Replace symbols with values without evaluation

Question

If I have assigned values to a, b and c:

a=7;
b=5;
c=6;

and I select the expression:

(a+b)/c

and I press cmd + enter in the Mathematica frontend, the expression will be replaced by 2.

I would like to do something similar which would fill in the values of the symbols without evaluating the result. So (a+b)/c would be replaced by (7+5)/6.

Is there a way of doing this? If so, can you also assign this to a keyboard shortcut that would replace the selected expression?

Note: Assume you don't know the values or the names of the variables. Inputting (a+b)/c to a function, should be able to produce something like (7+5)/6 without having to input a, b and c or 7, 5 and 6.

Answer 1

Using OwnValues and HoldForm:

{a = 7, b = 5, c = 6};

HoldForm[(a + b)/c] /. OwnValues@a /. OwnValues@b /. OwnValues@c

(7 + 5)/6

With[{a = a, b = b, c = c}, HoldForm[(a + b)/c]]

(7 + 5)/6

Assuming that variables are already defined and you don't want to bother with listing the symbols, and you want to have a function that does it in one go:

Attributes[hold] = {HoldAll};
hold[x_] := HoldForm@x /. Cases[Hold@x, s_Symbol :> (HoldPattern@s -> s), Infinity];

hold[(a + b)/c]

(7 + 5)/6

Answer 2

Hold[(a + b)/c] /. {a -> 1, b -> 3, c -> 4} /. Hold -> Defer

(*   (1 + 3)/4   *)

Hold[(a + b)/c] /. {a -> 1, b -> -3, c -> 4} /. Hold -> Defer

(*   (1 - 3)/4   *)

    Hold[(a + b)/c] /. {a -> 1, b -> -3, c -> -4} /. Hold -> Defer

(*   -(1/4) (1 - 3)   *)

Answer 3

Edit: answer rewrite

This question in its base form is a duplicate, but since I cannot find the original, and since you extended the question to something more unique than what I recall, I shall provide a short answer.

You asked:

I know this a probably even harder, but would it be possible to output (7+5)/(-6) instead of -(1/6)*(7+5) when c=-6?

For that kind of control see for example: Returning an unevaluated expression with values substituted in

Combining that with RuleCondition, described in Replacement inside held expression, we can use:

SetAttributes[{defer, fill}, HoldAll]

MakeBoxes[defer[args__], fmt_] := Block[{Times}, MakeBoxes[Defer[args], fmt]]

fill[expr_] := defer[expr] /. x_Symbol :> RuleCondition[x]

Now:

{a, b, c} = {7, 5, -6};

fill[(a + b)/c]

(7+5)*-(1/6)

Because this uses Defer you can use the output of fill as input and it will fully evaluate.

Answer 4

In the same search that found your post, I also found: https://mathematica.stackexchange.com/a/100152/76328

I tailored Edmund's approach just slightly to fit my needs. I hadn't heard of Inactivate before, but it performs substitutions while not evaluating the main expression. It sounds like a simple solution to this post?

That said, I do like István Zachar's general-case approach as well.

Tailored version noted above:

InactiveAndActive = Function[expr, Column@Through@{Inactivate, Identity}@Unevaluated@expr, HoldAll];

EDIT: I just noticed that the above doesn't format things like 1/x in StandardForm. A better solution is:

HoldAndEvaluate = Function[expr, Column@Through@{Inactivate, Identity}@Unevaluated@expr /. Inactive[x_] -> HoldForm@*x, HoldAll];

Answer 5

Mathematica has the function Hold which prevents evaluation, until you Release that Hold.

Hold[(a+b)/c] /.{a->1,b->2,c->3}

gives

Hold[(1 + 2)/3]

Then

Release[Hold[(1 + 2)/3]]

will give

1