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.

  • 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 '19 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 '19 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 '19 at 16:29

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

| improve this answer | |

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}]


enter image description here

| improve this answer | |
  • $\begingroup$ The y-axis should represent the number of elements less than or equal a[[i]] not the element itself. $\endgroup$ – Hamza Apr 13 '19 at 13:57
  • $\begingroup$ @Hamza fixed.... $\endgroup$ – J42161217 Apr 13 '19 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.