How to retrieve the indices of the values in each bin?

조회 수: 40 (최근 30일)
Lu Da Silva
Lu Da Silva 2021년 10월 10일
댓글: Walter Roberson 2021년 10월 28일
I have a histogram with values from a vector A , spread into 30 bins. How can I get the indices of the values in A, that correspond to each bin?
Example:
A = [ 4 6 8 2 5 3 3]
bin number 4 contains [4 3 3]
so I want to have a vecor B containing the indices B = [1 6 7]

답변 (1개)

Walter Roberson
Walter Roberson 2021년 10월 10일
The code could be slightly simpler if all of the bins were only one value wide.
A = randi(60, 1, 50)
A = 1×50
36 10 40 43 35 52 52 43 21 44 23 47 8 31 46 19 59 56 47 52 46 47 58 59 25 48 31 59 53 13
edges = [1:2:60, inf]
edges = 1×31
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59
[counts, ~, bins] = histcounts(A, edges)
counts = 1×30
1 1 0 2 1 1 1 1 0 2 2 1 3 1 1 3 0 2 0 1 1 3 2 4 1 4 2 2 4 3
bins = 1×50
18 5 20 22 18 26 26 22 11 22 12 24 4 16 23 10 30 28 24 26 23 24 29 30 13 24 16 30 27 7
B = accumarray(bins(:), reshape(1:numel(bins), [], 1), [], @(V){V.'})
B = 30×1 cell array
{[ 48]} {[ 37]} {0×0 double} {[ 13 45]} {[ 2]} {[ 33]} {[ 30]} {[ 32]} {0×0 double} {[ 16 46]} {[ 9 50]} {[ 11]} {[25 31 43]} {[ 42]} {[ 39]} {[14 27 35]}
  댓글 수: 3
Walter Roberson
Walter Roberson 2021년 10월 28일
If accumarray() is giving you that error, then it implies that some value in your matrix A is less than the first value in your edges vector or greater than the last one. For example,
A = randi([-2 60], 1, 50)
A = 1×50
51 59 42 31 0 39 6 59 48 56 43 30 -2 8 25 45 48 2 59 0 49 57 0 17 9 16 11 56 46 50
edges = [1:2:60, inf]
edges = 1×31
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59
[counts, ~, bins] = histcounts(A, edges)
counts = 1×30
2 0 2 1 2 1 0 3 1 0 1 1 1 0 2 1 0 1 0 2 3 1 4 3 2 3 1 3 1 3
bins = 1×50
26 30 21 16 0 20 3 30 24 28 22 15 0 4 13 23 24 1 30 0 25 29 0 9 5 8 6 28 23 25
B = accumarray(bins(:), reshape(1:numel(bins), [], 1), [], @(V){V.'})
Error using accumarray
First input SUBS must contain positive integer subscripts.
See how the 5th bin number is 0, which corresponds to the bin for the input value 0 in A, and 0 is before the first value in edges . (It is not because A has negative or 0 entries: it is strictly to do with the fact that it has entries that are outside the range of the edges list.)
Walter Roberson
Walter Roberson 2021년 10월 28일
Note that in the following code, any value in A that is outside the range of the edges will not have its index appear anywhere in B.
A = randi([-2 60], 1, 50)
A = 1×50
58 21 4 42 56 -2 4 20 34 29 40 26 58 4 60 -2 29 6 -2 44 38 28 32 33 28 52 9 1 8 33
edges = [1:2:60, inf]
edges = 1×31
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59
[counts, ~, bins] = histcounts(A, edges)
counts = 1×30
2 5 2 2 1 1 2 1 0 1 1 0 1 2 3 2 5 0 2 1 1 1 2 1 1 1 0 2 2 1
bins = 1×50
29 11 2 21 28 0 2 10 17 15 20 13 29 2 30 0 15 3 0 22 19 14 16 17 14 26 5 1 4 17
valididx = reshape(find(bins), [], 1);
B = accumarray( reshape(bins(valididx), [], 1), valididx, [], @(V){V.'})
B = 30×1 cell array
{[ 28 39]} {[3 7 14 31 37]} {[ 18 38]} {[ 29 47]} {[ 27]} {[ 46]} {[ 34 50]} {[ 42]} {0×0 double } {[ 8]} {[ 2]} {0×0 double } {[ 12]} {[ 22 25]} {[ 10 17 44]} {[ 23 40]}

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Histograms에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by