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Given a function with several arguments 

func[a,b,g1[x],g1[x,y],g2[1],g2[g1[1]],3]

I would like to extract all arguments with the head g1 and g2 as a list. So the output I am looking for is 

{g1[x],g1[x,y],g2[1],g2[g1[1]]}

The way I used to extract one head, say g1, is simply using the rule

/.func[l___,x__g1,r___]:> {x}

However, with two heads, this method does not work. I could write a module to do that but I wonder if there is a simpler way like the above rule for one head. Thank you so much!

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Either 

takeHeads = Cases[#, _g1 | _g2] &;
func[a, b, g1[x], g1[x, y], g2[1], g2[g1[1]], 3] // takeHeads

{g1[x], g1[x, y], g2[1], g2[g1[1]]}

or define func itself as

func = Cases[{##}, _g1 | _g2] &;
func[a, b, g1[x], g1[x, y], g2[1], g2[g1[1]], 3]

{g1[x], g1[x, y], g2[1], g2[g1[1]]}

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  • $\begingroup$ thank you, the method using cases is very useful. I did not know that Cases works inside the function func $\endgroup$
    – mastrok
    Apr 17 '19 at 16:34
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    $\begingroup$ You could also use the 1-arg form of Cases, e.g. takeHeads = Cases[_g1 | _g2]. $\endgroup$
    – Carl Woll
    Apr 17 '19 at 18:09
  • $\begingroup$ @mastrok If that surprised you that Cases worked on your custom func expression have a look at Everything is an Expression. Hope this helps to understand why this worked. $\endgroup$
    – Thies Heidecke
    Apr 17 '19 at 18:12
  • $\begingroup$ @Thies Heidecke Yes, I know that everything is an expression. However I thought that Cases only works with List only but not a general head. Thanks! $\endgroup$
    – mastrok
    Apr 17 '19 at 18:16
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Select[{##&@@func[a,b,g1[x],g1[x,y],g2[1],g2[g1[1]],3]},h=#;Head@#==h&]&/@{g1,g2}

{{g1[x], g1[x, y]}, {g2[1], g2[g1[1]]}}

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