how to draw a negative exponential plot of this situation with python












0












$begingroup$


I have a dataset with binary output ($Y$) and I have a column (Duration) contains the duration of each task that is stored by "days" and varied from 1day to 350 days.



when I think logically in our situation, I can deduce that the probability of getting a positive output value ($Y = 1$) require to have small duration task.



But I need to justify my opinion with some plots



I have tried the following source code but It doesn't represent correctly my assumption. 



#LoadData



min_duration = plot_data['Duration'].min()

max_duration = plot_data['Duration'].max()

xr_ = list(range(min_duration,  max_duration,  5))

y_ = 



for i in range(0,(len(xr_)-1)):

    a_ = np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1])

    b_ = np.logical_and(np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1]), plot_data['output'].values==1)

    y_.append(sum(b_)/sum(a_))



import matplotlib

matplotlib.pyplot.plot(xr_[1:len(xr_)], y_, 'o')


Based on my previous assumption I must get a plot which contains an exponential form like :



enter image description here



But I have got contrary the following plot:



enter image description here



I want to know where I have a mistake and If there is any other method to justify my assumption






python dataset matplotlib







share|improve this question











$endgroup$

















    0












    $begingroup$


    I have a dataset with binary output ($Y$) and I have a column (Duration) contains the duration of each task that is stored by "days" and varied from 1day to 350 days.



    when I think logically in our situation, I can deduce that the probability of getting a positive output value ($Y = 1$) require to have small duration task.



    But I need to justify my opinion with some plots



    I have tried the following source code but It doesn't represent correctly my assumption. 



    #LoadData
    

    
min_duration = plot_data['Duration'].min()
    
max_duration = plot_data['Duration'].max()
    
xr_ = list(range(min_duration,  max_duration,  5))
    
y_ = 
    

    
for i in range(0,(len(xr_)-1)):
    
    a_ = np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1])
    
    b_ = np.logical_and(np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1]), plot_data['output'].values==1)
    
    y_.append(sum(b_)/sum(a_))
    

    
import matplotlib
    
matplotlib.pyplot.plot(xr_[1:len(xr_)], y_, 'o')


    Based on my previous assumption I must get a plot which contains an exponential form like :



    enter image description here



    But I have got contrary the following plot:



    enter image description here



    I want to know where I have a mistake and If there is any other method to justify my assumption






    python dataset matplotlib







    share|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have a dataset with binary output ($Y$) and I have a column (Duration) contains the duration of each task that is stored by "days" and varied from 1day to 350 days.



      when I think logically in our situation, I can deduce that the probability of getting a positive output value ($Y = 1$) require to have small duration task.



      But I need to justify my opinion with some plots



      I have tried the following source code but It doesn't represent correctly my assumption. 



      #LoadData
      

      
min_duration = plot_data['Duration'].min()
      
max_duration = plot_data['Duration'].max()
      
xr_ = list(range(min_duration,  max_duration,  5))
      
y_ = 
      

      
for i in range(0,(len(xr_)-1)):
      
    a_ = np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1])
      
    b_ = np.logical_and(np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1]), plot_data['output'].values==1)
      
    y_.append(sum(b_)/sum(a_))
      

      
import matplotlib
      
matplotlib.pyplot.plot(xr_[1:len(xr_)], y_, 'o')


      Based on my previous assumption I must get a plot which contains an exponential form like :



      enter image description here



      But I have got contrary the following plot:



      enter image description here



      I want to know where I have a mistake and If there is any other method to justify my assumption






      python dataset matplotlib







      share|improve this question











      $endgroup$




      I have a dataset with binary output ($Y$) and I have a column (Duration) contains the duration of each task that is stored by "days" and varied from 1day to 350 days.



      when I think logically in our situation, I can deduce that the probability of getting a positive output value ($Y = 1$) require to have small duration task.



      But I need to justify my opinion with some plots



      I have tried the following source code but It doesn't represent correctly my assumption. 



      #LoadData
      

      
min_duration = plot_data['Duration'].min()
      
max_duration = plot_data['Duration'].max()
      
xr_ = list(range(min_duration,  max_duration,  5))
      
y_ = 
      

      
for i in range(0,(len(xr_)-1)):
      
    a_ = np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1])
      
    b_ = np.logical_and(np.logical_and(plot_data['Duration'].values >= xr_[i], plot_data['Duration'].values < xr_[i+1]), plot_data['output'].values==1)
      
    y_.append(sum(b_)/sum(a_))
      

      
import matplotlib
      
matplotlib.pyplot.plot(xr_[1:len(xr_)], y_, 'o')


      Based on my previous assumption I must get a plot which contains an exponential form like :



      enter image description here



      But I have got contrary the following plot:



      enter image description here



      I want to know where I have a mistake and If there is any other method to justify my assumption






      python dataset matplotlib




      python dataset matplotlib






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 5 mins ago









      Siong Thye Goh

      1,438620




      1,438620










      asked 7 hours ago









      NirmineNirmine

      537




      537






















          1 Answer
          1






          active

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          0












          $begingroup$

          The model you are looking for is this:



          $Y=A e^{-Bx} + c$, I know the implementation in R, because I don't know of a nonlinear estimation in Python



          This code in R might work:



          R=data.frame(X=c(1,2,3,4,5,6,7,8,9),Y=c(1,2,3,3,3,3,3,3,3)) # Data in which X, Y are your data
          
model=nls(formula = Y~A*exp(-B*X)+C,data=R)
          
summary(model)


          There is a limitation to take into account, is explained in this link, is summarized in the impossibility for all possible models to exist, the "most inside" model should be linear.



          First steps with Non-Linear Regression in R



          Singular Gradient Error in nls with correct starting values






          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you juan, I will try to look for the equivalent function in python .
            $endgroup$
            – Nirmine
            4 hours ago












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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          The model you are looking for is this:



          $Y=A e^{-Bx} + c$, I know the implementation in R, because I don't know of a nonlinear estimation in Python



          This code in R might work:



          R=data.frame(X=c(1,2,3,4,5,6,7,8,9),Y=c(1,2,3,3,3,3,3,3,3)) # Data in which X, Y are your data
          
model=nls(formula = Y~A*exp(-B*X)+C,data=R)
          
summary(model)


          There is a limitation to take into account, is explained in this link, is summarized in the impossibility for all possible models to exist, the "most inside" model should be linear.



          First steps with Non-Linear Regression in R



          Singular Gradient Error in nls with correct starting values






          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you juan, I will try to look for the equivalent function in python .
            $endgroup$
            – Nirmine
            4 hours ago
















          0












          $begingroup$

          The model you are looking for is this:



          $Y=A e^{-Bx} + c$, I know the implementation in R, because I don't know of a nonlinear estimation in Python



          This code in R might work:



          R=data.frame(X=c(1,2,3,4,5,6,7,8,9),Y=c(1,2,3,3,3,3,3,3,3)) # Data in which X, Y are your data
          
model=nls(formula = Y~A*exp(-B*X)+C,data=R)
          
summary(model)


          There is a limitation to take into account, is explained in this link, is summarized in the impossibility for all possible models to exist, the "most inside" model should be linear.



          First steps with Non-Linear Regression in R



          Singular Gradient Error in nls with correct starting values






          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you juan, I will try to look for the equivalent function in python .
            $endgroup$
            – Nirmine
            4 hours ago














          0












          0








          0





          $begingroup$

          The model you are looking for is this:



          $Y=A e^{-Bx} + c$, I know the implementation in R, because I don't know of a nonlinear estimation in Python



          This code in R might work:



          R=data.frame(X=c(1,2,3,4,5,6,7,8,9),Y=c(1,2,3,3,3,3,3,3,3)) # Data in which X, Y are your data
          
model=nls(formula = Y~A*exp(-B*X)+C,data=R)
          
summary(model)


          There is a limitation to take into account, is explained in this link, is summarized in the impossibility for all possible models to exist, the "most inside" model should be linear.



          First steps with Non-Linear Regression in R



          Singular Gradient Error in nls with correct starting values






          share|improve this answer









          $endgroup$



          The model you are looking for is this:



          $Y=A e^{-Bx} + c$, I know the implementation in R, because I don't know of a nonlinear estimation in Python



          This code in R might work:



          R=data.frame(X=c(1,2,3,4,5,6,7,8,9),Y=c(1,2,3,3,3,3,3,3,3)) # Data in which X, Y are your data
          
model=nls(formula = Y~A*exp(-B*X)+C,data=R)
          
summary(model)


          There is a limitation to take into account, is explained in this link, is summarized in the impossibility for all possible models to exist, the "most inside" model should be linear.



          First steps with Non-Linear Regression in R



          Singular Gradient Error in nls with correct starting values







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          Juan Esteban de la CalleJuan Esteban de la Calle

          76822




          76822












          • $begingroup$
            Thank you juan, I will try to look for the equivalent function in python .
            $endgroup$
            – Nirmine
            4 hours ago


















          • $begingroup$
            Thank you juan, I will try to look for the equivalent function in python .
            $endgroup$
            – Nirmine
            4 hours ago
















          $begingroup$
          Thank you juan, I will try to look for the equivalent function in python .
          $endgroup$
          – Nirmine
          4 hours ago




          $begingroup$
          Thank you juan, I will try to look for the equivalent function in python .
          $endgroup$
          – Nirmine
          4 hours ago


















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