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Suppose I have an expression of the form

a b c d + x y z

The FullForm of this is

Plus[Times[a,b,c,d], Times[x,y,z]]

Now suppose I want to set all expressions containing the product b*c to zero. The naive way of doing this is

a b c d + x y z /. Times[___, b, c, ___] -> 0

However, this doesn't work - nothing is replaced.

(Background info: I'm expanding some term using Series, and would like to cancel out all products dx*dy, so that only first-order terms remain.)

Even more confusing: if I don't use Times but a generic function f it works - even when I assign all the attributes of Times to f!

f;
SetAttributes[f, #]& /@ Attributes[Times];
Attributes[f]
f[a, b, c, d] /. f[___, b, c, ___] -> 0

==>
{Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected}
0

(If you evaluate this use a new kernel, as f will be Protected.)

So: how do I replace expressions containing some product*?

*: Assuming the order of the arguments is known. Don't worry about the fact that the expression for b might be on the left of the one for a; solving this would be another question.

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  • $\begingroup$ Feel free to roll back the question title... $\endgroup$
    – Yves Klett
    Dec 3, 2012 at 11:08

2 Answers 2

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Ok, here is the why: patterns do evaluate, just like other expressions. Therefore:

Times[___, b, c, ___]

gives

(*  b c ___^2  *)

even before the pattern-matching is attempted. In this form, the pattern of course does not match. Use HoldPattern to prevent that:

a b c d+x y z/.HoldPattern[Times[___,b,c,___]]->0

(* x y z *)
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  • $\begingroup$ @YvesKlett Yes, because the two named underscores are syntactically different. So, the pattern does evaluate, but no non-trivial transformation (like raising to a power) can be applied,thus it remains valid for the replacement. $\endgroup$
    – Leonid Shifrin
    Dec 3, 2012 at 11:17
  • $\begingroup$ just saw the ___^2 part and deleted the comment - reading helps! $\endgroup$
    – Yves Klett
    Dec 3, 2012 at 11:18
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Is this ok ?

expr = a b c d + x y z;
Replace[expr, Times[before___, b, c, after___] -> zero, Infinity] 
(* x y z + zero *)

The main point is that the rest of the factors should be named; this works too :

a b c d + x y z /. Times[before___, b, c, after___] -> 0
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  • $\begingroup$ Can you explain why the named pattern works? $\endgroup$
    – Yves Klett
    Dec 3, 2012 at 10:58
  • $\begingroup$ No, never had this issue, I always put names. I hope the experts will have an explanation. $\endgroup$
    – b.gates.you.know.what
    Dec 3, 2012 at 11:03

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