How to construct and plot the following accumulated list?












2












$begingroup$


I have a list, say it is 



a = {1,3,3.2,3.9,4,4.4,4.9,5,7,8}


To get the accumulate:



acc = Accumlate[a]


I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.



Then I want to take the derivative of the plot to plot it as well.






plotting list-manipulation







share|improve this question











$endgroup$








  • 1




    $begingroup$
    The plotting is not a problem see ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
    $endgroup$
    – Hugh
    Apr 13 at 11:26










  • $begingroup$
    The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[{Do[Print[{acc[[i]], Count[a[[i]]}], {i, 10}]}] but it only prints the pairs for me with an empty plot.
    $endgroup$
    – Hamza
    Apr 13 at 12:04










  • $begingroup$
    Your combined use of Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.
    $endgroup$
    – Roman
    Apr 13 at 16:29
















2












$begingroup$


I have a list, say it is 



a = {1,3,3.2,3.9,4,4.4,4.9,5,7,8}


To get the accumulate:



acc = Accumlate[a]


I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.



Then I want to take the derivative of the plot to plot it as well.






plotting list-manipulation







share|improve this question











$endgroup$








  • 1




    $begingroup$
    The plotting is not a problem see ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
    $endgroup$
    – Hugh
    Apr 13 at 11:26










  • $begingroup$
    The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[{Do[Print[{acc[[i]], Count[a[[i]]}], {i, 10}]}] but it only prints the pairs for me with an empty plot.
    $endgroup$
    – Hamza
    Apr 13 at 12:04










  • $begingroup$
    Your combined use of Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.
    $endgroup$
    – Roman
    Apr 13 at 16:29














2












2








2





$begingroup$


I have a list, say it is 



a = {1,3,3.2,3.9,4,4.4,4.9,5,7,8}


To get the accumulate:



acc = Accumlate[a]


I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.



Then I want to take the derivative of the plot to plot it as well.






plotting list-manipulation







share|improve this question











$endgroup$




I have a list, say it is 



a = {1,3,3.2,3.9,4,4.4,4.9,5,7,8}


To get the accumulate:



acc = Accumlate[a]


I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.



Then I want to take the derivative of the plot to plot it as well.






plotting list-manipulation




plotting list-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 13 at 16:26









Roman

5,62611131




5,62611131










asked Apr 13 at 11:22









HamzaHamza

225




225








  • 1




    $begingroup$
    The plotting is not a problem see ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
    $endgroup$
    – Hugh
    Apr 13 at 11:26










  • $begingroup$
    The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[{Do[Print[{acc[[i]], Count[a[[i]]}], {i, 10}]}] but it only prints the pairs for me with an empty plot.
    $endgroup$
    – Hamza
    Apr 13 at 12:04










  • $begingroup$
    Your combined use of Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.
    $endgroup$
    – Roman
    Apr 13 at 16:29














  • 1




    $begingroup$
    The plotting is not a problem see ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
    $endgroup$
    – Hugh
    Apr 13 at 11:26










  • $begingroup$
    The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[{Do[Print[{acc[[i]], Count[a[[i]]}], {i, 10}]}] but it only prints the pairs for me with an empty plot.
    $endgroup$
    – Hamza
    Apr 13 at 12:04










  • $begingroup$
    Your combined use of Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.
    $endgroup$
    – Roman
    Apr 13 at 16:29








1




1




$begingroup$
The plotting is not a problem see ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
$endgroup$
– Hugh
Apr 13 at 11:26




$begingroup$
The plotting is not a problem see ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
$endgroup$
– Hugh
Apr 13 at 11:26












$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[{Do[Print[{acc[[i]], Count[a[[i]]}], {i, 10}]}] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04




$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[{Do[Print[{acc[[i]], Count[a[[i]]}], {i, 10}]}] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04












$begingroup$
Your combined use of Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.
$endgroup$
– Roman
Apr 13 at 16:29




$begingroup$
Your combined use of Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.
$endgroup$
– Roman
Apr 13 at 16:29










2 Answers
2






active

oldest

votes


















1












$begingroup$

Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:



ListLinePlot[Transpose[{Accumulate[a], Range[Length[a]]}]]


If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].



enter image description here



If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], {x, 1, Total[a]}] and the first-derivative plot with



b = Interpolation[Transpose[{Accumulate[Sort[a]], Range[Length[a]]}], 

      InterpolationOrder -> 1];

Plot[b'[x], {x, 1, Total[a]}]


enter image description here






share|improve this answer











$endgroup$





















    3












    $begingroup$

    try this 



    a = {1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8};
    
acc = Accumulate@a;
    
aa[x_] := Length@Select[a, # <= a[[x]] &];
    
list = Transpose[{acc, aa /@ Range@Length@a}]
    
ListLinePlot@list



    {{1,1},{4,2},{7.2,3},{11.1,4},{15.1,5},{19.5,6},{24.4,7},{29.4,8},{36.4,9},{44.4,10}}




    enter image description here






    share|improve this answer











    $endgroup$













    • $begingroup$
      The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
      $endgroup$
      – Hamza
      Apr 13 at 13:57










    • $begingroup$
      @Hamza fixed....
      $endgroup$
      – J42161217
      Apr 13 at 14:12












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:



    ListLinePlot[Transpose[{Accumulate[a], Range[Length[a]]}]]


    If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].



    enter image description here



    If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], {x, 1, Total[a]}] and the first-derivative plot with



    b = Interpolation[Transpose[{Accumulate[Sort[a]], Range[Length[a]]}], 
    
      InterpolationOrder -> 1];
    
Plot[b'[x], {x, 1, Total[a]}]


    enter image description here






    share|improve this answer











    $endgroup$


















      1












      $begingroup$

      Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:



      ListLinePlot[Transpose[{Accumulate[a], Range[Length[a]]}]]


      If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].



      enter image description here



      If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], {x, 1, Total[a]}] and the first-derivative plot with



      b = Interpolation[Transpose[{Accumulate[Sort[a]], Range[Length[a]]}], 
      
      InterpolationOrder -> 1];
      
Plot[b'[x], {x, 1, Total[a]}]


      enter image description here






      share|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:



        ListLinePlot[Transpose[{Accumulate[a], Range[Length[a]]}]]


        If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].



        enter image description here



        If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], {x, 1, Total[a]}] and the first-derivative plot with



        b = Interpolation[Transpose[{Accumulate[Sort[a]], Range[Length[a]]}], 
        
      InterpolationOrder -> 1];
        
Plot[b'[x], {x, 1, Total[a]}]


        enter image description here






        share|improve this answer











        $endgroup$



        Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:



        ListLinePlot[Transpose[{Accumulate[a], Range[Length[a]]}]]


        If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].



        enter image description here



        If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], {x, 1, Total[a]}] and the first-derivative plot with



        b = Interpolation[Transpose[{Accumulate[Sort[a]], Range[Length[a]]}], 
        
      InterpolationOrder -> 1];
        
Plot[b'[x], {x, 1, Total[a]}]


        enter image description here







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Apr 13 at 16:30

























        answered Apr 13 at 16:18









        RomanRoman

        5,62611131




        5,62611131























            3












            $begingroup$

            try this 



            a = {1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8};
            
acc = Accumulate@a;
            
aa[x_] := Length@Select[a, # <= a[[x]] &];
            
list = Transpose[{acc, aa /@ Range@Length@a}]
            
ListLinePlot@list



            {{1,1},{4,2},{7.2,3},{11.1,4},{15.1,5},{19.5,6},{24.4,7},{29.4,8},{36.4,9},{44.4,10}}




            enter image description here






            share|improve this answer











            $endgroup$













            • $begingroup$
              The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
              $endgroup$
              – Hamza
              Apr 13 at 13:57










            • $begingroup$
              @Hamza fixed....
              $endgroup$
              – J42161217
              Apr 13 at 14:12
















            3












            $begingroup$

            try this 



            a = {1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8};
            
acc = Accumulate@a;
            
aa[x_] := Length@Select[a, # <= a[[x]] &];
            
list = Transpose[{acc, aa /@ Range@Length@a}]
            
ListLinePlot@list



            {{1,1},{4,2},{7.2,3},{11.1,4},{15.1,5},{19.5,6},{24.4,7},{29.4,8},{36.4,9},{44.4,10}}




            enter image description here






            share|improve this answer











            $endgroup$













            • $begingroup$
              The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
              $endgroup$
              – Hamza
              Apr 13 at 13:57










            • $begingroup$
              @Hamza fixed....
              $endgroup$
              – J42161217
              Apr 13 at 14:12














            3












            3








            3





            $begingroup$

            try this 



            a = {1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8};
            
acc = Accumulate@a;
            
aa[x_] := Length@Select[a, # <= a[[x]] &];
            
list = Transpose[{acc, aa /@ Range@Length@a}]
            
ListLinePlot@list



            {{1,1},{4,2},{7.2,3},{11.1,4},{15.1,5},{19.5,6},{24.4,7},{29.4,8},{36.4,9},{44.4,10}}




            enter image description here






            share|improve this answer











            $endgroup$



            try this 



            a = {1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8};
            
acc = Accumulate@a;
            
aa[x_] := Length@Select[a, # <= a[[x]] &];
            
list = Transpose[{acc, aa /@ Range@Length@a}]
            
ListLinePlot@list



            {{1,1},{4,2},{7.2,3},{11.1,4},{15.1,5},{19.5,6},{24.4,7},{29.4,8},{36.4,9},{44.4,10}}




            enter image description here







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Apr 13 at 14:12

























            answered Apr 13 at 12:45









            J42161217J42161217

            4,598324




            4,598324












            • $begingroup$
              The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
              $endgroup$
              – Hamza
              Apr 13 at 13:57










            • $begingroup$
              @Hamza fixed....
              $endgroup$
              – J42161217
              Apr 13 at 14:12


















            • $begingroup$
              The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
              $endgroup$
              – Hamza
              Apr 13 at 13:57










            • $begingroup$
              @Hamza fixed....
              $endgroup$
              – J42161217
              Apr 13 at 14:12
















            $begingroup$
            The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
            $endgroup$
            – Hamza
            Apr 13 at 13:57




            $begingroup$
            The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
            $endgroup$
            – Hamza
            Apr 13 at 13:57












            $begingroup$
            @Hamza fixed....
            $endgroup$
            – J42161217
            Apr 13 at 14:12




            $begingroup$
            @Hamza fixed....
            $endgroup$
            – J42161217
            Apr 13 at 14:12


















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