perform a lookup in the dictionary twice: once to check if it contains some key, and second time to get the value assigned to the key if it exists) then normally you would use nothing as a sentinel and then perform the check get_return_value !== nothing (note two = here - they are important for the compiler to generate an efficient code). So your code would look like this:

function myparse(expr::Array{Any}, owl) # better pass `owl` as a parameter to the function
   v = get(expr[1], owl, nothing)
   if v !== nothing
       return BinopNode(v, myparse(expr[2]), myparse(expr[3]))
   end
   # and what do we do if v === nothing?
end

Note that I use myparse name, as parse is a function defined in Base, so we do not want to have a name clash. Finally your myparse is recursive so you should define a second method to this function handling the case when expr is not an Array{Any}.

Upvotes: 7

Jet.B.Pope
Jet.B.Pope

Reputation: 760

I feel like an idiot for finding this so fast, but I came up with the following solution: (Willing to hear more efficient answers however)

yes = 1
yes = get(owl,expr[1],0)
if yes != 0
   #do return statement here

"yes" should get set equal to 0 if the expression is not found in the dictionary "owl". So a simple != if statement to see if it's zero fixes my problem.

Upvotes: -1

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