How to count number of terms in a symbolic expression?

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Mehdi
Mehdi 2023년 12월 2일
댓글: Walter Roberson 2023년 12월 3일
I have a long symbolic expression composed of many terms. How is it possible to count the total number of terms of my expression and then choose the ith term?
syms x y
z=x*y+cos(x*y)*x+x*sin(x)+sin(x)*cos(x)-x-y+2;
for example here the z is composed of 7 terms and the 3rd term is x*sin(x).

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Walter Roberson
Walter Roberson 2023년 12월 3일
편집: Walter Roberson 2023년 12월 3일
syms x y
z=x*y+cos(x*y)*x+x*sin(x)+sin(x)*cos(x)-x-y+2;
terms = children(z)
terms = 1×7 cell array
{[-x]} {[-y]} {[cos(x)*sin(x)]} {[x*y]} {[x*cos(x*y)]} {[x*sin(x)]} {[2]}
nterms = length(terms)
nterms = 7
terms{3}
ans = 
children(z, 3)
ans = 
  댓글 수: 2
Walter Roberson
Walter Roberson 2023년 12월 3일
Notice the "third" term as far as MATLAB is concerned is not x*sin(x)
MATLAB re-arranges terms according to algebraic equivalences, into its own preferred order. The rules for ordering are not published, and can be contextual.
Walter Roberson
Walter Roberson 2023년 12월 3일
Further example:
syms x
f = 6 + x - 3 - 2 - 1
as far as the symbolic toolbox is concerned this is the same as
f = x
The toolbox always folds rational and symbolic floating point numbers
5 + 2*sqrt(sym('3.2'))
will get completely executed to scalar
5 + 2*sqrt(sym(3.2))
will execute to an expression. sym(3.2) is converted to sym(32)/sym(10) and sqrt of a rational is not fully evaluated

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추가 답변 (1개)

VBBV
VBBV 2023년 12월 3일
편집: VBBV 2023년 12월 3일
syms x y
z='x*y+cos(x*y)*x+x*sin(x)+sin(x)*cos(x)-x-y+2'
z = 'x*y+cos(x*y)*x+x*sin(x)+sin(x)*cos(x)-x-y+2'
C = strsplit(z,{'+','-'})
C = 1×7 cell array
{'x*y'} {'cos(x*y)*x'} {'x*sin(x)'} {'sin(x)*cos(x)'} {'x'} {'y'} {'2'}
length(C)
ans = 7
C{3} % 3rd term
ans = 'x*sin(x)'
  댓글 수: 4
Walter Roberson
Walter Roberson 2023년 12월 3일
편집: Walter Roberson 2023년 12월 3일
Consider
syms x y
z='x*y+cos(x-y)*x+x*sin(x)+sin(x)*cos(x)-x-y+2'
z = 'x*y+cos(x-y)*x+x*sin(x)+sin(x)*cos(x)-x-y+2'
C = strsplit(z,{'+','-'})
C = 1×8 cell array
{'x*y'} {'cos(x'} {'y)*x'} {'x*sin(x)'} {'sin(x)*cos(x)'} {'x'} {'y'} {'2'}
length(C)
ans = 8
C{3} % 3rd term
ans = 'y)*x'
The pattern matching has to be a lot more complicated to extract expressions. Indeed, it can be proven that it cannot be done using traditional regular expressions -- not unless you are willing to impose a maximum nesting depth (and use terribly messy expressions.) Some programming languages such as perl add regexp constructs that extend the pattern capabilities to make it possible; if MATLAB has those facilities I am overlooking them.

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