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How do I exclude NaN values when calculating mean of each row in a matrix?

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gpd=[2014,3.320,3.364,3.532,3.659,3.691,3.695,3.633,3.481,3.403,...
3.182,2.887,2.560;2015,2.110,2.249,2.483,2.485,2.775,2.832,...
2.832,2.679,2.394,2.289,2.185,2.060;2016,1.967,1.767,1.958,...
2.134,2.264,2.363,2.225,2.155,2.208,2.243,2.187,2.230;2017,...
2.351,2.299,2.323,2.418,2.386,2.337,2.281,NaN,NaN,NaN,NaN,NaN];
for a=gpd(:,end)
y=mean(a,2);
end
The output for the last row of y returns NaN (a copy of the command window is shown below). How do you exclude the NaN values in the last row in order to output an average of all the real number values?
>>y =
2.5600
2.0600
2.2300
NaN

Risposte (4)

Guillaume
Guillaume il 8 Set 2017
Since version 2015a, the max, min, mean, median, sum, var, std, and cov function have included a flag to ignore nans
y = mean(gpd, 2, 'omitnan')
Note that your loop makes no sense at all. It's averaging just the last column, so not doing any averaging at all. The line above will average all the columns.

José-Luis
José-Luis il 8 Set 2017
y = nanmean(a,2)
  1 Commento
Jason Stockton
Jason Stockton il 9 Ago 2021
Note: Matlab Help says nanmean() is not recommended. They recommend using mean() with the 'omitnan' flag like Guillaume shows.
They may plan to remove it in the future. Personally, I like nanmean better.

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OCDER
OCDER il 8 Set 2017
Modificato: OCDER il 8 Set 2017
Remove the for loop, as it only does the last column, which can't be averaged.
To take mean with NaN's in it, use José-Luis' suggestion of nanmean (voted your answer up :) ).
y = nanmean(gpd, 2)
This will return a 5x1 matrix of average of gdp for each row.
y =
158.0313
157.2595
157.0539
254.1744
  2 Commenti
Jillian Sweatt
Jillian Sweatt il 8 Set 2017
Will that still calculate the mean of the remaining numbers in the last row of gpd?
OCDER
OCDER il 8 Set 2017
Modificato: OCDER il 8 Set 2017
Actually, I'm looking at your code and this for loop may not be needed:
for a = gpd(:, end)
... %This only does one iteration, where a = last column.
end
In this case, José-Luis' suggestion of nanmean is correct.

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Changoleon
Changoleon il 5 Mar 2018
https://www.mathworks.com/help/stats/nanmean.html

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