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Is there a function that replaces the first occurence of the expression instead of replacing all? For example, if I have HoldForm[x + 2 + 4 + x] /. x -> 4, is there a way to return 4 + 2 + 4 + x?

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    $\begingroup$ $f_x^x$ Which one is the first x ? $\endgroup$
    – Dr. belisarius
    Commented Sep 23, 2014 at 22:34
  • $\begingroup$ I'm not really sure, but I guess if you hit Ctrl+Shift+E, and the first occurence of x is what I meant $\endgroup$
    – Yituo
    Commented Sep 23, 2014 at 22:36
  • $\begingroup$ What I mean is that the order of the symbols in a general math expression doesn't mean anything "in general". Perhaps if you restrict the domain of the expressions ... $\endgroup$
    – Dr. belisarius
    Commented Sep 23, 2014 at 22:38
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    $\begingroup$ fun ClearAll[r]; r[4] := (r[4] = x; 4); HoldForm[x + 2 + 4 + x] /. x :> RuleCondition@r[4] and for less fun take a look at Position 4th arg + ReplacePart. $\endgroup$
    – Kuba
    Commented Sep 23, 2014 at 22:41

4 Answers 4

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Another way:

hf = HoldForm[x + 2 + 4 + x]
i = 0
hf /. (x :> 4 /; i++ == 0)

4 + 2 + 4 + x

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Well here's a way. Find the position of the first occurrence of x:

expr = HoldForm[x + 2 + 4 + x];

pos = Position[expr, x, -1, 1];

Then:

ReplacePart[expr, pos -> 4]

4 + 2 + 4 + x

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    $\begingroup$ +1 With version 10, we can use FirstPosition, e.g. HoldForm[x + 2 + 4 + x] // ReplacePart[#, FirstPosition[#, x] -> 4] &. $\endgroup$
    – WReach
    Commented Sep 24, 2014 at 5:22
  • $\begingroup$ @WReach I was just updating the answer with that optimization. I considered using FirstPosition but it doesn't seem to bring much benefit here (a bit of clarity I guess) so I used the general equivalent. $\endgroup$
    – Mr.Wizard
    Commented Sep 24, 2014 at 5:24
  • $\begingroup$ @Mr.Wizard, thanks for the update, certainly cleaner :) $\endgroup$
    – RunnyKine
    Commented Sep 24, 2014 at 5:38
  • $\begingroup$ @WReach, good point, I forgot about FirstPosition $\endgroup$
    – RunnyKine
    Commented Sep 24, 2014 at 5:39
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Edit

For order preserving as Jens says, I changed Attributes

ClearAttributes[Plus, Orderless]

HoldForm[7 + x + 2 + 4 + x + 5] /. f___ + x + l___ :> f + 4 + l

7 + 4 + 2 + 4 + x + 5

And you can revert by SetAttributes[Plus, Orderless]

Origin

How about this

HoldForm[x + 2 + 4 + x] /. x + a___ -> 4 + a

4 + 2 + 4 + x

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  • $\begingroup$ It won't work with this example: HoldForm[7 + x + 2 + 4 + x] /. x + a___ -> 4 + a because the order isn't preserved. $\endgroup$
    – Jens
    Commented Sep 24, 2014 at 2:25
  • $\begingroup$ @Jens : I got it, and I changed attributes. $\endgroup$
    – Junho Lee
    Commented Sep 24, 2014 at 4:37
  • $\begingroup$ Yes, that works. But now you'd have to do the same if it were Times instead of Plus... anyway, you put your finger exactly on the reason why the original didn't work, and changing the attributes is certainly a way to make it work. $\endgroup$
    – Jens
    Commented Sep 24, 2014 at 5:21
  • $\begingroup$ Related: (30152) $\endgroup$
    – Mr.Wizard
    Commented Sep 24, 2014 at 5:26
  • $\begingroup$ @Jens : I see. thank you. $\endgroup$
    – Junho Lee
    Commented Sep 24, 2014 at 5:31
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$\begingroup$ 
expr = HoldForm[x + 2 + 4 + x];
expr[[##& @@ FirstPosition[expr, x]]] = 4; expr

4 + 2 + 4 + x

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