Why std function returns NaN?

I have a set of 100 data points that I wish to find the standard deviation of. For some reason the std function returns NaN for certain chunks of the data. Particularly, I get normal values when going through using std on the data individually until I get to value 40. Why might this be? Thanks!

Answer 1

[John D'Errico]

LOOK AT WHAT YOU ARE DOING.

R(7)
ans =
          49.8875023348093
R(40)
ans =
           47.332112186806
R(7):R(40)
ans =
  1×0 empty double row vector

What is the standard deviation of an empty vector? NaN.

When you use colon with the first argument greater than the second, you get an empty vector. So you got exactly what is expected, based on what you did.

Why you are doing what you are doing, that nobody else can know, only you. Should I point out that std(R(7):R(40)) does not in fact generate the standard deviation of the numbers in R in that range?

For example, suppose I do this?

R(1):R(5)
ans =
  Columns 1 through 31
     0     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16    17    18    19    20    21    22    23    24    25    26    27    28    29    30
  Column 32
    31
>> std(R(1):R(5))
ans =
          9.38083151964686

But in fact, what I did was to generate the standard deviation of the linearly increasing sequence from 0 to 31. Is that really what you wanted to do?

[Matthew Wong]

you are correct in your assumption of what I was thinking. What I really wanted was to, for this example, find the standard deviation of the values between R(7) and R(40). Since the syntax I used appears not to be the correct way to do this, how might I go about doing it? Just to be clear, my goal is to find the standard deviation of a given section of the column vector R. In the end, I’d like to end up segmenting the vector R by using the standard deviation as a criteria for defining the bounds for a section (i.e. take standard deviation of first three points, then define a new section once the succeeding point falls more than one standard deviation away from the preceding point. then take standard deviation of next three points and repeat)

[Image Analyst]

Then you want to use stdfilt(), which does exactly that. It's in the Image Processing Toolbox.

[Stephen23]

"Since the syntax I used appears not to be the correct way to do this, how might I go about doing it?"

You should do the introductory tutorials, which teach many basic concepts that all beginners need to learn to start using MATLAB (such as how to use indexing):

https://www.mathworks.com/help/matlab/getting-started-with-matlab.html

As John D'Errico wrote, you also need to learn to look at what your code is doing. Code does not care what you think/believe/want/assure/hope that it is doing. It is your task to actually look at what it is really doing (which include checking output values against another method, sanity checks on intermediate values, reading the documentation, trying simpler examples to ensure that an operator does what you require, checking edge cases, etc.)

[John D'Errico]

When you see an error, take apart the pieces of what you did. Don't just assume that it does what you want it to do. In this case, you should havedone exactly what I did. Then you would realize that you were doing something wrong.

Here, it looks like what you really wanted was to compute

std(R(7:40))

Of course, that is very different from std(R(7):R(40)).

It may be that stdfilt does what you need. I don't have the IPT, but I do know that the Image Animalist knows what he is doing. ;-)

As Stephen suggested, this sort of indicates that what you really need to do is start with the turorials, to give yourself a basic understanding of MATLAB, instead of diving into the deep end of the pool without knowing how to swim.

Answer 2

[Star Strider]

Please explain ‘certain chunks of the data’. When I load ‘R’, there are no NaN values, and:

Rstd = std(R)

Rstd =
          65.2138227774813

[Matthew Wong]

Try running the code

stdev=std(R(7):R(40));

When I try to find the standard deviation of this section the function returns NaN when there are no NaN values in that section.

[Star Strider]

Try indexing it correctly, as John D’Errico notes in his Answer:

stdev=std(R(7:40))

stdev =
          84.1380930755812
          

See the documentation section on Matrix Indexing (link), and more to the point, the documentation section on Getting Started (link).

Answer 3

[Image Analyst]

I get the same results as Star. R(40) = 47.332112186806 which should be fine. You will get a nan if any values of R are nan. Are you taking the stdev of the whole array, or in a moving window, like with stdfilt()?

s = load('Radii.mat')
R = s.R
sd = std(R)