Something important when dealing with outliers is that one should try to use estimators as robust as possible. The mean of a distribution will be biased by outliers but e.g. the median will be much less.

Building on eumiro's answer:

```
def reject_outliers(data, m = 2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d/mdev if mdev else 0.
return data[s<m]
```

Here I have replace the mean with the more robust median and the standard deviation with the median absolute distance to the median. I then scaled the distances by their (again) median value so that `m`

is on a reasonable relative scale.

Note that for the `data[s<m]`

syntax to work, `data`

must be a numpy array.