Count number of specific values in matrix

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Corey Bullard
Corey Bullard 2012 年 5 月 2 日
コメント済み: Walter Roberson 2023 年 10 月 18 日
I have a large matrix, m, and am trying to count the number of a specific value (i.e. How many indexes are of the value 4?)
I tried using
val = sum(m == 4);
but I end up with val being a matrix/vector of numbers. I assume these numbers are from each column and should be added together for the total, so I tried another
num = sum(val == 4);
but then I just end up with another vector/matrix.
How can I do it?
  2 件のコメント
Dev
Dev 2023 年 10 月 18 日
What if the value is zero?, how do I count then?
Walter Roberson
Walter Roberson 2023 年 10 月 18 日
num = sum(val == 0, 'all'); %r2018b or later
num = sum(val(:) == 0); %any version
num = nnz(~val); %any version

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採用された回答

Walter Roberson
Walter Roberson 2012 年 5 月 2 日
sum(m(:) == 4)
  2 件のコメント
Romy Wolstencroft
Romy Wolstencroft 2019 年 8 月 22 日
This worked perfectly for me. Thank you
MathWorks Support Team
MathWorks Support Team 2020 年 9 月 2 日
An alternative syntax available in R2018b or later is sum(m==4,'all'). But for this simple problem colonizing the input with m(:) is likely to be faster.

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その他の回答 (6 件)

Kye Taylor
Kye Taylor 2012 年 5 月 2 日
Try this:
numberOfNonZeros = nnz(m==4);
Using nnz is more efficient than converting logicals to numeric, which is required to apply sum()
  1 件のコメント
Walter Roberson
Walter Roberson 2019 年 8 月 22 日
編集済み: Walter Roberson 2019 年 8 月 22 日
In the test I just did, the timings of sum() vs nnz() could not consistently tell the two cases apart. nnz() might possibly have been slightly faster, but the range of timings showed so much overlap that no real conclusion could be reached. It would make sense that nnz() could be faster, but I can't prove it at the moment. sum() on a large enough array could be dispatched to LAPACK after all.

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Sean de Wolski
Sean de Wolski 2012 年 5 月 2 日
This could be done easily with histc() and unique() to get the number of each value:
uv = unique(x);
n = histc(x,uv);
Or with unique() and accumarray():
[uv,~,idx] = unique(x);
n = accumarray(idx(:),1)
  2 件のコメント
Royi Avital
Royi Avital 2022 年 10 月 10 日
Pay attention that histcount() won't have the same result as histc() above for this case (Difference at the end).
Walter Roberson
Walter Roberson 2022 年 10 月 11 日
Royi is correct.
At the time the question was asked, histcounts did not exist.
The newer histcounts is recommended instead of histc()
In the where you pass in the bin edges, then histc() counts values that exactly match the upper limit separately, but histcounts counts them together with the previous bin.

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ntsh kr
ntsh kr 2017 年 10 月 12 日
編集済み: ntsh kr 2017 年 10 月 12 日
>> a
a =
5 5 5 5 5 5 5 6 9 96
5 3 9 5 2 7 5 6 2 1
8 3 6 9 8 7 5 1 6 9
>> ans1=sum(a==5)
ans1 =
2 1 1 2 1 1 3 0 0 0
>> b=sum(ans1)
b =
11
  1 件のコメント
Manoj Payani
Manoj Payani 2018 年 5 月 16 日
Many Thanks - It works perfect.

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dipanka tanu sarmah
dipanka tanu sarmah 2017 年 11 月 11 日
along with this if you want to count the number of NaN ,(if there any) use nnz(isnan(m))

vimal kumar chawda
vimal kumar chawda 2020 年 5 月 18 日
But if we want ot do for NaN and any numeric value in large matrix then ?
ans1=sum(a==5) so at this my value is numerical (which is not same all time) and other is NaN which is common. But i need to count only numerical value at particular value of x.,x2,x3...............x7000 which is on y axis.
-How many times y appear on the at particular value of x?

Patrick Benz
Patrick Benz 2021 年 4 月 2 日
How can I count the values in the second column of an array depending on the values in the column?
I've got an array that looks something like that:
400 0
396 0
392 1
400 0
396 1
400 1
and I want to know how often there is a "1" or a "0" next to a "400" or next to the other values.
but this only gives me the total numbers of "1" and "0" and how often there is a 392 in the first column.
  4 件のコメント
Corey Bullard
Corey Bullard 2021 年 4 月 2 日
Another year, another answer to this decade old question.

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