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Here's what I know how to do:

f[x_] := x^2 - 4 x + 4
Ordering[f /@ {1, 2, 3}]

This will evaluate f at each point and rank their sizes.

Here's what I really have:

g[x_,y_]:=(x-y)^2 - 4(x-y)+4

I'd like to do the same type of Ordering as above but with a fixed value of y. Currently, I'm doing it this way

yfixed=0;
f[x_]:=g[x,yfixed]
Ordering[f /@ {1, 2, 3}]

There's got to be a better way without creating the f function. Something like Ordering[g[_,0]/@{1,2,3}]. What's the technique here that I don't know?

share|improve this question
    
Ordering[Function[x, g[x, yfixed]] /@ {1, 2, 3}]? –  Guess who it is.♦ yesterday
    
@Guess: beat me to it. Unfortunately, I already posted the answer! (Without paying attention to the fact that you'd already commented, of course. Does that mean I should remove it? I'm not sure of the etiquette, here.) –  march yesterday
    
It's good that you wrote an answer, @march (I have upvoted it), but I believe OP will benefit greatly from you showing versions with and without Slot. :) –  Guess who it is.♦ yesterday
    
I'm new here, @Guess, so I'm unsure of etiquette in general. Thanks for the recommendation: I'll edit the post. –  march yesterday

1 Answer 1

up vote 3 down vote accepted

Updated to provide more details (per a suggestion in the comments). In addition, the implementation using Outer was wrong

A simple extension of what you've written will do the trick:

Ordering[Function[x,g[x,yFixed]] /@ {1,2,3}]

or more simply,

Ordering[g[#,yFixed]& /@ {1,2,3}]

where g[#,yFixed]& is essentially shorthand for Function[x,g[x,yFixed]].

The reason your original idea doesn't work is that Map (or /@) applies the left-hand side to each element of the right-hand in a literal fashion. For instance, the first element of the list is g[_,0][1]. In this case, g[_,0] gets evaluated first by plugging in _ for x and 0 for y. The resulting expression is then applied to 1, yielding (4 - 4 _ + _^2)[1], which is a very strange expression.

Alternatively, you could use Outer, although maybe this is overkill for this problem. On the other hand, it would allow you to do many values of y at once, by replacing the third argument of Outer with the list of y's and mapping Ordering over the result. For your example:

Ordering[Outer[g,{1,2,3},{yFixed}]]

Outer supplies all pairs of elements---the first from the first list, and the second from the second---as inputs to g and forms a list. If you have a list ys={0,4}, you could do

Ordering/@ Transpose[Outer[g,{1,2,3},ys]]
share|improve this answer
    
Thank you! Thank you! –  Aeryk yesterday

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