Re: Finding a function within an arbitrary expression

*To*: mathgroup at smc.vnet.net*Subject*: [mg131635] Re: Finding a function within an arbitrary expression*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Sat, 14 Sep 2013 06:04:19 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <CAEtRDSdto1VGsF3EdFnE+yuhhTaci98KC6Pc=CjtstSUQWgkLA@mail.gmail.com>

On closer reading of the documentation, use any of the following datav = {u[1, 0], u[2, 0], 1.1*u[1, 0]}; Either include level 0 so that the expresion does not need a head Select[datav, MemberQ[#, u[1, 0], {0, Infinity}] &] {u[1, 0], 1.1 u[1, 0]} Cases[datav, _?(MemberQ[#, u[1, 0], {0, Infinity}] &)] {u[1, 0], 1.1 u[1, 0]} or give it a head, say of List Select[datav, MemberQ[{#}, u[1, 0], Infinity] &] {u[1, 0], 1.1 u[1, 0]} Cases[datav, _?(MemberQ[{#}, u[1, 0], Infinity] &)] {u[1, 0], 1.1 u[1, 0]} or of any other arbitrary head since the head need not be List Select[datav, MemberQ[arbitraryHead[#], u[1, 0], Infinity] &] {u[1, 0], 1.1 u[1, 0]} Cases[datav, _?(MemberQ[arbitraryHead[#], u[1, 0], Infinity] &)] {u[1, 0], 1.1 u[1, 0]} Bob Hanlon On Fri, Sep 13, 2013 at 5:27 PM, Mannucci, Anthony J (335G) < anthony.j.mannucci at jpl.nasa.gov> wrote: > Bob, > > This does not quite work, for reasons that I do not understand. I add > the errant case here that is not one you considered, which is why your > tests all passed. The following code snippet will suffice to show what I > mean. It comes down to the difference between FreeQ and MemberQ. Any > insight you have here is greatly appreciated. > > u[1,0] and u[2,0] evaluate to themselves (no function evaluation occurs. > These are undefined symbols.) > > Try this: > > datav = {u[1, 0], u[2, 0], 1.1*u[1, 0]}; > Select[datav, MemberQ[#, u[1, 0], Infinity] &] > Cases[datav, _?(MemberQ[#, u[1, 0], Infinity] &)] > Select[datav, ! FreeQ[#, u[1, 0]] &] > > > You will see that only the last version gives the correct answer. The > Infinity level in MemberQ is needed for other reasons (this differs from > your example and was covered in a separate email). > > > Basically, the stand-alone u[1,0] expression (first element) is not > found by MemberQ, but is found by FreeQ. The documentation for these two > functions differs: > > MemberQ[list,form] > > returns True if an element of list matches form, and False otherwise. > > FreeQ[expr,form] > > yields True if no subexpression in expr matches form, and yields False > otherwise. > > > The documentation does not immediately tell me why these two functions > should behave as in this example. FreeQ wants an expression and gets a > list. That puts a head on the expression and somehow allows FreeQ to find > the stand-alone u[1,0]. MemberQ wants a list, so looks at the input > differently than FreeQ. > > > And yet, this yields True: > > MemberQ[{u[1, 0]}, u[1, 0], Infinity] > > > Mathematica 8.0.4 > > > -Tony > > > -- > Tony Mannucci > Supervisor, Ionospheric and Atmospheric Remote Sensing Group > Mail-Stop 138-308, Tel > (818) 354-1699 > Jet Propulsion Laboratory, Fax > (818) 393-5115 > California Institute of Technology, Email > > Tony.Mannucci at jpl.nasa.gov > 4800 Oak Grove Drive, > http://scienceandtechnology.jpl.nasa.gov/people/a_mannucci/ > Pasadena, CA 91109 > > > From: Bob Hanlon <hanlonr357 at gmail.com> > Date: Sunday, June 30, 2013 7:01 AM > To: Tony Mannucci <Anthony.J.Mannucci at jpl.nasa.gov> > Cc: MathGroup <mathgroup at smc.vnet.net> > Subject: Re: Finding a function within an arbitrary expression > > data = {3*u[1, 0], u[0, 0]/10., 1/u[1, 0], f[u[1, 0]]}; > > > data2 = Select[data, MemberQ[#, u[1, 0]] &] > > > {3*u[1, 0], 1/u[1, 0], f[u[1, 0]]} > > > Some other methods: > > > data2 == > Cases[data, _?(MemberQ[#, u[1, 0]] &)] == > DeleteCases[data, _?(! MemberQ[#, u[1, 0]] &)] == > Select[data, ! FreeQ[#, u[1, 0]] &] == > Cases[data, _?(! FreeQ[#, u[1, 0]] &)] == > DeleteCases[data, _?(FreeQ[#, u[1, 0]] &)] == > data[[First /@ Position[data, u[1, 0]]]] > > > True > > > > Bob Hanlon > > > > > On Sun, Jun 30, 2013 at 3:29 AM, amannucci < > Anthony.J.Mannucci at jpl.nasa.gov> wrote: > >> I am trying to figure out what that pattern is for the following. The >> difficulty here is that I can think textually, but not in Mathematica >> patterns. >> >> I am looking for a function u[1,0] within a list of expressions. The >> function might not be defined yet. For example, a list of expressions might >> be: >> >> {3*u[1,0], u[0,0]/10., 1/u[1,0], ...} >> >> I want to find all the expressions that have the function u[1,0] in it. >> There is certainly a way to do this "textually". How would one do this with >> patterns? Is it possible? Thank you. >> >> >> >