min/max values in 3 dimensional arrays

5 次查看（过去 30 天）
Mike Wilson
Mike Wilson 2017-3-7
评论： Walter Roberson 2017-3-9
I have a 3D array in this form. [date, occurrence count per date, observed data]
example:
[15, 1, 5.23; 15, 2, 6.33; 15, 3, 4.67; 15, 4, 5.63; 16, 1, 4.98; 16, 2, 7.12; ... ]
For each date, i need to find the max number of occurrences. I can find the global min/max easily and can write loop logic to go through the matrix to find the local (per day) maximum number of occurrence, but so far can't figure out how to use find() or max() to do this. I've gotta think there's an easy/fast way.
Thanks for any help.
Mike
  4 个评论
显示 2更早的评论
Jan
Jan 2017-3-7
But when all columns are treated equally by using A(:), the detail "For each date, i need to find the max number of occurrences" is not considered, but you build statistics over "[date, occurrence count per date, observed data]" together.
Star Strider
Star Strider 2017-3-7
@Jan — You are absolutely correct.
Retaining the structure of the original matrix is essential to the correct solution to this problem, as Stephen’s and my simultaneous and nearly identical approaches illustrate.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

John BG
John BG 2017-3-7
Hi Mike
the command histogram generates all you need for this question:
1. your data
A=[15, 1, 5.23; 15, 2, 6.33; 15, 3, 4.67; 15, 4, 5.63; 16, 1, 4.98; 16, 2, 7.12]
.
2. extrude data from 3D to 1D
A2=A(:)
.
3. finding out amount of different occurrences
nbins=numel(unique(A2))
.
4.
h1=histogram(A2,nbins)
.
the data, as pulled out from A into A2 is contained in
h1.Data
=
15.0000
15.0000
15.0000
15.0000
16.0000
16.0000
1.0000
2.0000
3.0000
4.0000
1.0000
2.0000
5.2300
6.3300
4.6700
5.6300
4.9800
7.1200
checking amounts
isequal(numel(unique(h1.Data)),numel(unique(A2)))
=
1
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
<mailto:jgb2012@sky.com jgb2012@sky.com>
  6 个评论
显示 4更早的评论
Walter Roberson
Walter Roberson 2017-3-9
The required output is
15 4
16 2
That is, there are two different dates, and the maximum occurrence number is needed for each.
Your code does not make appropriate computations. It would, for example, fail on
[15 16 0.123;
16 3 0.123
the required output for which would be
15 16
16 3
whereas your code would produce a histogram showing two copies of 0.123, one copy of 3, one copy of 15, and two copies of 16 -- information that is of no relevance to the question.
Walter Roberson
Walter Roberson 2017-3-9
"the max number of occurrences is right the first value repeated the max amount of time of the vector h1.Data"
No, not with your code it is not.
First off, you assume that the maximum number of occurrences is the same as the number of occurrences of the same number that is the date. That is not given in any respect: we are not told that the occurrence numbers will be 1, 2, 3, ... up to the maximum. We are asked to find the maximum of the numbers that do occur, per date.
Even if it is the case that the occurrence numbers occur in sequence starting from 1 for any given date, consider data that looks like,
x = -1:
[3 1 x;
3 2 x;
3 3 x;
3 4 x;
4 1 x;
4 2 x;
4 3 x]
The histogram that would be produced for this would be
-1 -- 7 times
1 -- 2 times
2 -- 2 times
3 -- 6 times
4 -- 4 times
when the desired output would be
3 -- maximum occurrence 4
4 -- maximum occurrence 3
Your code confused occurrence numbers and dates, and that's not even considering the possibility of non-sequential occurrence numbers.

请先登录，再进行评论。

更多回答（2 个）

Stephen23
Stephen23 2017-3-7
编辑：Stephen23 2017-3-7
Perhaps using accumarray:
>> X = [15, 1, 5.23; 15, 2, 6.33; 15, 3, 4.67; 15, 4, 5.63; 16, 1, 4.98; 16, 2, 7.12];
>> [uni,~,idx] = unique(X(:,1));
>> uni % these are the days
uni =
15
16
>> accumarray(idx,X(:,2),[],@max) % occurrences for each day
ans =
4
2
you can calculate other "per day" statistics easily, e.g.:
>> accumarray(idx,X(:,3),[],@mean) % mean of each day
ans =
5.4650
6.0500
etc
Star Strider
Star Strider 2017-3-7
It’s a 2D array. It has three columns.
One approach:
A = [15, 1, 5.23; 15, 2, 6.33; 15, 3, 4.67; 15, 4, 5.63; 16, 1, 4.98; 16, 2, 7.12;];
[Au,ia,ic] = unique(A(:,1));
DateMax = accumarray(ic, A(:,2), [], @max);
Result = [Au DateMax]
Result =
15 4
16 2
  2 个评论
Mike Wilson
Mike Wilson 2017-3-7
Roger that (on 3 column, 2D array). Great answer. Didn't know of those useful sorting/stats functions. Thanks!
Star Strider
Star Strider 2017-3-7
My pleasure!
A vote would be appreciated.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Data Distribution Plots 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by