Re: Finding a function within an arbitrary expression

*To*: mathgroup at smc.vnet.net*Subject*: [mg131463] Re: Finding a function within an arbitrary expression*From*: amannucci <Anthony.J.Mannucci at jpl.nasa.gov>*Date*: Mon, 29 Jul 2013 23:20:48 -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*: <20130630072942.2AF536A20@smc.vnet.net> <kqrj00$4ao$1@smc.vnet.net>

On Monday, July 1, 2013 2:43:28 AM UTC-7, Bob Hanlon wrote: > 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: [snip] > > > Bob Hanlon > > Your top example is a good way to do it. I am completely mystified, however. I originally tried to do this with Cases, but that did not work because of expression levels. I did not specify an expression level when using Cases. You do not specify the level in your MemberQ call, but it works. Yet, this does not work: MemberQ[{3*u[1, 0]}, u[1, 0]] gives False but MemberQ[{3*u[1, 0]}, u[1, 0],Infinity] gives True So why don't you need a level spec in your example (definition of data2)? It's as if creating a lambda function automatically inserts the level spec.