问题
I'm puzzled by Mathematica's responses to the following:
ClearAll[n]
#^2 & /@ Range[n]
#^2 & /@ Range[n] // StandardForm
It seems that even Mathematica (8.0) doesn't believe what it has just said:
#^2 & /@ Range[5]
Range[5^2]
Any thoughts about what is happening?
Edit:
The original context for this question was the following. I had written
PrimeOmega[Range[n]] - PrimeNu[Range[n]]
and since n was going to be very large (2^50), I thought I might save time by rewriting it as:
PrimeOmega[#] - PrimeNu[#] &/@Range[n]
Thinking back, that probably wasn't such a good idea. (I could have used Module to 'compute' the Range only once.)
回答1:
Since n is undefined, Range[n] evaluated to itself. Therefore, Map acts on it as on any other symbolic head, mapping your function on its elements - here it is just n
In[11]:= #^2 & /@ someHead[n]
Out[11]= someHead[n^2]
EDIT
Addressing the question in your edit - for numeric n, Range evaluates to a list all right, and you get the expected result (which is, Range[5]^2. It is all about the order of evaluation. To get Range[5^2], you could have used #^2&/@Unevaluated[Range[5]], in which case everything happens just like for symbolic n above) . In fact, Range issues an error message on non-numeric input. Also, it is tangential to the question, but functions like #^2& are Listable, and you don't have to map them.
回答2:
Slightly off topic, but you can improve the speed by redefining in terms of FactorInteger, which then is only called once per input.
f1[n_] := PrimeOmega[Range[n]] - PrimeNu[Range[n]]
f2[n_] := With[{fax=FactorInteger[#]}, Total[fax[[All,2]]]-Length[fax]]& /@ Range[n]
Example:
In[27]:= Timing[pdiff1 = f1[2^20];]
Out[27]= {37.730264, Null}
In[28]:= Timing[pdiff2 = f2[2^20];]
Out[28]= {9.364576, Null}
In[29]:= pdiff1===pdiff2
Out[29]= True
Daniel Lichtblau
来源：https://stackoverflow.com/questions/7769613/mathematicas-puzzling-interpretation-of-2-rangen