问题
This is my very first Erlang project. Basically, I have to write a function that takes a list of integers as a parameter & returns the number of integers from the list that are less than 1. What I have so far is a function that just returns how many integers in the list. I am not sure where/if I'm supposed to put a if statement and counter to only return how many integers are less than 1.
-export([num/1]).
num([]) -> 0 ;
num(L) -> num(L,0).
num([],Len) -> Len;
num([_|T],Len) ->
num(T,Len+1).
回答1:
Your code is almost there. Key skill to learn: guard
-export([num/1]).
num([]) -> 0;
num(NUMS) ->
num(NUMS, 0).
num([H|L], Count) when H < 1 -> %% use of guard
num(L, Count+1);
num([_|L], Count) ->
num(L, Count);
num([], Count) ->
Count.
回答2:
You can use length() to find the length of a list, and can use list comprehensions to filter your list.
num(L) -> length([X || X <- L, X < 1]).
Working example:
% list counter program
-module(listcounter).
-export([printnum/0, num/1]).
printnum() ->
L = [1,2,3,0,0],
io:fwrite("List size: ~p\n",[num(L)]).
num(L) ->
length([X || X <- L, X < 1]).
回答3:
This one avoid to build an intermediate list. roughly the same than Anthony proposal using erlang library and anonymous function.
lists:foldl(fun(X,Count) when X < 1 -> Count+1; (_,Count) -> Count end,0,L).
来源：https://stackoverflow.com/questions/26111962/erlang-how-to-return-count-number-of-elements