Replace NaN with mean of previous values

Hello, I have a vector that contains some Nan's. How can I replace them with a mean of the previous 5 values? This is what I have tried until now:

 for i = 254:1:size(vector)
  vector(isnan(vector(i))) = mean(vector(i-5:i-1));
 end

and

 for i = 254:1:size(vector)
  if vector(i) == NaN
    vector(i) = mean(vector(i-5:i-1));
  end
 end

Thank you!

[Jan]

What do you want for the case of neighboring NaNs?

     [1 2 3 4 5 NaN NaN]
  => [1 2 3 4 5  3  3.5]  % Last is mean([2 3 4 5])
  or [1 2 3 4 5  3  3.4]  % Last is mean([2 3 4 5 3])

[Tomás Nunes]

the second!

[Tomás Nunes]

do you know what is wrong with the formula? Thank you

[Jan]

If the 2nd one is wanted, Ameer Hamza's answer does not, what you want.

Answer 1

[Ameer Hamza]

You can avoid for loop altogether

ind = find(isnan(vector));
vector(ind) = arrayfun(@(x) nanmean(vector(x-5:x-1)), ind);

a vectorized code is usually faster and at worst equivalent to loops. Also using arrayfun() makes your code more readable.

[Tomás Nunes]

I have tried this and it did work well but only for the first 5 Nan's. after that I still have a some rows with Nan's.

Thank you

[Ameer Hamza]

Can you give an example of your dataset? This method should replace all nan.

[Tomás Nunes]

yes! it is a 7567*1 vector. from 0 to 253 it is all zeros. from then onwards it is supposed to have non-zeros. however, from 5792 until 5843, and from 7023 until 7209, it contains Nan's. What that code did was populate from 5792 until 5797 and the same in the second period of nans.

[Tomás Nunes]

i mean, this is kinda dumb, right? it would be easier to set the nans to a stupid value like 100000 (which has nothing to do with the remaining values of the vector) and then replace them with the mean of the previous days. it would be quite easier, correct?

[Ameer Hamza]

Why do you need to set nan to 100000. You can directly replace them as I showed in my answer. Does your vector contain more the 5 consecutive nan, in that case, this method will not work. Also, an @Jan pointed out in the comment, if more then one nan occur closely (less than 5 elements apart) my answer do average using first example in Jan's comment.

Answer 2

[Steven Lord]

Use the fillmissing function, specifically the F = fillmissing(A,movmethod,window) syntax. You're going to want to specify a two-element vector [b f] as the window input as described in the documentation for that input argument.

Answer 3

[Fangjun Jiang]

Use the second code, replace size() with numel()

Answer 4

[Jan]

You forgot to mention, which problem you have with the shown code.

1. size(vector) replies a vector. If vector is a row vector, the output is e.g. [1, 1000]. When this is used in

for i = 254:1:size(vector)

only the first element is used. So this is equivalent to:

for i = 254:1:1

As Fangjun Jiang has suggested already, use numel instead of size.

2. A comparison with NaN is FALSE in every case by definition. See:

NaN == NaN

Therefore

if vector(i) == NaN

must fail. Use this instead:

if isnan(vector(i))

A working version:

for k = 254:numel(vector)
  if isnan(vector(k))
    vector(k) = mean(vector(k-5:k-1));
  end
end

Or

for k = find(isnan(vector))
  vector(k) = mean(vector(k-5:k-1));
end