How do you calculate interquartile range (IQR) correctly using Python?





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3















I am trying to understand the way to compute iqr (interquartile range).



according this, this and this, I tried 3 solutions to do this.



solution_1



a = numpy.array([1, 2, 3, 4, 5, 6, 7])
q1_a = numpy.percentile(a, 25)
q3_a = numpy.percentile(a, 75)
q3_a - q1_a


solution_2



from scipy.stats import iqr
iqr(a)


solution_3



q1_am = np.median(numpy.array([1, 2, 3, 4]))
q3_am = np.median(numpy.array([4, 5, 6, 7]))
q3_am - q1_am


3 of them give the same result 3 which is correct.



when I tried another set of numbers, things were going weird.



both solution_1 and 2 output 0.95 which is not correct.



x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25)
q3_x = numpy.percentile(x, 75)
q3_x - q1_x


solution_3 gives 1.2 which is correct



q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
q3_xm - q1_xm


What am I missing with the solutions?



any clue would be appreciated.










share|improve this question































    3















    I am trying to understand the way to compute iqr (interquartile range).



    according this, this and this, I tried 3 solutions to do this.



    solution_1



    a = numpy.array([1, 2, 3, 4, 5, 6, 7])
    q1_a = numpy.percentile(a, 25)
    q3_a = numpy.percentile(a, 75)
    q3_a - q1_a


    solution_2



    from scipy.stats import iqr
    iqr(a)


    solution_3



    q1_am = np.median(numpy.array([1, 2, 3, 4]))
    q3_am = np.median(numpy.array([4, 5, 6, 7]))
    q3_am - q1_am


    3 of them give the same result 3 which is correct.



    when I tried another set of numbers, things were going weird.



    both solution_1 and 2 output 0.95 which is not correct.



    x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
    q1_x = numpy.percentile(x, 25)
    q3_x = numpy.percentile(x, 75)
    q3_x - q1_x


    solution_3 gives 1.2 which is correct



    q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
    q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
    q3_xm - q1_xm


    What am I missing with the solutions?



    any clue would be appreciated.










    share|improve this question



























      3












      3








      3


      1






      I am trying to understand the way to compute iqr (interquartile range).



      according this, this and this, I tried 3 solutions to do this.



      solution_1



      a = numpy.array([1, 2, 3, 4, 5, 6, 7])
      q1_a = numpy.percentile(a, 25)
      q3_a = numpy.percentile(a, 75)
      q3_a - q1_a


      solution_2



      from scipy.stats import iqr
      iqr(a)


      solution_3



      q1_am = np.median(numpy.array([1, 2, 3, 4]))
      q3_am = np.median(numpy.array([4, 5, 6, 7]))
      q3_am - q1_am


      3 of them give the same result 3 which is correct.



      when I tried another set of numbers, things were going weird.



      both solution_1 and 2 output 0.95 which is not correct.



      x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
      q1_x = numpy.percentile(x, 25)
      q3_x = numpy.percentile(x, 75)
      q3_x - q1_x


      solution_3 gives 1.2 which is correct



      q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
      q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
      q3_xm - q1_xm


      What am I missing with the solutions?



      any clue would be appreciated.










      share|improve this question
















      I am trying to understand the way to compute iqr (interquartile range).



      according this, this and this, I tried 3 solutions to do this.



      solution_1



      a = numpy.array([1, 2, 3, 4, 5, 6, 7])
      q1_a = numpy.percentile(a, 25)
      q3_a = numpy.percentile(a, 75)
      q3_a - q1_a


      solution_2



      from scipy.stats import iqr
      iqr(a)


      solution_3



      q1_am = np.median(numpy.array([1, 2, 3, 4]))
      q3_am = np.median(numpy.array([4, 5, 6, 7]))
      q3_am - q1_am


      3 of them give the same result 3 which is correct.



      when I tried another set of numbers, things were going weird.



      both solution_1 and 2 output 0.95 which is not correct.



      x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
      q1_x = numpy.percentile(x, 25)
      q3_x = numpy.percentile(x, 75)
      q3_x - q1_x


      solution_3 gives 1.2 which is correct



      q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
      q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
      q3_xm - q1_xm


      What am I missing with the solutions?



      any clue would be appreciated.







      python numpy scipy






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 16 '18 at 12:50









      tel

      7,52121431




      7,52121431










      asked Nov 16 '18 at 11:54









      czlswsczlsws

      1236




      1236
























          1 Answer
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          You'll get your expected result with numpy.percentile if you set interpolation=midpoint:



          x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
          q1_x = numpy.percentile(x, 25, interpolation='midpoint')
          q3_x = numpy.percentile(x, 75, interpolation='midpoint')
          print(q3_x - q1_x)


          This outputs:



          1.2000000000000002


          Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:



          from scipy.stats import iqr

          x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
          print(iqr(x, rng=(25,75), interpolation='midpoint'))


          which outputs:



          1.2000000000000002


          See the interpolation parameter in the linked docs for more info on what the option actually does.






          share|improve this answer
























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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            0














            You'll get your expected result with numpy.percentile if you set interpolation=midpoint:



            x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
            q1_x = numpy.percentile(x, 25, interpolation='midpoint')
            q3_x = numpy.percentile(x, 75, interpolation='midpoint')
            print(q3_x - q1_x)


            This outputs:



            1.2000000000000002


            Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:



            from scipy.stats import iqr

            x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
            print(iqr(x, rng=(25,75), interpolation='midpoint'))


            which outputs:



            1.2000000000000002


            See the interpolation parameter in the linked docs for more info on what the option actually does.






            share|improve this answer




























              0














              You'll get your expected result with numpy.percentile if you set interpolation=midpoint:



              x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
              q1_x = numpy.percentile(x, 25, interpolation='midpoint')
              q3_x = numpy.percentile(x, 75, interpolation='midpoint')
              print(q3_x - q1_x)


              This outputs:



              1.2000000000000002


              Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:



              from scipy.stats import iqr

              x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
              print(iqr(x, rng=(25,75), interpolation='midpoint'))


              which outputs:



              1.2000000000000002


              See the interpolation parameter in the linked docs for more info on what the option actually does.






              share|improve this answer


























                0












                0








                0







                You'll get your expected result with numpy.percentile if you set interpolation=midpoint:



                x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
                q1_x = numpy.percentile(x, 25, interpolation='midpoint')
                q3_x = numpy.percentile(x, 75, interpolation='midpoint')
                print(q3_x - q1_x)


                This outputs:



                1.2000000000000002


                Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:



                from scipy.stats import iqr

                x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
                print(iqr(x, rng=(25,75), interpolation='midpoint'))


                which outputs:



                1.2000000000000002


                See the interpolation parameter in the linked docs for more info on what the option actually does.






                share|improve this answer













                You'll get your expected result with numpy.percentile if you set interpolation=midpoint:



                x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
                q1_x = numpy.percentile(x, 25, interpolation='midpoint')
                q3_x = numpy.percentile(x, 75, interpolation='midpoint')
                print(q3_x - q1_x)


                This outputs:



                1.2000000000000002


                Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:



                from scipy.stats import iqr

                x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
                print(iqr(x, rng=(25,75), interpolation='midpoint'))


                which outputs:



                1.2000000000000002


                See the interpolation parameter in the linked docs for more info on what the option actually does.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 16 '18 at 12:44









                teltel

                7,52121431




                7,52121431
































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