Hello Everyone! Recently, I have discussed this issue on Slack as well, but I could not find a solution. I am trying to iterate through a dictionary elements and compare them to another dictionary and check if one of them or all or nothing exist in the second dictionary, and then count all the occ…

If you want to just count identical keys you can do something like this: module Main exposing (main) import Dict exposing (..) import Html exposing (..) firstDict = Dict.fromList [("and",1),("fox",1),("stars",1),("the",2)] otherDict = Dict.fromList [("and",1),("fox",1),("now",1),("way",2)] …

Hi @hanifhefaz , how about turning the keys into 2 sets and then finding the intersection? dict1 = Dict.fromList [("and",1),("fox",1),("stars",1),("the",2)] dict2 = Dict.fromList [("and",1),("fox",1),("now",1),("way",2)] sharedKeys = Set.intersect (dict1 |> Dict.keys …

[image] Atlewee: The most powerful way is to use Dict.merge function. With that one you know what is only in the first, only in the second, and in both… And you can return all you need from a single function… (like AllMatches, 2 matches, or nothing matches type of result) Like so: Dict.merge…

This is a great solution I think. and it works fine for my dictionaries, but I have forgotten to say, that I want to compare my first dictionary to multiple dictionaries, count identical keys, just as you did and then return the one with most occurrences. something like finding maximum count from th…

Hello Chris, That is a great idea too. but what if i compare the first dictionary to some multiple dictionaries? and then return the one as result, where the identical words are the most?

I found the simplest way to count matches in both Dicts is this: Dict.size ( Dict.intersect dict1 dict2 ) Finding the Dict with the most similar keys: You need to choose what happens if you give the function an empty list of dicts to check… Return the first Dict (checkingDict) Return Dict.empty R…

[image] hanifhefaz: Dict.fromList [(“and”,1),(“fox”,1),(“stars”,1),(“the”,2)] Good solutions above, so I don’t have one to add to that. What I wanted to say is that if you really are counting words and you are scanning large texts, it could make sense to use a Trie implementation instead of a…

Thank you for the suggestion. but I don’t have a large amount of data in my text, so I think dict will be fine.

How to compare dictionary elements to another dictionary?

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How to compare dictionary elements to another dictionary?

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