Compute distance in Cartesian Coordinate System in

问题:

Analyzing Eye-movements on a screen, I set my origin to the bottom left corner of it (Hard to modify at that point).

Trying to compute distance between some points and the center of the screen I use the simple formula displayed below. Problem is that using this in conditional statement, it gets ugly.

Sqrt[
(
(fixationX - centerX)^2 + (fixationY - centerY)^2
)
]

Is there a way to customize Norm to compute distance between points and not between a point and the origin ?

Or in my case, set the origin to be at the "center" of the current coordinate system ?

回答1:

A slight variation of Simon's method is to use a default value in the function, rather than a global variable ($Center).

Suppose your default origin is (5, 5), then:

myNorm[pos:{_, _}, center_:{5, 5}] := EuclideanDistance[pos, center]

Notice the use of _:{5, 5} to define the default value.

Now you can do:

myNorm[{5, 7}]

(* Out[]= 2 *)

Or temporarily use a different the center with:

myNorm[{5, 7}, {8, 8}]

(* Out[]= Sqrt[10] *)

For this simple function, you could use EuclideanDistance in the second case instead, but I hope you can see the utility of this method were the definition of myNorm more complex.

The downside to this method is that you cannot easily change the default center.

---

Another variation that does allow one to easily change the default center, but is more verbose, is to use Options:

Options[myNorm2] = {Center -> {5, 5}};

myNorm2[pos : {_, _}, OptionsPattern[]] := 
 EuclideanDistance[pos, OptionValue[Center]]

Syntax is:

myNorm2[{5, 7}]

myNorm2[{5, 7}, Center -> {8, 8}]

   2

   Sqrt[10]

Changing the default center:

SetOptions[myNorm2, Center -> {8, 8}];

myNorm2[{5, 7}]

   Sqrt[10]

回答2:

Can you just use EuclideanDistance

In[1]:= EuclideanDistance[{x,y}, {cx,cy}]
Out[1]= Sqrt[Abs[-cx +x ]^2 + Abs[-cy + y]^2]

Or define a $Center and a new CNorm, e.g.

$Center = {cx, cy};
CNorm[pos:{x_, y_}] := EuclideanDistance[pos, $Center]