How to count continuous appearance of numeric values

Question

Mekala balaji 2016-11-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/313094-how-to-count-continuous-appearance-of-numeric-values

编辑： Guillaume 2016-11-20

Hi,

I have below matrix, and I want count (a) what is the maximum count that numerical appeared continuously, (2) total count of numerical (by column wise).

2      N/A
4      8
3      7
N/A    1
N/A    N/A
5      9
4      2
3      2
1      5
N/A    8
N/A   N/A

My desired output is (do not consider N/A):

maxCount: 
4 5 (column1 maximum numerical continuous appearance is 4, column2 maximum continuous numerical appearance is: 5)
totalCount:
7 8

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2016-11-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/313094-how-to-count-continuous-appearance-of-numeric-values#answer_243970

在 MATLAB Online 中打开

ncol = size(YourArray, 2);
mask = [true(1, ncol); isnan(YourArray); true(1, ncol)] .';
maxcount = zeros(1, ncol);
totalcount = zeros(1, ncol);
for col = 1 : ncol
   starts = strfind( mask(col, :), [1 0]);
   ends   = strfind( mask(col, :), [0 1]);
   maxcount(col) = max(ends-starts);
end
totalcount(:) = sum(~mask, 2);

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Guillaume 2016-11-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/313094-how-to-count-continuous-appearance-of-numeric-values#answer_243971

编辑：Guillaume 2016-11-20

在 MATLAB Online 中打开

I'm assuming the N/A are actually NaN, otherwise your question does not make sense as your input is not valid syntax in matlab.

m = [2 NaN; 4 8; 3 7; NaN 1; NaN; NaN; 5 9; 4 2; 3 2; 1 5; NaN 8; NaN NaN];

totalCount is easy:

totalCount = sum(~isnan(m));

For maxCount it's a little bit more complicated but you can obtain the information from a combination of isnan, find and diff. Unfortunately, find does not work by column so you have to loop over the whole matrix:

maxCount = cellfun(@(nancolumn) max(diff([0; find(nancolumn); size(m, 1)+1]) - 1), num2cell(isnan(m), 1))

The cellfun above is equivalent to:

maxCount = zeros(1, size(m, 2));
nanm = isnan(m);
for column = 1 : size(m, 2)
   nancolumn = nanm(:, column); 
   nanidx = find(nancolumn);
   runlengths = diff([0; nanidx; size(m, 1)+1] - 1;
   maxCount(1, column) = max(runlengths);
end