Zip 3列入Tuple-Erlang

Question

是否有按顺序压缩3个列表的优化方法

zip(L1, L2, L3)

这样它们就会产生一个元组列表,例如:

L1 = [1, 2, 3, 4],
L2 = [a, b, c, d],
L3 = [1, 2, 3, 4],

结果的结果列表应如下所示:

[{1,a,1}, {2,b,2}, {3,c,3}, {4,d,4}]

Answer 1

看来你需要使用list:zip3/3:

1> L1 = [1, 2, 3, 4].
[1,2,3,4]
2> L2 = [a, b, c, d].
[a,b,c,d]
3> L3 = [1, 2, 3, 4].
[1,2,3,4]
4> lists:zip3(L1, L2, L3).
[{1,a,1},{2,b,2},{3,c,3},{4,d,4}]

Answer 2

1> lists:zip3([1,2,3,4],[a,b,c,d],[1,2,3,4]).
[{1,a,1},{2,b,2},{3,c,3},{4,d,4}]
2>

或者您可以手动实现它:

myzip3([], _, _) ->
  [];
myzip3([X|Xs], [Y|Ys], [Z|Zs]) ->
  [{X,Y,Z}|myzip3(Xs,Ys,Zs)].

或使用累加器:

myzip3acc(Xs, Ys, Zs) ->
  myzip3acc_do([], Xs, Ys, Zs).

myzip3acc_do(Acc, [], _, _) ->
  lists:reverse(Acc);
myzip3acc_do(Acc, [X|Xs], [Y|Ys], [Z|Zs]) ->
  myzip3acc_do([{X,Y,Z}|Acc], Xs,Ys,Zs).