How to calculate Average without receiving 'Inf' as a result.

I am trying to find the average efficiency of this 1 column in MATLAB. I have tried N = mean(EstimatedEff,"omitnan") but get 'Inf' asa result. How can I change my code so I recieve a numerical figure for average efficiency? I have attached the Excel sheet below for reference and Using 2020b. Thankyou!

Answer 1

[Sameer]

Hi Mahnoor

It could be due to the presence of Inf values in your "EstimatedEff" array. The "omitnan" option only ignores NaN values, not Inf. To handle Inf values, you need to remove or replace them before calculating the mean.

You can modify the code as follows:

% Remove Inf values
EstimatedEff = EstimatedEff(~isinf(EstimatedEff));
% Calculate the mean, omitting NaN values
N = mean(EstimatedEff, 'omitnan');

Please refer to the below MathWorks documentation link:

https://www.mathworks.com/help/releases/R2020b/matlab/ref/isinf.html

Hope this helps!

[Steven Lord]

That would work if EstimatedEff is a vector or if you want the mean over the whole array. I'd probably use standardizeMissing to replace the Inf values with NaN and then use mean with 'omitnan' on that array (including specifying a dimension if necessary.)

EstimatedEff = randi(10, 5, 5);
nonfinitelocations = randperm(25, 10);
EstimatedEff(nonfinitelocations(1:4)) = NaN;
EstimatedEff(nonfinitelocations(5:7)) = Inf;
EstimatedEff(nonfinitelocations(8:10)) = -Inf

EstimatedEff = 5×5

6 9 7 10 9 6 5 NaN 10 1 NaN 5 -Inf Inf 8 Inf NaN 1 -Inf NaN 10 -Inf 1 4 Inf

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Replace all the Inf and -Inf values with NaN

E = standardizeMissing(EstimatedEff, [Inf, -Inf])

E = 5×5

6 9 7 10 9 6 5 NaN 10 1 NaN 5 NaN NaN 8 NaN NaN 1 NaN NaN 10 NaN 1 4 NaN

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mean(E, "omitnan")

ans = 1×5

7.3333 6.3333 3.0000 8.0000 6.0000

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mean(E, 2, "omitnan")

ans = 5×1

8.2000 5.5000 6.5000 1.0000 5.0000

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mean(E, "all", "omitnan")

ans = 6.1333

Answer 2

[Star Strider]

Uz = unzip('Matlab Copy.zip')

Uz = 1x1 cell array
    {'Matlab Copy.csv'}

T1 = readtable(Uz{1}, 'VariableNamingRule','preserve')

T1 = 674835x1 table
    Estimated Eff.
    ______________

          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       
          0       

NonZeros = nnz(T1{:,1})

NonZeros = 580446

NrInf = nnz(isinf(T1{:,1}))

NrInf = 29

NrNaN = nnz(isnan(T1{:,1}))

NrNaN = 0

NrFinite = nnz(isfinite(T1{:,1}))

NrFinite = 674806

mean_of_finite_values = mean(T1{isfinite(T1{:,1}),:})

mean_of_finite_values = 0.8364

locs = isfinite(T1{:,1});                                   % Alternative Approach
NrNonfinite  = nnz(~locs)

NrNonfinite = 29

mean_of_finite_values = mean(T1{locs,:})

mean_of_finite_values = 0.8364

Of the 674835 values in the file, there are 0 NaN values, 29 are Inf and 674506 are finite.

.

[Star Strider]

Using the 'omitnan' flag is absolutely pointless in this instance!

The reason is that there are no NaN values at all, only Inf and finite values!

Answer 3

[Aquatris]

Here is one way:

% create dummy matrix 
A = rand(100,1);
A([5 10 24 55 23 60]) = inf;
A([12 56 29 35]) = nan;

mean(A) % simple mean

ans = NaN

mean(A,'omitnan') % omitting nan s

ans = Inf

mean(A(~isinf(A)),'omitnan') % omitting nan and inf -inf

ans = 0.4871

[Mahnoor]

Thankyou all so much for the support!