Suppose I have an expression like so:

$\qquad {\rm expr}=2\,i\,{\rm Gamma}(4)\,\pi\, 9.93 f(x)\,g(y) \frac{h(x, 4y)^3}{11 q}$

What is best (most succinct) way to extract out only the product of numbers?

The answer I expect is:

$\qquad 2\,i\,{\rm Gamma}(4)\,\pi\,9.93\,\frac{1}{11}$

I would want this the code to work for as generic an expression as possible in the sense that it should pick all the numbers(Real or Complex or Transcendental or Special Constants).


(*0. + 10.8327 I*)

almost seems to work but not quite as it performs the product and simplifies. Any way to HoldForm here?

  • 2
    $\begingroup$ Look up FactorTerms[]. $\endgroup$ – J. M.'s discontentment Mar 11 '18 at 18:09
  • $\begingroup$ @J.M. Any suggestions following the edit? Wasn't able to make much out of FactorTerms[] $\endgroup$ – Subho Mar 13 '18 at 19:05
  • $\begingroup$ Well, did you look at what expr looked like after storing that expression? Why were you expecting that the arithmetic won't be performed? $\endgroup$ – J. M.'s discontentment Mar 13 '18 at 19:10
  • $\begingroup$ @J.M. Sure, the expression has been evaluated in the assignment itself. I wanted to find a way to extract out all the numeric factors in the form they appear in the expression. Is there a way to do that? So, if my input expression was the one I wrote as in the example, the output should have the unevaluated form of the numeric subexpression. $\endgroup$ – Subho Mar 13 '18 at 19:42


Following function is more efficient and it captures terms in the denominator:



In[364]:= extractCoefficient[2I Gamma[4] Pi 9.93 f[x] g[y] h[x,4 y]^3/(11q)]
Out[364]= 2*I*Gamma[4]*\[Pi]*9.93*1/11


Following function, albeit ugly, does the job.

    Function[input,  HoldForm@@((Times@@Select[Thread[Replace[Hold[input],Hold[Times[a_,b__]]:>Hold[{a,b}]]],(NumericQ@@#&)])
    //.Hold[a_]Hold[b_]:> Hold[a b])

Your case:

In[152]:= extractCoefficient[2I Gamma[4] Pi 9.93 f[x] g[y] h[x,4 y]^3/(11q)]

Out[152]= (((2 9.93) I) \[Pi]) Gamma[4]
| improve this answer | |
  • $\begingroup$ It misses the $1/11$ factor $\endgroup$ – Subho Mar 14 '18 at 8:33
  • $\begingroup$ Corrected it now. $\endgroup$ – Soner Mar 14 '18 at 8:47

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