Breaking a sorted list into bins of a specified size

Question

Can somebody please explain me the following:

If I try to split data every increase in 0.5:

list = Range[1.2, 3.2, 0.15]

{1.2, 1.35, 1.5, 1.65, 1.8, 1.95, 2.1, 2.25, 2.4, 2.55, 2.7, 2.85, 3., 3.15}

I would like to get the following:

{{1.2, 1.35, 1.5, 1.65}, { 1.8, 1.95, 2.1}, {2.25, 2.4, 2.55, 2.7}, {2.85,3., 3.15}

If I use the example given I here:

Splitting a list using SplitBy, by comparing adjacent elements

I can split either not at all

splittedlist = Split[list, (#1 < ( #2 - 0.1)) &]

 {{1.2, 1.35, 1.5, 1.65, 1.8, 1.95, 2.1, 2.25, 2.4, 2.55, 2.7, 2.85, 3., 3.15}}

or I can split each of them

splittedlist = Split[list, (#1 < ( #2 - 0.5)) &]

{{1.2}, {1.35}, {1.5}, {1.65}, {1.8}, {1.95}, {2.1}, {2.25}, {2.4}, {2.55}, {2.7}, {2.85}, {3.}, {3.15}}

I found this package,

http : // www.theophys.kth.se/~phl/Mathematica/

which uses exactly this method (on 2D data, but still), why does it work in their method? Obviously, I am missing something.

I found this way of doing it,

Calculate mean of values in bins

but is this really the simplest way to split into intervals? I could of course write a loop, but I tried to not use a loop.

Answer 1

This is built-in approach:

BinLists[#, {First@#, Last@# + .5, .5}] &@list

{{1.2, 1.35, 1.5, 1.65}, {1.8, 1.95, 2.1}, {2.25, 2.4, 2.55}, {2.7, 2.85, 3., 3.15}}

Answer 2

This question is a simpler variation of each of these questions:

• How to partition a list into sublists in a similar way to Histogram
• Partitioning a list of numbers the Mathematica way

I will not fight it being closed as a duplicate but perhaps there is value in the clarify of the simple answers this one permits.

Showing each of the methods from my earlier answer simplified as appropriate:

GatherBy

split[data_, width_] :=
  With[{offset = Mod[First @ data, width]},
    GatherBy[data, Floor[# - offset, width] &]
  ]

{{1.2, 1.35, 1.5, 1.65}, {1.8, 1.95, 2.1}, {2.25, 2.4, 2.55}, {2.7, 2.85, 3., 3.15}}

BinLists

BinLists[#, {First@#, Last@# + #2, #2}] &[list, 0.5]

{{1.2, 1.35, 1.5, 1.65}, {1.8, 1.95, 2.1}, {2.25, 2.4, 2.55}, {2.7, 2.85, 3., 3.15}}

This of course is the same as Kuba's answer, but it's hard to avoid this as it is the built-in function. Unlike the first method it will return empty bins, which may or may not be desirable. (The first method can be extended to handle this case as well if required, but it makes it significantly less clean which is why I did not write it that way by default.)

Answer 3

splitEvenly[list_List, step_] := SplitBy[list, Floor[Divide[# - First[list], step]] &]

splitEvenly[list, 0.5]
(* {{1.2, 1.35, 1.5, 1.65}, {1.8, 1.95, 2.1}, {2.25, 2.4, 2.55}, {2.7, 2.85, 3., 3.15}} *)

As Kuba already mentioned in his comment, 2.7 should be in the last sublist as it is 3 * 0.5 from the starting value.

If the starting value is not necessarily the value of the first element you could define it as:

splitEvenly[list_List, step_, start_: First[list]] := 
   SplitBy[list, Floor[Divide[# - start, step]] &]

giving you an optional third argument to specify a starting value which defaults to the first value of the list.

Answer 4

I;m guessing you want the bins end points to defined as I have defined binEnds. By doing so, I got the answer you say you want.

data = Range[1.2, 3.2, 0.15];
binWidth = .5;
binEnds = Range[data[[1]], data[[-1]], binWidth];
Module[{i = 1}, 
  Split[data, If[#2 - binEnds[[i]] <= binWidth, True, i++; False] &]]

{{1.2, 1.35, 1.5, 1.65}, {1.8, 1.95, 2.1}, {2.25, 2.4, 2.55, 2.7}, {2.85, 3., 3.15}}