Reasoning about consecutive data points without using iteration

I am doing SPC analysis using numpy/pandas.

Part of this is checking data series against the Nelson rules and the Western Electric rules.

For instance (rule 2 from the Nelson rules): Check if nine (or more) points in a row are on the same side of the mean.

Now I could simply implement checking a rule like this by iterating over the array.

But before I do that, I'm checking here on SO if numpy/pandas has a way to do this without iteration?
In any case: What is the "numpy-ic" way to implement a check like the one described above?

Solution

As I mentioned in a comment, you may want to try using some stride tricks.

First, let's make an array of the size of your anomalies: we can put it as np.int8 to save some space
```
anomalies = x - x.mean()
signs = np.sign(anomalies).astype(np.int8)
```
Now for the strides. If you want to consider N consecutive points, you'll use
```
from np.lib.stride_tricks import as_strided
strided = as_strided(signs, 
                     strides=(signs.itemsize,signs.itemsize), 
                     shape=(signs.shape,N))
```
That gives us a (x.size, N) rollin array: the first row is x[0:N], the second x[1:N+1]... Of course, the last N-1 rows will be meaningless, so from now on we'll use
```
strided = strided[:-N+1]
```
Let's sum along the rows
```
consecutives = strided.sum(axis=-1)
```
That gives us an array of size (x.size-N+1) of values between -N and +N: we just have to find where the absolute values are N:
```
(indices,) = np.nonzero(consecutives == N)
```
indices is the array of the indices i of your array x for which the values x[i:i+N] are on the same side of the mean...

Example with x=np.random.rand(10) and N=3

>>> x = array([ 0.57016436,  0.79360943,  0.89535982,  0.83632245,  0.31046202,
            0.91398363,  0.62358298,  0.72148491,  0.99311681,  0.94852957])
>>> signs = np.sign(x-x.mean()).astype(np.int8)
array([-1,  1,  1,  1, -1,  1, -1, -1,  1,  1], dtype=int8)
>>> strided = as_strided(signs,strides=(1,1),shape=(signs.size,3))
array([[  -1,    1,    1],
       [   1,    1,    1],
       [   1,    1,   -1],
       [   1,   -1,    1],
       [  -1,    1,   -1],
       [   1,   -1,   -1],
       [  -1,   -1,    1],
       [  -1,    1,    1],
       [   1,    1, -106],
       [   1, -106,  -44]], dtype=int8)
>>> consecutive=strided[:-N+1].sum(axis=-1)
array([ 1,  3,  1,  1, -1, -1, -1,  1])
>>> np.nonzero(np.abs(consecutive)==N)
(array([1]),)