Compute how many samples have been improved, according to a minimum threshold or confidence interval, in a...












0















I have the following dataframe:



ID      VAL1    VAL2
Q2241 0.3333 0.3353
Q2242 0.5 0.5
Q2243 0.3333 0.3333
Q2244 0.2137 0.4792
Q2245 0.1429 0.2
Q2246 0.5 0.5
Q2247 0.4167 0.6667
Q2248 1 1
Q2249 0.125 0.0909
Q2250 0.2 0.2
Q2251 0.325 0.2667
Q2252 0.1667 0.2
Q2253 0.3333 0.25
Q2254 0.45 0.8333
Q2255 0.3333 0.5
Q2256 1 1
Q2257 0.5 0.51
Q2258 0.3929 0.3333
Q2259 0.3611 0.625


Is there a way to correctly compute the number of samples (ID) where VAL2 is significantly higher/lower than VAL1 in a given dataframe.
I'm looking for something like t-test, where a measure gives results like the following example:



Win Tie Loss        
64 36 137


where:




Win: number of IDs where VAL2 is higher than VAL1 with some confidence interval  
Tie: number of IDs where VAL2 ~ VAL1 (no significant difference, 0.0001 for example)
Loss: number of IDs where VAL2 is lower than VAL1 with some confidence interval










share|improve this question



























    0















    I have the following dataframe:



    ID      VAL1    VAL2
    Q2241 0.3333 0.3353
    Q2242 0.5 0.5
    Q2243 0.3333 0.3333
    Q2244 0.2137 0.4792
    Q2245 0.1429 0.2
    Q2246 0.5 0.5
    Q2247 0.4167 0.6667
    Q2248 1 1
    Q2249 0.125 0.0909
    Q2250 0.2 0.2
    Q2251 0.325 0.2667
    Q2252 0.1667 0.2
    Q2253 0.3333 0.25
    Q2254 0.45 0.8333
    Q2255 0.3333 0.5
    Q2256 1 1
    Q2257 0.5 0.51
    Q2258 0.3929 0.3333
    Q2259 0.3611 0.625


    Is there a way to correctly compute the number of samples (ID) where VAL2 is significantly higher/lower than VAL1 in a given dataframe.
    I'm looking for something like t-test, where a measure gives results like the following example:



    Win Tie Loss        
    64 36 137


    where:




    Win: number of IDs where VAL2 is higher than VAL1 with some confidence interval  
    Tie: number of IDs where VAL2 ~ VAL1 (no significant difference, 0.0001 for example)
    Loss: number of IDs where VAL2 is lower than VAL1 with some confidence interval










    share|improve this question

























      0












      0








      0








      I have the following dataframe:



      ID      VAL1    VAL2
      Q2241 0.3333 0.3353
      Q2242 0.5 0.5
      Q2243 0.3333 0.3333
      Q2244 0.2137 0.4792
      Q2245 0.1429 0.2
      Q2246 0.5 0.5
      Q2247 0.4167 0.6667
      Q2248 1 1
      Q2249 0.125 0.0909
      Q2250 0.2 0.2
      Q2251 0.325 0.2667
      Q2252 0.1667 0.2
      Q2253 0.3333 0.25
      Q2254 0.45 0.8333
      Q2255 0.3333 0.5
      Q2256 1 1
      Q2257 0.5 0.51
      Q2258 0.3929 0.3333
      Q2259 0.3611 0.625


      Is there a way to correctly compute the number of samples (ID) where VAL2 is significantly higher/lower than VAL1 in a given dataframe.
      I'm looking for something like t-test, where a measure gives results like the following example:



      Win Tie Loss        
      64 36 137


      where:




      Win: number of IDs where VAL2 is higher than VAL1 with some confidence interval  
      Tie: number of IDs where VAL2 ~ VAL1 (no significant difference, 0.0001 for example)
      Loss: number of IDs where VAL2 is lower than VAL1 with some confidence interval










      share|improve this question














      I have the following dataframe:



      ID      VAL1    VAL2
      Q2241 0.3333 0.3353
      Q2242 0.5 0.5
      Q2243 0.3333 0.3333
      Q2244 0.2137 0.4792
      Q2245 0.1429 0.2
      Q2246 0.5 0.5
      Q2247 0.4167 0.6667
      Q2248 1 1
      Q2249 0.125 0.0909
      Q2250 0.2 0.2
      Q2251 0.325 0.2667
      Q2252 0.1667 0.2
      Q2253 0.3333 0.25
      Q2254 0.45 0.8333
      Q2255 0.3333 0.5
      Q2256 1 1
      Q2257 0.5 0.51
      Q2258 0.3929 0.3333
      Q2259 0.3611 0.625


      Is there a way to correctly compute the number of samples (ID) where VAL2 is significantly higher/lower than VAL1 in a given dataframe.
      I'm looking for something like t-test, where a measure gives results like the following example:



      Win Tie Loss        
      64 36 137


      where:




      Win: number of IDs where VAL2 is higher than VAL1 with some confidence interval  
      Tie: number of IDs where VAL2 ~ VAL1 (no significant difference, 0.0001 for example)
      Loss: number of IDs where VAL2 is lower than VAL1 with some confidence interval







      python dataframe statistics difference






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 24 '18 at 10:28









      Belkacem ThiziriBelkacem Thiziri

      69111




      69111
























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          tol = 0.0001
          win = (df.VAL2 > (df.VAL1 + tol)).sum()
          loss = (df.VAL2 < (df.VAL1 - tol)).sum()
          tie = ((df.VAL1 - df.VAL2).abs() <= tol).sum()

          df = pd.DataFrame([{'Win': win, 'Tie':tie, 'Loss': loss}])
          print (df)
          # Loss Tie Win
          # 0 4 6 9





          share|improve this answer
























          • Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

            – Belkacem Thiziri
            Nov 24 '18 at 12:31











          • There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

            – Ghilas BELHADJ
            Nov 24 '18 at 12:37













          • The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

            – Belkacem Thiziri
            Nov 24 '18 at 12:51











          • Probably not. but I'll let you know if I find something.

            – Ghilas BELHADJ
            Nov 24 '18 at 13:38











          • Okay, thanks. I'll discuss that with my advisors then leave a comment here.

            – Belkacem Thiziri
            Nov 24 '18 at 13:41











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          tol = 0.0001
          win = (df.VAL2 > (df.VAL1 + tol)).sum()
          loss = (df.VAL2 < (df.VAL1 - tol)).sum()
          tie = ((df.VAL1 - df.VAL2).abs() <= tol).sum()

          df = pd.DataFrame([{'Win': win, 'Tie':tie, 'Loss': loss}])
          print (df)
          # Loss Tie Win
          # 0 4 6 9





          share|improve this answer
























          • Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

            – Belkacem Thiziri
            Nov 24 '18 at 12:31











          • There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

            – Ghilas BELHADJ
            Nov 24 '18 at 12:37













          • The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

            – Belkacem Thiziri
            Nov 24 '18 at 12:51











          • Probably not. but I'll let you know if I find something.

            – Ghilas BELHADJ
            Nov 24 '18 at 13:38











          • Okay, thanks. I'll discuss that with my advisors then leave a comment here.

            – Belkacem Thiziri
            Nov 24 '18 at 13:41
















          1














          tol = 0.0001
          win = (df.VAL2 > (df.VAL1 + tol)).sum()
          loss = (df.VAL2 < (df.VAL1 - tol)).sum()
          tie = ((df.VAL1 - df.VAL2).abs() <= tol).sum()

          df = pd.DataFrame([{'Win': win, 'Tie':tie, 'Loss': loss}])
          print (df)
          # Loss Tie Win
          # 0 4 6 9





          share|improve this answer
























          • Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

            – Belkacem Thiziri
            Nov 24 '18 at 12:31











          • There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

            – Ghilas BELHADJ
            Nov 24 '18 at 12:37













          • The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

            – Belkacem Thiziri
            Nov 24 '18 at 12:51











          • Probably not. but I'll let you know if I find something.

            – Ghilas BELHADJ
            Nov 24 '18 at 13:38











          • Okay, thanks. I'll discuss that with my advisors then leave a comment here.

            – Belkacem Thiziri
            Nov 24 '18 at 13:41














          1












          1








          1







          tol = 0.0001
          win = (df.VAL2 > (df.VAL1 + tol)).sum()
          loss = (df.VAL2 < (df.VAL1 - tol)).sum()
          tie = ((df.VAL1 - df.VAL2).abs() <= tol).sum()

          df = pd.DataFrame([{'Win': win, 'Tie':tie, 'Loss': loss}])
          print (df)
          # Loss Tie Win
          # 0 4 6 9





          share|improve this answer













          tol = 0.0001
          win = (df.VAL2 > (df.VAL1 + tol)).sum()
          loss = (df.VAL2 < (df.VAL1 - tol)).sum()
          tie = ((df.VAL1 - df.VAL2).abs() <= tol).sum()

          df = pd.DataFrame([{'Win': win, 'Tie':tie, 'Loss': loss}])
          print (df)
          # Loss Tie Win
          # 0 4 6 9






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 24 '18 at 11:20









          Ghilas BELHADJGhilas BELHADJ

          5,31962961




          5,31962961













          • Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

            – Belkacem Thiziri
            Nov 24 '18 at 12:31











          • There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

            – Ghilas BELHADJ
            Nov 24 '18 at 12:37













          • The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

            – Belkacem Thiziri
            Nov 24 '18 at 12:51











          • Probably not. but I'll let you know if I find something.

            – Ghilas BELHADJ
            Nov 24 '18 at 13:38











          • Okay, thanks. I'll discuss that with my advisors then leave a comment here.

            – Belkacem Thiziri
            Nov 24 '18 at 13:41



















          • Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

            – Belkacem Thiziri
            Nov 24 '18 at 12:31











          • There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

            – Ghilas BELHADJ
            Nov 24 '18 at 12:37













          • The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

            – Belkacem Thiziri
            Nov 24 '18 at 12:51











          • Probably not. but I'll let you know if I find something.

            – Ghilas BELHADJ
            Nov 24 '18 at 13:38











          • Okay, thanks. I'll discuss that with my advisors then leave a comment here.

            – Belkacem Thiziri
            Nov 24 '18 at 13:41

















          Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

          – Belkacem Thiziri
          Nov 24 '18 at 12:31





          Thanks @Ghilas BELHADJ I already tried something like that, but I was wondering if there is some specific method in data science that enable to do such a statistic, some official function like "t-test" ?

          – Belkacem Thiziri
          Nov 24 '18 at 12:31













          There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

          – Ghilas BELHADJ
          Nov 24 '18 at 12:37







          There is a list of Software implementations in the bottom of your Wikipedia page. So basically, you can use scipy.stats.ttest_ind in Python.

          – Ghilas BELHADJ
          Nov 24 '18 at 12:37















          The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

          – Belkacem Thiziri
          Nov 24 '18 at 12:51





          The t-test evaluates the significance of the differences between 2 distributions but does not give how many samples that are significantly different. One better solution will combine the significance probability given by the t-test with something else to compute the number of these samples. Do you think that computing the t-test value for each line in my data frame will make sense?

          – Belkacem Thiziri
          Nov 24 '18 at 12:51













          Probably not. but I'll let you know if I find something.

          – Ghilas BELHADJ
          Nov 24 '18 at 13:38





          Probably not. but I'll let you know if I find something.

          – Ghilas BELHADJ
          Nov 24 '18 at 13:38













          Okay, thanks. I'll discuss that with my advisors then leave a comment here.

          – Belkacem Thiziri
          Nov 24 '18 at 13:41





          Okay, thanks. I'll discuss that with my advisors then leave a comment here.

          – Belkacem Thiziri
          Nov 24 '18 at 13:41




















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