Creating a third list from two given lists（通过以给出的两个列表创建第三个列表）–[5]

Creating a third list from two given lists（通过以给出的两个列表创建第三个列表）

问题

原文链接：https://mathematica.stackexchange.com/questions/155953/creating-a-third-list-from-two-given-lists

原文

I imported 2 lists of the form {{x1,y1}, {x2,y2}, ...} and {{x1,y'1}, {x2,y'2}, ...} into Mathematica. I want to make a ListPlot the list {{x1, y1 - y'1}, {x2, y2 - y'2}, ...}.

How can I create the third list by subtracting only the second column values of the first two lists keeping the first column intact.

翻译

我在mathematica输入了两个形式如下的列表， {{x1,y1}, {x2,y2}, ...} ， {{x1,y'1}, {x2,y'2}, ...}. 我想要画如下列表的散点图 {{x1, y1 - y'1}, {x2, y2 - y'2}, ...}.

我应该如何通过减前两个列表的第二列二保持他们第一列不变，来创建第三个列表

高票回答

原文

Another solution is:

p = {{x1, y1}, {x2, y2}}; q = {{x1, Y1}, {x2, Y2}};
p - ({0, 1} # & /@ q)
(* {{x1, y1 - Y1}, {x2, y2 - Y2}} *)

This works by simply replacing the first column of one of the lists with 0 before subtracting.

翻译

另一种解决办法如下：

p = {{x1, y1}, {x2, y2}}; q = {{x1, Y1}, {x2, Y2}};
p - ({0, 1} # & /@ q)
(* {{x1, y1 - Y1}, {x2, y2 - Y2}} *)

其工作原理是用0替换 之前两列数字中一个的第一列，然后再做减法。

还有一种方法如下，使用Associate

a1 = AssociationThread @@ Transpose[d1];
a2 = AssociationThread @@ Transpose[d2];
r = a1 - a2;

一些单词及句子

• intact：完整的，原封不动的
• duplicate：复制的，重复的