Re: Function to detect presence of a variable in
- To: mathgroup at smc.vnet.net
- Subject: [mg109398] Re: Function to detect presence of a variable in
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sun, 25 Apr 2010 06:24:22 -0400 (EDT)
The default levelspec for Position is {0,Infinity}, but level 0 doesn't
work (in any way that makes sense):
expr = x;
Position[expr, x, {0, Infinity}]
{{}}
FAIL! This contradicts, I think, tutorial/LevelsInExpressions, where Help
says:
"You can think of levels in expressions in terms of trees. The level of a
particular part in an expression is simply the distance down the tree at
which that part appears, with the top of the tree considered as level 0."
x is obviously the top of TreeForm[x]... but according to Position, x
appears nowhere in x, at all.
Cases works, however.
expr = 1/x;
Length@Cases[expr, x, {0, Infinity}]
1
expr = 1*x;
Length@Cases[expr, x, {0, Infinity}]
1
Bobby
On Sat, 24 Apr 2010 03:01:50 -0500, <carlos at colorado.edu> wrote:
> I would like to have a function
>
> k = CountMeIn[expr,var]
>
> that returns the number of times var appears in expr, eg.
> CountMeIn[(1/x)^x,x] returns 2. var could be a list, in which case
> k would be a conforming list: k = CountMeIn[(2/x)^(x*w^4*z),{x,y,w,z}]
> should return {2,0,1,1}. expr is never a list.
>
> Tried to use Position for this, but the behavior is finicky:
> Position[1/x,x] returns {{1,1}} but Position[x,x] or Position[1*x,x]
> returns {{}} (why?) May be expr should be converted to {expr}
> inside the function for safety?
>
--
DrMajorBob at yahoo.com