
Re: Pattern with powers
Posted:
Aug 2, 2013 2:49 AM


On Wednesday, July 31, 2013 11:14:07 PM UTC5, Dr. Wolfgang Hintze wrote: > Am Mittwoch, 31. Juli 2013 10:49:04 UTC+2 schrieb Alexei Boulbitch: > > > I must admit that I am an absolute beginner in patterns, as I cannot cope with a little problem with patterns consisting of powers of variables x and y. > > > > > Specifically, I would like to select from a list all terms of the form > > > c x^u y^v (numerical coefficient c times x to the power u times y to the power v) > > > > where u and v are allowed to take the values 0 and 1. > > > How can I do this using Cases? > > > > > > I have already accomplished the first non trivial step using _. (blank followed by a dot) in order to get first powers of the variables: > > > > > > ls = List@@Expand[5 (x + y)^3] > > > > > > {5*x^3, 15*x^2*y, 15*x*y^2, 5*y^3} > > > > > > Example 1 > > > a = 2; Cases[ls, (_.)*x^(u_.)*y^(v_.) /; u >= a && v < a] > > > gives > > > {15*x^2*y} > > > but misses the term > > > 5*x^3 > > > Example 2: this would be the form I would like most > > > Cases[ls, (_.)*x^_?(#1 >= a & )*y^_?(#1 < a & )] > > > gives > > > {} > > > Here even I didn't get the dot behind the blank before the test, so it misses first powers. > > > Thanks in advance for any help. > > > Best regards, > > > > > > Wolfgang > > > > > > Hi, Wolfgang, > > > Your explanation is not quite clear. Have a look: > > > > > > Clear[a, u, v, b]; > > > a = 2; > > > Cases[ls, (Times[_, x, Power[y, v_]] /; > > > v <= a)  (Times[_, Power[x, u_]] /; u >= a)] > > > {5 x^3, 15 x^2 y, 15 x y^2} > > > Is it, what you are after? Or this: > > > Clear[a, u, v, b]; > > > a = 2; > > > Cases[ls, (Times[_, x, Power[y, v_]] /; > > > v < a)  (Times[_, Power[x, u_]] /; u >= a)] > > > {5 x^3, 15 x^2 y} > > > ?? > > > Have fun, Alexei > > > Alexei BOULBITCH, Dr., habil. > > > Alexei, > > > > thanks for your message. > > As I tried to explain, I wish to extract from the list ls all terms of the form > > c x^u y^v, with c a numerical factor, u and v integers subject to the conditions u>=a and v<a with integer a>0. > > I have no problem as long as all terms in the list are "true" powers, i.e. as long as u>=2, v>=2. > > Therefore I asked for a solution which also covers the values 0 and 1 for the powers. > > Your first solution contains the wrong conditions, and the second one fails for a term x^2 y^3 which is selected by your proposal but it shouldn't be selected because v>2. > > Best regards, > > Wolfgang
Powers of zero, that is, missing factors, will be difficult to handle via patterns. A more natural choice, in my view, would be to define predicate functions. Here is an example.
In[19]:= s = List @@ Expand[5 (x + y)^3]
Out[19]= {5 x^3, 15 x^2 y, 15 x y^2, 5 y^3}
In[20]:= a = 1;
In[21]:= Select[s, Exponent[#, x] >= a && Exponent[#, y] <= a &]
Out[21]= {5 x^3, 15 x^2 y}
Obviously this can also be done with Cases, but that seems a bit more awkward to me.
In[22]:= Cases[s, aa_ /; Exponent[aa, x] >= a && Exponent[aa, y] <= a]
Out[22]= {5 x^3, 15 x^2 y}
Since "natural" and "awkward" are in the mind of the beholder, one might well hold an opinion the reverse of my own.
Daniel Lichtblau Wolfram Research

