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Why do I get Null when using my function?

Luis Ledesma
Posted 10 years ago

I've built my own function, but using it returns me Null which should not be,the function which I have built is as follows

RCL[sup_, inf_, alpha_, lista_] := Module[{down, up, w, sel},
  down = Apply[Times, sup] - 
    0.4*(Apply[Times, sup] - Apply[Times, inf]);
  up = Apply[Times, sup];
  Table[
   If[
    down <= Apply[Times, lista[[w]]] <= up, lista[[w]]],
   {w, 1, Length[lista]}]
  ]

When used with the following data arises the problem that said, I hope someone can help me to understand this behavior

listin={{65, 80}, {65, 80}, {40, 40}, {40, 40}, {40, 40}, {25, 30}, {25,30}, {25, 30}, {25, 30}, {25, 30}};
RCL[{65, 80}, {25, 30}, 0.4, listin]
Out[2]:={{65, 80}, {65, 80}, Null, Null, Null, Null, Null, Null, Null, Null}

Thanks in advance
POSTED BY: Luis Ledesma
5 Replies
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Luis Ledesma
Luis Ledesma
Posted 10 years ago

Thank you Frank,Ii'm going to use Select as you suggested, thanks again,look after ,greetings
POSTED BY: Luis Ledesma 
In[9]:= p = Partition[Range[10], 2]

Out[9]= {{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}}

In[15]:= Select[p, 3 <= Times @@ # <= 45 &]

Out[15]= {{3, 4}, {5, 6}}
POSTED BY: Frank Kampas

You could use Select to get rid of the Nulls

In[7]:= Select[{2, Null, 3, 4, Null}, # =!= Null &]

Out[7]= {2, 3, 4}

or use Select instead of If to get the results you want, which is probably the better approach.

In[8]:= Select[Range[5], 2 <= # <= 4 &]

Out[8]= {2, 3, 4}
POSTED BY: Frank Kampas
Luis Ledesma
Luis Ledesma
Posted 10 years ago

Hi Frank, thanks for your quick response, now my question is, how do I do this function unless you get the Null? I I thought that was the way to do it, I hope you can help me, greetings
POSTED BY: Luis Ledesma

Null is what you get from your If statement when the criterion isn't satisfied.

In[3]:= If[2 > 3, "yes"] // InputForm

Out[3]//InputForm=
Null
POSTED BY: Frank Kampas
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