Skip to main content

Filter list based on date



What I would like to do is remove entries from a list of instrument data when it is in maintenance. The maintenance data I have is a series of dates which look like this;


{{{2009, 6, 29, 10, 41, 0.}, {2009, 6, 30, 15, 26, 0.}}, {{2009, 6, 
   30, 16, 52, 0.}, {2009, 7, 1, 6, 0, 0.}}, {{2009, 7, 1, 6, 0, 
   0.}, {2009, 7, 1, 6, 2, 0.}}}

So between 29/6/2009 10:41 and 30/6/2008 15:26 the instrument was in maintenance.


The instrument data looks like this;


{{{2010, 1, 1, 6, 15, 0.}, 0.04375}, {{2010, 1, 1, 6, 30, 0.}, 
  0.04375}, {{2010, 1, 1, 6, 45, 0.}, 
  0.04375}, {{2010, 1, 1, 7, 0, 0.}, 

  0.04375}, {{2010, 1, 1, 7, 15, 0.}, 0.04375}}

With the first column being the date/time and the second the value.


What I would like is a quick (the instrument data is 100,000 records) way to remove rows from the instrument data when their date falls inside a maintenance period.



Answer



Starting with:


gaps = {{{2009, 6, 29, 10, 41, 0.}, {2009, 6, 30, 15, 26, 
     0.}}, {{2009, 6, 30, 16, 52, 0.}, {2009, 7, 1, 6, 0, 
     0.}}, {{2009, 7, 1, 6, 0, 0.}, {2009, 7, 1, 6, 2, 0.}}};


data = {{{2010, 1, 1, 6, 15, 0.}, 0.04375}, {{2010, 1, 1, 6, 30, 0.}, 
    0.04375}, {{2010, 1, 1, 6, 45, 0.}, 
    0.04375}, {{2010, 1, 1, 7, 0, 0.}, 
    0.04375}, {{2010, 1, 1, 7, 15, 0.}, 
    0.04375}, {{2009, 6, 30, 7, 26, 0.}, 0.1}};

I recommend:


maint = Interval @@ Map[AbsoluteTime, gaps, {2}];

Cases[data, {date_, _} /; ! IntervalMemberQ[maint, AbsoluteTime@date]]


Or if you prefer a form with Select like wxffles shows you could write:


makeTest[gaps_] :=
 With[{maint = Interval @@ Map[AbsoluteTime, gaps, {2}]},
   ! IntervalMemberQ[maint, AbsoluteTime @ #[[1]]] &
 ]

Select[data, makeTest @ gaps]

This should be more convenient to use for multiple "gaps" lists.




This is what I believe Murta wanted to write:


removeRanges[data_, gaps_] :=
 Module[{absData, absGaps},
  absData = AbsoluteTime /@ data[[All, 1]];
  absGaps = Interval @@ Map[AbsoluteTime, gaps, {2}];
  Pick[data, ! IntervalMemberQ[absGaps, #] & /@ absData]
 ]

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more