Confusing title, I know. But the question is, if we have two patterns which have the same general structure but different names used in the patterns and different names:
a = HoldPattern[f[x_, y_, g_, h_]] :> g[x] + h[y];
b = HoldPattern[g[y_, z_, m_, l_]] :> m[y] + l[z];
And I would like to be able to define a pattern for these two pattens, letting {f,x,y,g,h} take arbitrary values. How would I go about this?
To clearify. If I had a=4;b=5. I could define a common pattern through: _Integer and get MatchQ[a,_Integer](*=>True*) and MatchQ[b,_Integer](*=>True*).
But for my above two patterns, I cannot simply base my pattenr on a and substitute out {f,x,y,g,h} with _ eg:
badpattern = HoldPattern[f_[Pattern[x_,_], Pattern[y_,_],
Pattern[g_,_], Pattern[h_,_]]] :> h_[x_] + h_[y_];
I should not that what I want as a result is a working pattern not just a method that accomplishes this. Why? Well because the I would have to reimpliment MatchQ, Cases, Position and so forth for everything that expects a pattern as it's input to still work. The code below accomplishes this, however in an ad-hoc fasion.
This is wrong since the result does not distinguish between structural Blanks, and pattern blanks.
My initial code
Just to get a pattern to match to itself I need to get rid of HoldPattern which I do as:
MatchQ[a, a /. HoldPattern -> hp_ /; hp === HoldPattern]
(* True *)
Of cause I could just use Verbatim however then I won't be able to do the next part. Where I extend the same type of pattern of switching out pattern components such as HoldPattern.
To check b against a pattern based on a, I switch out a collection of heads inject new patterns and condition the pattern on the switched out heads:
myPatternPatternA=(a/.{HoldPattern->hp_,Pattern->p_,Blank->b_,RuleDelayed->rd_}
/.{f->f_,x-> x_,y-> y_,g-> g_,h-> h_ }
)/;And[hp===HoldPattern,p===Pattern,b===Blank,rd===RuleDelayed];
MatchQ[b,myPatternPatternA]
(* True *)
Note that I'm just using a as a template of the new pattern I construct, in the new pattern {f,x,y,g,h} can all take arbitrary values since I'm injecting a new pattern for them after removing {HoldPattern,Pattern,Blank,RuleDelayed}.
It seems however that I'll occasionally have problems with other symbols, which means I need to extend the list for instance to include Set and SetDelayed, however only when they actually appear in the expression, since otherwise the check fails. As such it feels like a rather cumbersome method. So I'm wondering if others have dealt with such cases and how they have carried this out.
Update
I should add that I'm relying on the matching to return values aswell, eg:
myPatternPattern = (a/.{HoldPattern->hp_,Pattern->p_,Blank->b_,RuleDelayed->rd_}
/.{f->f_,x-> x_,y-> y_,g-> g_,h-> h_ }
)/;And[hp===HoldPattern,p===Pattern,b===Blank,rd===RuleDelayed]:>f;
Cases[{a,b},myPatternPattern]
(*{f,g}*)
Mr. Wizard inquired as to what exactly I mean by the two definitions having the same structure. So I'll clarify. If we have two different expressions: f[3,4] and g[2,5] and I wanted to describe their structure, then I could write name_[_Integer,_Integer] as one possible common structure. When I have two patterns this approach breaks down for obvious reasons, Consider; _[varA_] and _[varB_]. I would like to define a "pattern^2" that matches both of these patterns and assigns varA or varB to a name, so I just follow the same recipe as before and substitute varA in the first with name_ and get: `_[_name_]. This obviously fails.
So what I have above is a way to define a pattern for patterns such that I can match elements inside them. Confusing, I know.
f =!= g. How can these be considered the same? If they are, what else is too? – Mr.Wizard Jan 16 '13 at 15:32fandgare never considered the same. When matching the patternf->f_replaces the symbolfwith a pattern that (maybe confusingly so) is calledf. You could equally well writef->x_and then havexmatchfandgin each respectively. – jVincent Jan 16 '13 at 15:36fandgare not patterns, they are literals. – Mr.Wizard Jan 16 '13 at 15:51Pattern,Blankand such. – jVincent Jan 16 '13 at 15:56fandgin the two patterns are supposed to be equivalent. – Mr.Wizard Jan 16 '13 at 15:59MatchQ[b, (a /. {HoldPattern -> hp_,...} /. {f -> f_,...}) /; And[hp === HoldPattern,...]]Now the "f-> f_" replaces the literalfwith a pattern. The method goes, 1: break down all patterns, 2: inject new patterns; 3: Verify that all the structure of the broken down patterns is still correct. – jVincent Jan 16 '13 at 16:14Cases, however I wouldn't call that transformations using the patterns. The question is really just about using pattern matching to match things that are themselves patterns. – jVincent Jan 16 '13 at 16:20letting {f,x,y,g,h} take arbitrary values, and attempted to answer the question ignoring this part even after some effort on my part to clarify this for him. Having one answer that misinterprets the question is hardly makes it vague. I have attempted to clarify by restructuring. – jVincent Jan 16 '13 at 21:23