distribution of consecutive numbers

Danielle Leblance さんによって質問されました 2018 年 10 月 30 日

最新アクティビティ John D'Errico さんによってコメントされました 2018 年 10 月 30 日

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Hi ,

I have a column of 9 states (attached the file) . How can I know in this column how many times each state is followed by another state? in other words, the values ranges between 1 and 9. how can i know how many times 1 is followed by 1, 1 followed by 2, ...1 followed by 9 (the same goes for each state)?

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回答者: Guillaume 2018 年 10 月 30 日

編集済み: Guillaume 2018 年 10 月 30 日

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I don't think there's a way t do that without a loop (either explicit or with arrayfun).

%build list of patterns to search:
[n1, n2] = ndgrid(1:9);
patterns = [n1(:), n2(:)];
%search for each pattern with a loop:
x1 = reshape(x1, 1, []);  %need a row vector
count = zeros(size(patterns, 1), 1);
for row = 1:size(patterns, 1)
   count(row) = numel(strfind(x1, patterns(row, :)));  %despite its name strfind can be used to find patterns of numbers
end
%pretty display
table(patterns, count)

edit: actually I was completely wrong, it can easily be done without a loop, but temporarily requires much more memory:

[n1, n2] = ndgrid(1:9);
patterns = [n1(:), n2(:)];  
%count of pattern
transitions = permute([x1(1:end-1), x1(2:end)], [3 2 1]);
count = sum(all(patterns == transitions, 2), 3)
%pretty display
table(patterns, count)

edit again: Another option is to build the graph of the transitions and use the edgecount function:

g = graph(x1(1:end-1), x1(2:end));
[n1, n2] = ndgrid(1:9);
count = edgecount(g, n1(:), n2(:));
patterns = [n1(:), n2(:)];  
table(patterns, count)

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回答者: John D'Errico 2018 年 10 月 30 日

編集済み: John D'Errico 2018 年 10 月 30 日

Actually, easy. No loop needed, and just one line of code. You are essentially looking to find an array, I'll call it Counts, such that Counts(i,j) is the number of times the number i was followed by the number j in the vector x1. Pretty much two lines of code.

Counts = accumarray([x1(1:end-1),x1(2:end)],ones(numel(x1)-1,1))
Counts =
   5     2     3     0     0     0     0     0
  29     5     1     3     0     0     0     0
   3    33     0     0     3     0     0     1
   0     0    25     1     3    11     1     0
   4     0     5    32     6     0     3     0
   1     5     2     6    29     1     0     3
   0     0     5     2     0    36     6     6
   0     0     1     7     0     4    24     2
   0     0     0     0     6     3     4    28

So, a 1 was followed by a 1 exactly 25 times, but a 3 was followed by 9 EXACTLY once. The most common occurrence was the pair [7 7], which we saw 36 times. We can test quickly to verify that fact.

strfind(x1',[7 7])
ans =
Columns 1 through 31
  14    15    62    68    69    70    71    72    81    95    96   101   102   121   122   201   213   286   287   288   289   290   291   292   308   309   310   311   312   330   331
Columns 32 through 36
 332   333   338   339   347

Yep. 36 occasions where [7 7] was seen. Where did that lonely [3 9] pair arise?

strfind(x1',[3 9])
ans =
    33

If the vector x1 included 0 or negative numbers, we would need a second line, utilizing the function unique.

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Image Analyst 2018 年 10 月 30 日

Very nice and clever.

I think graycomatrix() could also do the same thing since this is it's reason for being - what the function was made for.

John D'Errico 2018 年 10 月 30 日

Yeah, I don't have the IPT, but it has many utilities that can be applied to non-imaging problems, so that I could arguably justify the IPT even if I never did image processing.

The only problem with the above code is if the vector included zeros or negatives. But then you just use unique first on it.

x = round(rand(1000,1)*8 - 4);
[xu,~,xrefs] = unique(x);
xu
xu =
    -4
    -3
    -2
    -1
     0
     1
     2
     3
     4
Counts = accumarray([xrefs(1:end-1), ...
   xrefs(2:end)],ones(numel(xrefs)-1,1))
Counts =
     2     8     4     6     9     6     6    11     2
     6    30    28    14     8    22    15    18     9
     9    17    11    20    18    15    16    13     1
     5    19    15    10    18    23    12    14     6
     7    22    10    21    15    13    18    17     9
    12    16    17    15    22    15    20     7     7
     5    17    17    19    20    16    19    11     6
     4    14    11    11    11    15    21    12     9
     4     7     7     6    11     6     3     5     3

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回答者: madhan ravi 2018 年 10 月 30 日

編集済み: madhan ravi 2018 年 10 月 30 日

load matlab.mat
u = unique(x1)
Expected_result = histc(x1,u)
consecutive_numbers_and_their_repetition = [u Expected_result]

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Guillaume 2018 年 10 月 30 日

Danielle Leblance comment mistakenly posted as an answer moved here:

? unique shows unique values and not how many times 1 is followed by 2 ,etc..

madhan ravi 2018 年 10 月 30 日

check edited answer

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