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Re: How can I totalize by month, by year?

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  • Subject: [mg111910] Re: How can I totalize by month, by year?
  • From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
  • Date: Wed, 18 Aug 2010 07:06:52 -0400 (EDT)

Hi,

I'm new with Mathematica, I need some help to start it.

I've got a date series like a cashflow, how can I totalize at each date 
change? And how can I totalize by month, by year, etc?

For example:

In[]=

{{{{2010, 8, 3, 0, 0, 0.}, 
   65.}, {{2010, 8, 3, 0, 0, 0.}, -72.}, {{2010, 8, 3, 0, 0, 0.}, 
   45.}, {{2010, 8, 2, 0, 0, 0.}, 67.}, {{2010, 8, 2, 0, 0, 0.}, 
   83.}, {{2010, 8, 2, 0, 0, 0.}, -42.}, {{2010, 7, 30, 0, 0, 
    0.}, -32.}, {{2010, 7, 30, 0, 0, 0.}, 
   48.}, {{2010, 7, 30, 0, 0, 0.}, 12.}, {{2010, 7, 30, 0, 0, 0.}, 
   34.}, {{2010, 7, 29, 0, 0, 0.}, -52.}, {{2010, 7, 29, 0, 0, 0.}, 
   78.}, {{2010, 7, 29, 0, 0, 0.}, -26.}, {{2010, 7, 28, 0, 0, 0.}, 
   45.}, {{2010, 7, 28, 0, 0, 0.}, -23.}, {{2010, 7, 28, 0, 0, 
    0.}, -32.}, {{2010, 7, 28, 0, 0, 0.}, 
   74.}, {{2010, 7, 27, 0, 0, 0.}, 53.}, {{2010, 7, 27, 0, 0, 0.}, 
   58.}, {{2010, 7, 27, 0, 0, 0.}, 25.}}}

Out[]=

{{{{2010, 8, 3, 0, 0, 0.}, 38.}, {{2010, 8, 2, 0, 0, 0.}, 
   108.}, {{2010, 7, 30, 0, 0, 0.}, 
   28.}, {{2010, 7, 29, 0, 0, 0.}, -10.}, {{2010, 7, 28, 0, 0, 0.}, 
   64.}, {{2010, 7, 2, 0, 0, 0.}, 136.}}}

Thanks
Leandro Tenfen 




Hi, Leandro,
if I understood you right, this may help:

(* That is your list: *)
lst = {{{{2010, 8, 3, 0, 0, 0.}, 
     65.}, {{2010, 8, 3, 0, 0, 0.}, -72.}, {{2010, 8, 3, 0, 0, 0.}, 
     45.}, {{2010, 8, 2, 0, 0, 0.}, 67.}, {{2010, 8, 2, 0, 0, 0.}, 
     83.}, {{2010, 8, 2, 0, 0, 0.}, -42.}, {{2010, 7, 30, 0, 0, 
      0.}, -32.}, {{2010, 7, 30, 0, 0, 0.}, 
     48.}, {{2010, 7, 30, 0, 0, 0.}, 12.}, {{2010, 7, 30, 0, 0, 0.}, 
     34.}, {{2010, 7, 29, 0, 0, 0.}, -52.}, {{2010, 7, 29, 0, 0, 0.}, 
     78.}, {{2010, 7, 29, 0, 0, 0.}, -26.}, {{2010, 7, 28, 0, 0, 0.}, 
     45.}, {{2010, 7, 28, 0, 0, 0.}, -23.}, {{2010, 7, 28, 0, 0, 
      0.}, -32.}, {{2010, 7, 28, 0, 0, 0.}, 
     74.}, {{2010, 7, 27, 0, 0, 0.}, 53.}, {{2010, 7, 27, 0, 0, 0.}, 
     58.}, {{2010, 7, 27, 0, 0, 0.}, 25.}}};


(* That is your operator giving the total expenses staying in the last position *)
(* e.g. if the data element is {{2010, 8, 2, 0, 0, 0.}, 67.}, then 67 is summed up *)
(* together with other figures staying at the same position to calculate the total:  *)

Total[Transpose[lst[[1]]][[2]]]

408.

or otherwise, you can apply this operator:

Total[Table[lst[[1, i, 2]], {i, 1, Length[lst[[1]]]}]]

408.

In both cases 408 is the answer.

Have fun, Alexei



-- 
Alexei Boulbitch, Dr. habil.
Senior Scientist
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