# How can I identify the most important independent variable in an equation?

조회 수: 2(최근 30일)
Ni Made Ayu Sinta Dewi 2021년 6월 30일
답변: John D'Errico 2021년 6월 30일
Help! I have this equation. I would like to know the contribution of each independent variables to the dependent variable (for identifying the most important independent variable). Is this possible in MATLAB?
I tried doing this by taking the partial derivative of the equation with respect to each independent variable, but I am not sure how to identify which independent variable is most important/dominant.
syms x y z
f=sqrt(((x+y+y*z)-sqrt((x+y+y*z)^2-4*x*y*z))/((1+y+y*z)-sqrt((1+y+y*z)^2-4*y*z)));
This is the results:
ans =
-((2*x + 2*y - 2*y*z)/(2*((x + y + y*z)^2 - 4*x*y*z)^(1/2)) - 1)/(2*((x + y + y*z - ((x + y + y*z)^2 - 4*x*y*z)^(1/2))/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1))^(1/2)*(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1))
((z + (4*x*z - 2*(z + 1)*(x + y + y*z))/(2*((x + y + y*z)^2 - 4*x*y*z)^(1/2)) + 1)/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1) - ((z + (4*z - 2*(z + 1)*(y + y*z + 1))/(2*((y + y*z + 1)^2 - 4*y*z)^(1/2)) + 1)*(x + y + y*z - ((x + y + y*z)^2 - 4*x*y*z)^(1/2)))/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1)^2)/(2*((x + y + y*z - ((x + y + y*z)^2 - 4*x*y*z)^(1/2))/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1))^(1/2))
((y + (4*x*y - 2*y*(x + y + y*z))/(2*((x + y + y*z)^2 - 4*x*y*z)^(1/2)))/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1) - ((y + (4*y - 2*y*(y + y*z + 1))/(2*((y + y*z + 1)^2 - 4*y*z)^(1/2)))*(x + y + y*z - ((x + y + y*z)^2 - 4*x*y*z)^(1/2)))/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1)^2)/(2*((x + y + y*z - ((x + y + y*z)^2 - 4*x*y*z)^(1/2))/(y + y*z - ((y + y*z + 1)^2 - 4*y*z)^(1/2) + 1))^(1/2))
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
dpb 2021년 6월 30일
Unless you standardize the inputs, the magnitudes of the partial derivatives will be dependent upon the scale of the various variables and so won't mean anything in comparsion to each other.

댓글을 달려면 로그인하십시오.

### 채택된 답변

John D'Errico 2021년 6월 30일
Is it possible? Not really. Not using MATLAB or anything else. You have ONE expression. You might describe the "most important" variable as the one where that expression has the largest partial derivative. But depending on where in space you look, those partial derivatives will vary in size. I recall this being called a sensitivity.
But just consider a very simple expression.
syms x y
Z = x^2 + 2*y^2 + x*y;
You might think that y looks a little "more" important, because of the factor of 2 in front.
gz = But this idea about sensitivity is only relevant if you know the location where you are looking. For example, in the vicinity of the point (x,y) == (3,1), then each variable equally important.
subs(gz,[x,y],[3,1])
ans = At the point (-3,1), x is significantly more important.
subs(gz,[x,y],[-3,1])
ans = And at the location (-4,1), the importance of y is ZERO.
subs(gz,[x,y],[-4,1])
ans = Move a little ways away in a different direction, and we see x now has zero importance.
subs(gz,[x,y],[1,-2])
ans = In fact, we can determine a locus for this simple equation where x has zero importance. That is, the locus
gz(1) == 0
ans = is the set of all points where x is locally of zero importance to Z. Likewise, we can find the locus of points where y has zero sensitivity, as
gz(2) == 0
ans = And of course, for this simple expression, the locus of points where they are of equal importance.
abs(gz(1)) == abs(gz(2))
ans = So is what you want to do possible? Yes, and no. Unless you know where to look, the most important variable can easily change.

댓글을 달려면 로그인하십시오.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!