String manipulation of type String substitution in mathematical expression

Question

Imagine something like

exp(49/200)+(x-49/200)

I want to pass as argument of the function "roundn" whatever operation that is not a addtion or a subtraction So my expression became

roundn(exp(roundn(49/200, n)), n) + (x - roundn(49/200, n)

Well the expression I want to manipulate is this:

exp(49/200)+exp(49/200)*(x-49/200)+1/2*exp(49/200)*(x-49/200)^2+1/6*exp(49/200)*(x-49/200)^3+1/24*exp(49/200)*(x-49/200)^4+1/120*exp(49/200)*(x-49/200)^5+1/720*exp(49/200)*(x-49/200)^6+1/5040*exp(49/200)*(x-49/200)^7+1/40320*exp(49/200)*(x-49/200)^8+1/362880*exp(49/200)*(x-49/200)^9+1/3628800*exp(49/200)*(x-49/200)^10+1/39916800*exp(49/200)*(x-49/200)^11

I think you mean function `round()` described in the docs [here](http://docs.python.org/library/functions.html?highlight=round#round). — martineau, Nov 23 '10 at 11:38
Seems like you could just write your own functions for multiply, divide, exp, pow, etc and rewrite you expression using them. If it's something you'll do to a lot of different expressions then it might be worthwhile to automate the conversion with an expression parser. The Python `ast` (Abstract Syntax Trees) module should be helpful. — martineau, Nov 23 '10 at 14:17
As far as I can tell, your main expression is approximately `exp(x)`. Roughly, the first 12 terms of the [power series expansion](http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function). — Gary Kerr, Nov 23 '10 at 15:05

PaulMcG · Answer 1 · 2010-11-23T11:27:44.253

Maybe you think you want to do this, but you don't really want to do this. New Pythoners usually think they need to round floating point numbers because when evaluated they get unexpected results (like 1.0/10 = 0.100000000000001). Rather than do some goofy string substitution on your expression, I just created a variable for round(49/200,n), and did a little format cleanup. Also exp(49/200) does not need to be evaluated 13 times, just do it once and refer to the computed value.

zz = round(49/200,n)
e_zz = exp(zz)
ans = (e_zz + 
    e_zz * (x-zz) +
    1/2 * e_zz * (x-zz)**2 +
    1/6 * e_zz * (x-zz)**3 +
    1/24 * e_zz * (x-zz)**4 +
    1/120 * e_zz * (x-zz)**5 +
    1/720 * e_zz * (x-zz)**6 +
    1/5040 * e_zz * (x-zz)**7 +
    1/40320 * e_zz * (x-zz)**8 +
    1/362880 * e_zz * (x-zz)**9 +
    1/3628800 * e_zz * (x-zz)**10 +
    1/39916800 * e_zz * (x-zz)**11)

Raising e to a rounded number is almost never appropriate. Likewise for raising a rounded number to the 11'th power. (Note also that in Python, the exponentiation operator is **, not ^.)

Edited: If S.Lott hadn't suggested the algebraic simplification, I would have left this as-is. But the * e_zz can be factored out of every term, giving the simpler (and probably faster):

zz = round(49/200,n)
e_zz = exp(zz)
ans = e_zz * (1 + 
    (x-zz) +
    1/2 * (x-zz)**2 +
    1/6 * (x-zz)**3 +
    1/24 * (x-zz)**4 +
    1/120 * (x-zz)**5 +
    1/720 * (x-zz)**6 +
    1/5040 * (x-zz)**7 +
    1/40320 * (x-zz)**8 +
    1/362880 * (x-zz)**9 +
    1/3628800 * (x-zz)**10 +
    1/39916800 * (x-zz)**11)

Paul, I think the term for what they wanted to do is "premature rounding". ;-) — martineau, Nov 23 '10 at 11:27
Seems like the `x-zz` term also doesn't need to be calculated umpteen times either... — martineau, Nov 23 '10 at 11:50
I like your answer but the center of my question is not related with the background but also the task I´m asking for.I think is normal to have string inquietudes. — Peterstone, Nov 23 '10 at 12:24

score 2 · Answer 2 · answered Nov 30 '10 at 00:39

2

Use this

http://sympy.org/

answered Nov 30 '10 at 00:39

S.Lott

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score 1 · Answer 3 · answered Nov 23 '10 at 14:59

1

I wonder whether this is what you need:

If you original equation is in the string variable eq you can create your new equations using the replace method of strings:

eq.replace('49/200', 'roundn(49/200,n)')

and a similar expression could put roundn around the exp() function (possibly need some nifty regex here).

answered Nov 23 '10 at 14:59

Ber

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eq.replace will not do inplace modification (strings being immutable and all). Change to `eq = eq.replace(etc.)` – PaulMcG Nov 30 '10 at 12:25
Maybe that's not what the OP wants. The expression in my answer creates a new string containing the modified equation. No need to change that. – Ber Nov 30 '10 at 16:14

score 0 · Answer 4 · answered Nov 23 '10 at 09:57

You could sub every match of say, p=re.compile(r'\d+/\d+') with the output of your roundn function. FYI, approximating so many terms in you expression will yield a result that is not necessarily very close to the actual result, depending on how many digits you round to.

String manipulation of type String substitution in mathematical expression

4 Answers4