关于对两个list的每个元素做减法操作的讨论。

 

> Hi All, 

> after a long time, here I am programming in Erlang again... 

> I have 2 lists: 

> L1=[1,2,3,4,5], L2=[1.1,2.2,3.3,4.4,5.5] 

> and I want to have a third list which is the difference of them: 

> L3=[ 

> L2[1] - L1[1], 

> L2[2] - L1[2], 

> L2[3] - L1[3], 

> L2[4] - L1[4], 

> L2[5] - L1[5] 

> ] 

> The result should be 

> L3=[0.1,0.2,0.3,0.4,0.5] 

> 

> I have tried: 

> 

> sub(L2,L1) -> [X2-X1||X1<-L1,X2<-L2].

 

This says 

    for each X1 in L1 { 

        for each X2 in L2 { 

            include X1 - X2 in the result 

        } 

    } 

just like it does in Haskell and Clean. 

 

Direct method: 

   sub([X|Xs], [Y|Ys]) -> [X-Y | sub(Xs, Ys)]; 

   sub([],     [])     -> []. 

 

Zip then enumerate pairs: 

   sub(Xs, Ys) -> [X-Y || {X,Y} <- lists:zip(Xs, Ys)]. 

 

Zip with a function other than pairing: 

   sub(Xs, Ys) -> lists:zipwith(fun (X, Y) -> X - Y end, Xs, Ys). 

 

One expects a Haskell compiler to expand calls to zip. 

One expects an Erlang compiler NOT to, and it doesn't. 

There's an EEP proposing that you should be allowed to write 

 ** sub(Xs, Ys) -> [X-Y || X <- Xs && Y <- Ys] 

but that's not allowed in Erlang yet. 

以上内容来自 erlang_quest mail

 

我看到上面提用 直接方法和lists:zip/2 及lists:zipwith/3 三种方法解决此问题。

带着疑问看来了一下源码，原来这三种方法是的思路是一样的。

 

 list:zip/2的源码 如下

zip([X | Xs], [Y | Ys]) -> [{X, Y} | zip(Xs, Ys)];

zip([], []) -> [].

 

zipwith 的如下：

zipwith(F, [X | Xs], [Y | Ys]) -> [F(X, Y) | zipwith(F, Xs, Ys)];

zipwith(F, [], []) when is_function(F, 2) -> [].