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I'd like to replace fractions with Missing[] as the denominator with Missing[]. This pattern in ReplaceAll does not do the job:

ReplaceAll[{1/Missing[], 2}, (_/Missing[] :> Missing[])]

But the exact match works fine:

ReplaceAll[{1/Missing[], 2}, (1/Missing[] :> Missing[])]

Why is this?

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  • $\begingroup$ In the interest of helping you help yourself: look at FullForm[{1/Missing[], _/Missing[]}]. It should be quickly apparent why you're not getting what you want. $\endgroup$ – J. M. is away Oct 22 '15 at 23:34
  • $\begingroup$ ...and since you're matching denominators only: you know there's a Denominator[] function, yes? $\endgroup$ – J. M. is away Oct 22 '15 at 23:36
  • $\begingroup$ As an extension to the comments above , {a/Missing[ ],2}/. _/Missing[]:>Missing[ ] will work for any a .... except 1 :) $\endgroup$ – Dr. belisarius Oct 22 '15 at 23:40
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By using the optional value in the pattern, there is no special case:

ReplaceAll[{1/Missing[], 2/Missing[], 3}, _./Missing[] :> Missing[]]

(* {Missing[], Missing[], 3} *)
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1/Missing[] is a special case, equivalent to Missing[]^-1. This is why the original pattern matches 2/Missing[]. I guess the best way is to handle it separately:

ReplaceAll[{1/Missing[], 2/Missing[], 3}, {_/Missing[] :> Missing[], 1/Missing[] :> Missing[]}]

(Thanks to J.M. and belisarius for the hints.)

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  • $\begingroup$ Actually, I'd have done {1/Missing[], 2/Missing[], 3} /. x_ /; Denominator[x] === Missing[] :> Missing[]. ;) $\endgroup$ – J. M. is away Oct 23 '15 at 0:07

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