why does data mean(data, 1) give differnt answer form finding mean of individual columns ,like mean_clm = mean(data( : ,1)) ?

why ddoes data mean(data) give differnt answer form finding mean of individual column ,like mean_clm = mean(data( : ,1)) ?

[Stephen23]

"why ddoes data mean(data) give differnt answer form finding mean of individual column ,like mean_clm = mean(data( : ,1)) ?"

Does it? Lets check:

data = rand(123,7);
M1 = mean(data);
M2 = mean(data(:,1));
isequal(M1(:,1),M2)

ans = logical
   1

You should always specify the dimension argument, because the dimension which MEAN() operates along changes depending on the size of the input array. Basically if you want robust code do not use MEAN() with one input.

Answer 1

[Dyuman Joshi]

"why does data mean(data, 1) give differnt answer form finding mean of individual columns, like meancl = mean(data( : ,1)) ?"

Because both the commands are different, and thus give different outputs.

Individual column(s) is/are different than a dimension

mean(data(:,k))

This is to calculate the mean of the kth column. k can be a scalar or a vector of values.

Whereas

mean(data,k)

returns the mean along the kth dimension.

Take a look at the documentation of mean to understand more.

[Celestie Gladys]

@Dyuman Joshi also when I put mean(data), it shows diferrent values along each column , what can I do? , should I calculate the mean vales along each column induvidually ?

[Dyuman Joshi]

You can see both commands return the same output, below

data = randi(20,4,5)

data = 4×5

20 6 5 20 1 5 10 8 19 4 16 12 8 13 20 1 20 8 3 11

mean(data)

ans = 1×5

10.5000 12.0000 7.2500 13.7500 9.0000

mean(data,1)

ans = 1×5

10.5000 12.0000 7.2500 13.7500 9.0000

"... what can I do?"

What are you trying to do?

[Celestie Gladys]

@Dyuman Joshi yes mean(data( :,1) showes diferrent value when compare to the first element of the whole mean(data) .

[Stephen23]

"When you don't input the dimension to the command, MATLAB assumes k as 1."

The documentation actually states "mean(A) returns the mean of the elements of A along the first array dimension whose size does not equal 1." This might be dimension 1, or might be another dimension.

" mean(data( :,1) showes diferrent value when compare to the first element of the whole mean(data) ."

Please show us exactly how you are checking this. (my guess is that this is a FORMAT issue)

[Dyuman Joshi]

You are correct, Stephen. My attempt at generalization is incorrect. I'll edit my comment.

[Dyuman Joshi]

"yes mean(data( :,1) showes diferrent value when compare to the first element of the whole mean(data) ."

Are you sure? @Celestie Gladys

data = randi(20,4,5)

data = 4×5

16 10 8 16 8 17 6 18 3 15 4 9 6 2 19 6 20 17 1 15

val1 = mean(data(:,1))

val1 = 10.7500

val2 = mean(data,1);
val2(1)

ans = 10.7500

[Celestie Gladys]

@Stephen23@Dyuman Joshi I have a 30720x5 data, when I do mean(data), the values are 24.008354 -0.50000000 907.66724 -1.5500033 412.95447 , but when i do mean(data( : , 1)), the data is -1.4285823, which is different

and so are the 2,3,4,5, which are 346.57669 , 2239.3198... and so on ..... am i missing somthing ?

[Dyuman Joshi]

Can you please attach your data, using the paperclip button?

[Dyuman Joshi]

Looks good to me, @Celestie Gladys

load('imu.mat')
out1 = mean(data1(:,1))

out1 = single 24.0084

out2 = mean(data1);
out2(1)

ans = single 24.0084

[Celestie Gladys]

@Dyuman Joshi thank you i was trying to run diferrent data , sorry for the time taken .

Answer 2

[Walter Roberson]

For future reference:

If mean() detects that the amount of data is below some internal size threshold, then it runs a straight forward internal loop summing the values along the dimension strictly from beginning to end, and then dividing by the size.

However if mean detects that the array is "large enough" it runs the request through a high performance multi-core library. The library might split the data up into pieces and compute the sum of each piece, and then the overall sum. Algebraically the results are the same, but there might be differences in floating point.

Suppose for example that you had an array that contained 1 exactly, then nine zeros, then 10 cases of 1e-17. If you run this forward in order, the 1 exactly carries through the first half. Then you have 1 exactly plus 1e-17, but 1e-17 is too small relative to 1 exactly to make a difference in double precision, so the sum would still be 1 exactly. And that would carry through the rest of the 1e-17, so the overall sum in strict order would be 1 exactly.

Now suppose you split the array into halves and sum each half. The 1 followed by zeros still adds to 1 exactly. But the ten 1e-17 are not negligible compared to each other, so that half would sum to approximately 1e-16. And 1e-16 is not negligible compared to 1 exactly, so when you add the two partial sums the result is not exactly 1.

When you mean() an entire array over the first dimension then the size of the entire array is used to decide whether to use the high performance routines. But when you select a single column and mean that, it is the size of the column by itself that is used, and since that is smaller it could lead to a different decision about whether to use the high performance routines.

In most cases, the difference between these possibilities is pretty small, but in worst case it can be arbitrarily large compared to the magnitude of the values, if there is a lot of accumulated round-off error.

From time to time it can lead to problems, with means that you "know" should be non-negative showing up as slightly negative, which gets to be a problem if you are using sqrt() for example. Different order of addition of floating point numbers can lead to confusing results because of round-off.

This is not the situation you happened to be facing, but Yes, you can end up with different results for the different calls.