How to count number of terms in a symbolic expression?

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Mehdi
Mehdi 2023-12-2
评论： Walter Roberson 2023-12-3
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 2023-12-3
编辑：Walter Roberson 2023-12-3
Ran in:
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 = 
  2 个评论
Walter Roberson
Walter Roberson 2023-12-3
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 2023-12-3
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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更多回答（1 个）

VBBV
VBBV 2023-12-3
编辑：VBBV 2023-12-3
Ran in:
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)'
  4 个评论
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Walter Roberson
Walter Roberson 2023-12-3
编辑：Walter Roberson 2023-12-3
Ran in:
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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