Position and Real Numbers
Jason Gill
I had a member of my department ask me this question, and I thought
I'd share it with the group.
In short he was using Position to return to the location of a number in
a list, and it wouldn't work in many cases. The example was
In[251]:=
x=Range[-3,2,.05]
Out[251]=
{-3,-2.95,-2.9,-2.85,-2.8,-2.75,-2.7,-2.65,-2.6,-2.55,-2.5,-2.45,-2.4,-2.35,-\
2.3,-2.25,-2.2,-2.15,-2.1,-2.05,-2.,-1.95,-1.9,-1.85,-1.8,-1.75,-1.7,-1.65,-1.\
6,-1.55,-1.5,-1.45,-1.4,-1.35,-1.3,-1.25,-1.2,-1.15,-1.1,-1.05,-1.,-0.95,-0.9,\
-0.85,-0.8,-0.75,-0.7,-0.65,-0.6,-0.55,-0.5,-0.45,-0.4,-0.35,-0.3,-0.25,-0.2,-\
0.15,-0.1,-0.05,0.,0.05,0.1,0.15,0.2,0.25,0.3,0.35,0.4,0.45,0.5,0.55,0.6,0.65,
0.7,0.75,0.8,0.85,0.9,0.95,1.,1.05,1.1,1.15,1.2,1.25,1.3,1.35,1.4,1.45,1.5,
1.55,1.6,1.65,1.7,1.75,1.8,1.85,1.9,1.95,2.}
For a variety or Values, Position returned the null set.
In[286]:=
Position[x,-.2]
Position[x,-1.15]
Position[x,-.15]
Position[x,-.05]
Position[x,.1]
Out[286]=
{}
Out[287]=
{{38}}
Out[288]=
{}
Out[289]=
{}
Out[290]=
{}
The use of InputForm sheds a little light on the situation, but even
that is not consistent, for example the values -1.15 vs. -.15.
I was able to offer a work around that seemed to work, but I am not sure
how reliable it is:
xList=Floor[x*100]/100//N
In anycase, even though I partially understand the behavior, it in
general seems wierd. Is there an easy way to make Position work as
expected for real numbers, or more accurately make Range work as
expected ???
Thanks,
Jason
P.S. Mathematica 3.0 on Win 95.
In[285]:=
x//InputForm
Out[285]//InputForm=
{-3, -2.95, -2.9, -2.85, -2.799999999999999, -2.75, -2.7,
-2.649999999999999, -2.6, -2.549999999999999, -2.5,
-2.45, -2.399999999999999, -2.35, -2.299999999999999,
-2.25, -2.2, -2.149999999999999, -2.1,
-2.049999999999999, -2., -1.95, -1.9, -1.849999999999999,
-1.799999999999999, -1.75, -1.7, -1.649999999999999,
-1.599999999999999, -1.549999999999999, -1.5, -1.45,
-1.399999999999999, -1.349999999999999,
-1.299999999999999, -1.25, -1.199999999999999,
-1.149999999999999, -1.099999999999999,
-1.049999999999999, -1., -0.9499999999999996, -0.9,
-0.85, -0.7999999999999998, -0.75, -0.6999999999999997,
-0.6499999999999999, -0.5999999999999996,
-0.5499999999999998, -0.5, -0.4499999999999997,
-0.3999999999999999, -0.3499999999999996,
-0.2999999999999998, -0.25, -0.1999999999999997,
-0.1499999999999999, -0.09999999999999964,
-0.04999999999999982, 0, 0.05000000000000026, 0.1,
0.1500000000000003, 0.2000000000000001, 0.25,
0.3000000000000002, 0.35, 0.4000000000000003,
0.4500000000000001, 0.5, 0.5500000000000002, 0.6,
0.6500000000000003, 0.7000000000000001, 0.75,
0.8000000000000003, 0.85, 0.9000000000000003,
0.9500000000000001, 1., 1.049999999999999, 1.1, 1.15,
1.2, 1.25, 1.299999999999999, 1.35, 1.4, 1.45, 1.5,
1.549999999999999, 1.6, 1.65, 1.7, 1.75, 1.8, 1.85, 1.9,
1.95, 2.}
Paul Abbott
> Is there an easy way to make Position work as expected for real numbers, or more
> accurately make Range work as
> expected ???
Both work "as expected". With
In[1]:= x=Range[-3,2,.05];
since the elements are distinct, to locate -0.05 you could use a test such as
In[2]:= Position[x,_?(-0.06<#<-0.04&)]
Out[2]= {{60}}
or you could Rationalize the entries
In[3]:= Position[Rationalize[x],Rationalize[-1/20]]
Out[3]= {{60}}
You can make Range give a set of exact values if you give it exact input:
In[4]:= x=Range[-3,2,Rationalize[0.05]];
In[5]:= Position[x,Rationalize[-1/20]]
Out[5]= {{60}}
Moral: If you give Mathematica exact input and it will generally give you exact
ouput. With approximate input you generally get approximate output.
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia
Nedlands WA 6907 mailto:pa...@physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
God IS a weakly left-handed dice player
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