List 将新列从一个表组合到另一个表的函数

我需要一些关于两个问题的帮助： 如何实现两个表的简单水平并集？ 像这样：

table2table :: Table -> Table -> Table

table1 = [["Zoe", "2", "1"], 
         ["Mark", "2", "5"]]

table2 = [["Zbir", "4", "7"], 
         ["Nels", "1", "3"]]

table2table = [["Zoe", "2", "1"], ["Zbir", "4", "7"], 
              ["Mark", "2", "5"], ["Nels", "1", "3"]]

似乎您想将两个列表混合在一起-一个简单的解决方案可能是

mix:：[a]->[a]->[a]
混合xs[]=xs
混合[]ys=ys
混合（x:xs）（y:ys）=x:y:mix-xs-ys
表2表格：：表格->表格->表格
表2表格=混合

---

注意这并没有真正考虑到“表”式表示，但实际上您只有一个列表列表，Haskell并不真正关心如何在编辑器中将该列表的部分写进行中。

对于执行这种组合的一般方法，您可以使用

concat。转置

：

> concat . transpose $ [table1, table2]
[["Zoe","2","1"],["Zbir","4","7"],["Mark","2","5"],["Nels","1","3"]]

如果您需要，它适用于两个以上的列表：

> concat . transpose $ [table1, table2, table1]
[["Zoe","2","1"],["Zbir","4","7"],["Zoe","2","1"],
 ["Mark","2","5"],["Nels","1","3"],["Mark","2","5"]]   -- manually formatted

transpose

将它们“按列”分组，

concat

将列表展平

混合列表的另一种方法是按照以下方式进行：

对于两个列表，结果与

concat相同。转置

，但列表越多，它们就不同了

> diagonal [table1, table2, table1]
[["Zoe","2","1"],["Zbir","4","7"],["Mark","2","5"],
 ["Zoe","2","1"],["Nels","1","3"],["Mark","2","5"]]

---

有一个可爱的技巧可以减少案例数量：在每次递归调用时交换参数<代码>混合[]ys=ys；混合（x:xs）ys=x:mix ys xs

> diagonal [table1, table2, table1]
[["Zoe","2","1"],["Zbir","4","7"],["Mark","2","5"],
 ["Zoe","2","1"],["Nels","1","3"],["Mark","2","5"]]