confusion during finding out the mean of a 3d vector

I have a 3d vector of size 1498x1498x24. I want to calculate the wholes mean of the vector. I have used two method and got two different result. If anyone please explain this part that will be really helpful. I am giving the code and corresponding result.

mean(mmr_timeseries,'all','omitnan')

Using this i got the value 970.8025. Then i used :

mean(mean(mean(mmr_timeseries,'omitnan'),'omitnan'),'omitnan')

In this case I got 884.2371. I think these two lines are equivalent but I got different result. If ant=yone could please explain this part that will be really helpful. Thank you.

[KSSV]

Thy should be same

mmr_timeseries = rand(1498,1498,24) ;
idx = randperm(numel(mmr_timeseries,1000)) ; 
mmr_timeseries(idx) = NaN ;

mean(mmr_timeseries,'all','omitnan')

ans = 0.5000

mean(mean(mean(mmr_timeseries,'omitnan'),'omitnan'),'omitnan')

ans = 0.5000

[SWARNENDU PAL]

Thanks for your reponse sir. But i got different. I dont know what is the problem. Is there any probable reason for this?

[Bruno Luong]

See Chunru's answer

Answer 1

[Chunru]

A 2-d example (applicable to 3d in principle).

a=[1    2;  3   3;  nan     4;    nan   nan]

a = 4×2

1 2 3 3 NaN 4 NaN NaN

% mean of [1 2 3 3 4] after omitting nans
mean(a,'all','omitnan')

ans = 2.6000

% mean for each column (omit nans)
mean(a,'omitnan')

ans = 1×2

2 3

% mean of the above vector , i.e. [2, 3]
mean(mean(a,'omitnan'),'omitnan')

ans = 2.5000

% The following is for 3d (not shown in this example)
mean(mean(mean(a,'omitnan'),'omitnan'),'omitnan')

ans = 2.5000

[Walter Roberson]

This is correct, the NaN changes the calculations.

[SWARNENDU PAL]

yeah here also the same thing happened. The results are different in different method. Then which one should be used?

[Chunru]

It depends on what you want, as long as you understand the difference.

[Bruno Luong]

. Then which one should be used?

A=[0 1; NaN 1; NaN 1; NaN 1; NaN 1; NaN 1; NaN 1; NaN 1; NaN 1]

A = 9×2

0 1 NaN 1 NaN 1 NaN 1 NaN 1 NaN 1 NaN 1 NaN 1 NaN 1

mean(mean(A,'omitnan'),'omitnan')

ans = 0.5000

mean(mean(A,'all','omitnan'))

ans = 0.9000

[SWARNENDU PAL]

I think 2nd one is the correct and more meaningful as it gives the mean of all numbers.

Answer 2

[Jonas]

the short answer:

your second approach takes mean values of means and the first approach takes the mean value of all elements

[Walter Roberson]

Those work out the same except when there are nan that are being omitted.

[SWARNENDU PAL]

technically these two things are same i think. Taking mean of mean and find out the whole mean of a multidimensional vector.