How do I use rmoutliers without using rmoutliers

I have to do the rmoutliers function for an assignment without using the rmoutliers function. I believe I need to use a while loop, but I am still unfamiliar with the syntax

Essentially for the while loop I want is to check each value in a vector and check to see if it is outside the threshold and an outlier, return the index and replace the value with []

The threshold is based off of the mean +- 3*std of the vector just like the rmoutliers function

This is the exact function I am trying to replicate:

[B,TF] = rmoutliers(___) also returns a logical vector corresponding to the rows or columns of A that were removed.

[Rik]

How would you determine the threshold based on your input data? What have you tried? It is your homework, what specific question do you have for us?

[Walter Roberson]

You might not need to use a loop at all. If you do use a loop, a for loop would be more typical for such an operation.

Exception: there is sometimes iterative outlier detection, where you measure a property of the current vector, determine whether any entries fit outside some ranged based upon the measured property, if so remove them and loop back to measure / remove again, stopping when there are not remaining entries outside the desired range. A situation such as that would call for a while loop instead of a for loop.

[Walter Roberson]

The threshold is based off of the mean +- 3*std of the vector just like the rmoutliers function

3*std of the original vector? If so then you do not need any loops.

But once those entries are removed, what remains has a new std and there might be entries more than 3*std of the new vector. Do you want to keep going, or do you want to stop after a single pass?

For example

format long g
x =  [1 2 3 4 5  100 1e100]

x = 1×7

1 2 3 4 5 100 1e+100

s = std(x)

s = 
      3.77964473009227e+99

x(end) > 3*s

ans = logical
   0

No entries more than 3*std. If you reduced it down to (say) 2.5*std then only the last entry would get removed.

s2 = std(x(1:end-1))

s2 = 
          39.6253286001096

but then the 100 would be more than 2.5 the std of what remained. Should it be trimmed out too?

(and remember to subtract out the mean, which I did not do here.)

[Connor Tamaki]

the outlier detection is iterative with the mean and std being calculated each time. I know i'm supposed to do a while loop, but I don't know which condition to loop with

[Walter Roberson]

while true
    calculate the current standard deviation
    determine whether there are any current points
    that are beyond the threshold based upon standard deviation
    if there are none, exit the loop
    otherwise there must be some. Delete them.
end