How can I count the number of elements of a given value in a matrix?
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I have a matrix such as.
44.147231 ; 24.358619
44.461415 ; 24.118248
44.258173 ; 23.185014
43.166729 ; 24.100443
43.159002 ; 26.708013
42.954059 ; 27.338612
44.139759 ; 23.927882
44.424679 ; 25.301393
42.957588 ; 25.530434
44.615223 ; 24.200719
44.384409 ; 25.486735
43.965701 ; 28.454219
43.400971 ; 21.492123
42.068687 ; 22.303336
43.245372 ; 28.932286
44.593751 ; 23.635179
The first column is longitude , the second column is latitude.
So, I want to count the number of region.
In the above matrix,

But I don't know how make the code.
Please help me...
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その他の回答 (3 件)
Yona
2015 年 3 月 5 日
0 投票
after this run double loop, one on latitude and the second on longitude, and by using length(find(A(:,1)==i & A(:,2)==j)) you will get the number of each combination.
puneet joshi
2015 年 3 月 5 日
0 投票
well u can compare the complete matrix with the num u want to count .... and each time there is a match u can increase the count
Andrei Bobrov
2015 年 3 月 5 日
編集済み: Andrei Bobrov
2015 年 3 月 5 日
n = floor(min(a));
m = ceil(max(a));
[~,~,lo] = histcounts(a(:,1),n(1):m(1));
[~,~,la] = histcounts(a(:,2),n(2):m(2));
out = flip(accumarray([la,lo],1),1)
2 件のコメント
Jos (10584)
2015 年 3 月 5 日
Did you mean
[~, lo] = histc(..)
histc has been deprecated in R2014b. The replacement is histcounts which uses a different syntax and of course behaves slightly differently while still not solving all the issues of histc.
The syntax for histcounts is indeed the one Andrei used in his answer.
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