CODEWITHSUNDEEP
listerlang
0xAX
0xAX

Reputation: 21817

Erlang list generation

I have 2 list:

["asd", "dsa"].

[[123, "asd"], [4534, "fgh"]].

How can i generate next list: I ned list that tail of each nested list =:= other element of 1 list.

In this example:

["asd", "dsa"].

[[123, "asd"], [4534, "fgh"]].

"asd" =:= "asd" ->

Output list:

[123, "asd"]

I try:

Here S = [[123, "asd"], [4534, "fgh"]]. D = ["asd", "dsa"].

List = lists:filter(fun(X) -> lists:last(X) =:= D end, S),

But D in this example list, and i need element of list.

How can do it?

Upvotes: 3

Views: 267

Answers (2)

rvirding
rvirding

Reputation: 20916

A slightly more direct way of writing it would be:

lists:filter(fun (X) -> lists:member(lists:last(X), D) end, S).

or with list comprehensions:

[ X || X <- S, lists:member(lists:last(X), D) ].

They are a little faster as they will not attempt to match against more elements in D if the element is found. Expanding D in the comprehension will do this.

Upvotes: 2

Yasir Arsanukayev
Yasir Arsanukayev

Reputation: 9676

Maybe something like:

1> [X || X<-[[1,2,4],[7,8,3],[2,5,4],[9,1,6]], Y<-[4,3], lists:last(X)=:=Y].    
[[1,2,4],[7,8,3],[2,5,4]]

Or, using your sample data:

2> [X || X<-[[123,"asd"], [4534,"fgh"]], Y<-["asd","dsa"], lists:last(X)=:=Y].
[[123,"asd"]]

Upvotes: 3

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