Compute average of a column in a cell without considering NaN values

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Maria
Maria 2014 年 7 月 28 日
編集済み: Maria 2014 年 7 月 28 日
I have a cell array with 3 columns and 130000 rows:
A={56276 2009 '-'
56278 2009 NaN
56281 2009 NaN
56285 2009 33.9
56301 2009 '-'
56313 2009 42.5}
I want to have the mean of the values of the third column. If I use the following code,
mean(cell2mat(A(:,3)))
I get an error because I am inlcuding 'NaN' and '-' elements to do the average and I only want to do the average of the numbers.
Can someone help me? Thanks
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Andrew Reibold
Andrew Reibold 2014 年 7 月 28 日
編集済み: Andrew Reibold 2014 年 7 月 28 日
Do you mean 3 columns and 130,000 rows as is begun in your example?
And I am assuming 42,5 should be 42.5
Maria
Maria 2014 年 7 月 28 日
Yes I am sorry. I will edit the question

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採用された回答

Azzi Abdelmalek
Azzi Abdelmalek 2014 年 7 月 28 日
編集済み: Azzi Abdelmalek 2014 年 7 月 28 日
A={56276 2009 '-'
56278 2009 NaN
56281 2009 NaN
56285 2009 33.9
56313 2009 42.5}
c3=cell2mat(A(2:end,3))
out=nanmean(c3)
%or
out=mean(c3(~isnan(c3)))
  2 件のコメント
Maria
Maria 2014 年 7 月 28 日
But the '-' appears several times along the column, not only in the first row!
Azzi Abdelmalek
Azzi Abdelmalek 2014 年 7 月 28 日
c3=A(cellfun(@(x) ~isnan(x) & x~='-',A(:,3)),3)
out=mean(cell2mat(c3))

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その他の回答 (2 件)

Image Analyst
Image Analyst 2014 年 7 月 28 日
It may not be the most compact way, but it's simple and easy to understand:
clc;
A={56276 2009 '-';...
56278 2009 NaN;...
56281 2009 NaN;...
56285 2009 33.9;...
56301 2009 '-';...
56313 2009 42.5}
col3 = A(:,3)
theSum = 0;
counter = 0;
for row = 1 : size(col3, 1)
if isnumeric(col3{row}) & ~isnan(col3{row})
theSum = theSum + col3{row};
counter = counter + 1;
end
end
theMean = theSum / counter

Andrew Reibold
Andrew Reibold 2014 年 7 月 28 日
編集済み: Andrew Reibold 2014 年 7 月 28 日
Hi Maria,
Here is a rough example of how to get the mean value for EVERY column gathered in your matrix. Its definitely not the best practice, but I think it will work for you.
for clmn = 1:130000
rowsum= 0;
count = 0;
for row = 1:3
if isnumeric(A{rw,clmn}) & ~isnan(A{rw,clmn})
rowsum = rowsum+ A{rw,clmn}
count= count+1
end
end
MeanOfEachColumn(clmn) = rowsum/count;
end
Now if you want the mean of column three, you can just ask for MeanOfEachColumn(3). You can do this with any column you want now.
Variable names are annoyingly long for clarity. I would play with this and let me know if you get errors
I wrote this very quickly and with minimal testing. Let me know if you find walls you can't overcome.

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