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From: "Joseph L. Casale" <jcasale@activenetwerx.com>
To: =?windows-1256?Q?Python_=FD=5Bpython-list=40python=2Eorg=5D=FD?= <python-list@python.org>
Subject: RE: Numpy outlier removal
Thread-Topic: Numpy outlier removal
Thread-Index: AQHN7F3oLivXMPS/qkW0/R0RnvrEGJg85zXY
Date: Sun, 6 Jan 2013 22:50:07 +0000
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Xref: csiph.com comp.lang.python:36300

>Assuming your data and the dictionary are keyed by a common set of keys:=
=A0=0A=
=0A=
>=0A=
>for key in descriptions:=0A=
> =A0 =A0if abs(data[key] - mean(data)) >=3D m * std(data):=0A=
> =A0 =A0 =A0 =A0del data[key]=0A=
> =A0 =A0 =A0 =A0del descriptions[key]=0A=
=0A=
=0A=
Heh, yeah sometimes the obvious is too simple to see. I used a dict comp to=
 rebuild=0A=
the results with the comparison.=0A=
=0A=
=0A=
Thanks!=0A=
jlc=