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Groups > comp.lang.python > #26002
| Date | 2012-07-24 19:51 +0100 |
|---|---|
| From | MRAB <python@mrabarnett.plus.com> |
| Subject | Re: no data exclution and unique combination. |
| References | <864b0104-daa8-4ea5-8003-6cfaab19fdea@googlegroups.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.2547.1343155889.4697.python-list@python.org> (permalink) |
On 24/07/2012 19:27, giuseppe.amatulli@gmail.com wrote: > Hi, > would like to take eliminate a specific number in an array and its correspondent in an other array, and vice-versa. > > given > > a=np.array([1,2,4,4,5,4,1,4,1,1,2,4]) > b=np.array([1,2,3,5,4,4,1,3,2,1,3,4]) > > no_data_a=1 > no_data_b=2 > > a_clean=array([4,4,5,4,4,4]) > b_clean=array([3,5,4,4,3,4]) > > after i need to calculate unique combination in pairs to count the observations > and obtain > (4,3,2) > (4,5,1) > (5,4,1) > (4,4,2) > > For the fist task i did > > a_No_data_a = a[a != no_data_a] > b_No_data_a = b[a != no_data_a] > > b_clean = b_No_data_a[b_No_data_a != no_data_b] > a_clean = a_No_data_a[a_No_data_a != no_data_b] > > but the results are not really stable. > mask = (a != no_data_a) & (b != no_data_b) a_clean = a[mask] b_clean = b[mask] > For the second task > The np.unique would solve the problem if it can be apply to a two arrays. >
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no data exclution and unique combination. giuseppe.amatulli@gmail.com - 2012-07-24 11:27 -0700
Re: no data exclution and unique combination. MRAB <python@mrabarnett.plus.com> - 2012-07-24 19:51 +0100
Re: no data exclution and unique combination. MRAB <python@mrabarnett.plus.com> - 2012-07-24 20:08 +0100
Re: no data exclution and unique combination. Terry Reedy <tjreedy@udel.edu> - 2012-07-24 15:32 -0400
Re: no data exclution and unique combination. giuseppe.amatulli@gmail.com - 2012-08-09 13:06 -0700
Re: no data exclution and unique combination. Dave Angel <d@davea.name> - 2012-08-09 17:23 -0400
Re: no data exclution and unique combination. Terry Reedy <tjreedy@udel.edu> - 2012-08-09 17:33 -0400
Re: no data exclution and unique combination. Hans Mulder <hansmu@xs4all.nl> - 2012-08-10 10:47 +0200
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