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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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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6$\begingroup$ +1 With version 10, we can use
FirstPosition
, e.g.HoldForm[x + 2 + 4 + x] // ReplacePart[#, FirstPosition[#, x] -> 4] &
. $\endgroup$ – WReach Sep 24 '14 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 Sep 24 '14 at 5:24 -
$\begingroup$ @Mr.Wizard, thanks for the update, certainly cleaner :) $\endgroup$ – RunnyKine Sep 24 '14 at 5:38
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$\begingroup$ @WReach, good point, I forgot about
FirstPosition
$\endgroup$ – RunnyKine Sep 24 '14 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 Sep 24 '14 at 2:25 -
$\begingroup$ @Jens : I got it, and I changed attributes. $\endgroup$ – Junho Lee Sep 24 '14 at 4:37
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$\begingroup$ Yes, that works. But now you'd have to do the same if it were
Times
instead ofPlus
... 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 Sep 24 '14 at 5:21 -
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expr = HoldForm[x + 2 + 4 + x];
expr[[##& @@ FirstPosition[expr, x]]] = 4; expr
4 + 2 + 4 + x
x
? $\endgroup$ – Dr. belisarius Sep 23 '14 at 22:34ClearAll[r]; r[4] := (r[4] = x; 4); HoldForm[x + 2 + 4 + x] /. x :> RuleCondition@r[4]
and for less fun take a look atPosition
4th arg +ReplacePart
. $\endgroup$ – Kuba♦ Sep 23 '14 at 22:41