I would like to write a function similar to Cases, but can search for a group of related patterns together. For example,

casesList[f[x] g[x] + f[y] + g[y]/f[z], {f[a_], g[a_]}]

{{f[x], g[x]}, {f[y], g[y]}}

By "related patterns", I mean that in the above example a_ matches the same x for both f and g at the first list, and matches y at the second list.

Note that here the input {f[a_], g[a_]} could be any other patterns. As another example,

caseList[D[f[x,y],x] + D[f[x,y],y] + D[g[x,y],x] + D[g[x,y],y],
    {D[f_[x_,y_],x_], D[f_[x_,y_],y_]}]

{{Derivative[1, 0][f][x, y], Derivative[0, 1][f][x, y]}, {Derivative[1, 0][g][x, y], Derivative[0, 1][g][x, y]}}

Is there a simple way to do this? Thanks!

EDIT: To further clarify the question, I'd like to compare the situation with Cases for a list. For example,

Cases[{{f[a], f[b]}, {f[c], f[c]}}, {f[a_], f[a_]}, Infinity]

{{f[c], f[c]}}

Cases[{{f[c], f[c]}}, {f[a_], f[b_]}, Infinity]

{{f[c], f[c]}}

In the above two examples, Cases does exactly what I want. However, more generally the expressions which match f[a_] does not necessarily stays in a list structure, but rather may be at elsewhere in the expression. This is the major difficulty I met.

  • $\begingroup$ Have you seen Alternatives[]? $\endgroup$ Feb 19, 2014 at 21:38
  • $\begingroup$ @belisarius : Thanks for the comment. I haven't been able to use Alternatives to do this. For example, Cases[f[x] g[x] + f[y] + g[y]/f[z], f[a_] | g[a_], Infinity] gives {f[y], f[x], g[x], f[z], g[y]}. This is not what I want, as I gave in the example output. The pattern a_ in f[a_] and g[a_] are treated separately here. However I want them to match for the same thing in each sublist. $\endgroup$
    – Yi Wang
    Feb 19, 2014 at 21:44
  • $\begingroup$ @YiWang You are right that Alternatives is not the direct solution to your problem. I still think that not everything is clear. E.g. the pattern list {f[a_],f[b_]} will only return something when there are two f in your expression with different argument, right? $\endgroup$
    – halirutan
    Feb 19, 2014 at 21:54
  • $\begingroup$ @YiWang Sorry, I misunderstood the question $\endgroup$ Feb 19, 2014 at 21:58
  • $\begingroup$ @halirutan : Thanks! I haven't specify this... The ideal return for pattern list {f[a_],f[b_]} is two f's, with either the same or different arguments. The situation can be compared with ReplaceAll. For example {f[a],f[a]}/.{f[a_],f[b_]}->0. Here a_ and b_ can stand for the same thing. (On the other hand, {f[a],f[b]}/.{f[a_],f[a_]}->0 will not trigger the replacement.) $\endgroup$
    – Yi Wang
    Feb 19, 2014 at 22:01

3 Answers 3


I'm not entirely sure whether this is right and works correctly, but the following could be an idea for a general rule-based approach. The idea is to use Cases to extract all matching expression separately for the given list of patterns. Let me illustrate this by your simple f example

expr = f[x] g[x] + f[y] + g[y]/f[z];

Cases[expr, #, Infinity, Heads -> True] & /@ {f[a_], g[a_]}
(* {{f[y], f[x], f[z]}, {g[x], g[y]}} *)

Now we have two result lists where all in the first list match f[a_] and all in the second list match g[a_]. Having this, the next step is kind of obvious: We need a replacement rule, where the a_ will match the same in both patterns. Given our result, this should be an easy rule of the following form

{{___,f[a_],___},{___,g[a_],___}} :> {f[a], g[a]}

The only tedious work is to build this rule from the input pattern list {f[a_], g[a_]}. Let's assume we have already build this, then we can use ReplaceList to get all possibilities

ReplaceList[{{f[y], f[x], f[z]}, {g[x], 
   g[y]}}, {{___, f[a_], ___}, {___, g[a_], ___}} :> {f[a], g[a]}]
(* {{f[y], g[y]}, {f[x], g[x]}} *)

Looks OK for me. With this in mind, we can write our CasesList combining all ideas. Here you see how the last replacement rule is built automatically

CasesList[expr_, pattern_List] := 
 With[{cases = Cases[expr, #, Infinity, Heads -> True] & /@ pattern,
   ruleLHS = {___, #, ___} & /@ pattern,
   ruleRHS = pattern /. Verbatim[Pattern][arg_, ___] :> arg
  ReplaceList[cases, ruleLHS :> ruleRHS]

Now, let's try this with your second example


{{Derivative[1, 0][f][x, y], Derivative[0, 1][f][x, y]}, {Derivative[1, 0][g][x, y], Derivative[0, 1][g][x, y]}}

  • $\begingroup$ Thanks a lot! I think it works, as far as I understand and tested. $\endgroup$
    – Yi Wang
    Feb 19, 2014 at 23:12

This question may be underspecified. My impression is that you want a Gather-type operation of subexpressions of a particular type, but this is complicated by your first example.

The FullForm of your first input expression is:

Plus[f[y], Times[f[x], g[x]], Times[Power[f[z], -1], g[y]]]

As you can see f[x], g[x] are adjacent, but f[y], g[y] are not. The first is easily found with:

Cases[f[x] g[x] + f[y] + g[y]/f[z], f[a_] g[a_], -1]
{f[x] g[x]}

Perhaps you want to consider all appearances of f[_] and g[_] and then find pairs. This could be done as follows:

Cases[f[x] g[x] + f[y] + g[y]/f[z], f[_] | g[_], -1]

GatherBy[%, First]
{f[y], f[x], g[x], f[z], g[y]}

{{f[y], g[y]}, {f[x], g[x]}, {f[z]}}

From there you could filter the results you want. (I'll wait for confirmation of what you want before attempting to implement this, as things like order must be considered.)

  • $\begingroup$ Thanks a lot for the answer! Indeed GatherBy is very useful here. @Kuba uses a similar method and essentially answers my question. I apologize that I haven't described my question clearly enough... $\endgroup$
    – Yi Wang
    Feb 19, 2014 at 22:22

Examples are good but I find hard to get what would be a general description for that function.

  Cases[f[x] g[x] + f[y] + g[y]/f[z], z : (f[x_] | g[x_]) :> {z, x}, ∞],
  Last][[;; , ;; , 1]]
, {_}]

The same idea works for second case

  Cases[D[f[x, y], x] + D[f[x, y], y] + D[g[x, y], x] + D[g[x, y], y]
        , z : (D[f_[x_, y_], x_] | D[f_[x_, y_], y_]) :> {z, f}, ∞],
 Last][[;; , ;; , 1]]
, {_}]
{{Derivative[0, 1][f][x, y], Derivative[1, 0][f][x, y]}, 
 {Derivative[0, 1][g][x, y], Derivative[1, 0][g][x, y]}}
  • $\begingroup$ Thanks a lot! This is exactly what I want (and sorry that I haven't been able to describe the question clearly). $\endgroup$
    – Yi Wang
    Feb 19, 2014 at 22:19
  • $\begingroup$ A minor comment: In the second case, I would need something like z : (D[f_[x_, y_], x_] | D[f_[x_, y_], y_]) :> {z, {f, x, y}}. This is easy to fix and I can write a general function myself following your idea. $\endgroup$
    – Yi Wang
    Feb 19, 2014 at 22:29
  • $\begingroup$ @YiWang No need to be sorry. You've stated it rather clearly, my point was that I'm not sure how to generalize the problem, e.g. what possible arguments such function might need. I'm glad you like it :) $\endgroup$
    – Kuba
    Feb 19, 2014 at 22:33

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