# [SciPy-User] Select rows according to cell value

Wes McKinney wesmckinn@gmail....
Tue Nov 13 14:56:55 CST 2012

```On Tue, Nov 13, 2012 at 12:53 PM, Scott Lasley <slasley@space.umd.edu> wrote:
> I don't know if this is any less ugly, but you could use
>
> magic_indices = lambda a: np.any([data[:,0] == x for x in a],axis=0)
> d = data[magic_indices(altitudes)]
>
> Or, as Oleksandr pointed out, if you're concerned about comparing floats you might use
>
> tol = 0.01
> magic_indices = lambda a: np.any([np.abs(data[:,0] - x) < tol for x in a],axis=0)
> d = data[magic_indices(altitudes)]
>
> or just
>
> tol = 0.01
> d = data[np.any([np.abs(data[:,0] - x) < tol for x in altitudes],axis=0)]
>
> Scott
>
> On Nov 13, 2012, at 12:41 PM, Juan Luis Cano Rodríguez <juanlu001@gmail.com> wrote:
>
>> Actually I arrived to a couple of one-liners:
>>
>>     d = np.take(data, [np.argwhere(data[:, 0] == a).flatten()[0] for a in altitudes], axis=0)
>>
>> or
>>
>>     d = np.array([data[data[:, 0] == a][0] for a in altitudes])
>>
>> I find them sort of ugly but maybe it's the only way. The same way I'd do
>>
>>     data[[1, 3, 8]]
>>
>> to retrieve the first, third and eighth I'd like to do
>>
>>     data[np.magic_indices(altitudes)]
>>
>>
>> On Tue, Nov 13, 2012 at 6:02 PM, Oleksandr Huziy <guziy.sasha@gmail.com> wrote:
>> Yeps, I admit with pandas it appears much easier
>>
>> import pandas
>> df = df.dropna(axis = 1)
>>
>> #df.index = df["alt"]
>> selection = df.select(lambda i: df.ix[i, "alt"] in altitudes)
>> print selection
>>
>>
>> cheers
>> --
>> Oleksandr (Sasha) Huziy
>>
>>
>>
>> 2012/11/13 Oleksandr Huziy <guziy.sasha@gmail.com>
>> I am not sure if this way is easier thsn yours, but here is what I wpuld do
>>
>> tol = 0.01
>> all_alts = data[:,0]
>> print all_alts
>> all_alts_temp = np.vstack([all_alts]*len(altitudes))
>> print all_alts_temp
>>
>> sel_alts_temp = np.vstack([altitudes]*len(all_alts)).transpose()
>> print sel_alts_temp
>> sel_pattern = np.any( np.abs(all_alts_temp - sel_alts_temp) < tol, axis = 0)
>> print sel_pattern
>> print data
>> print data[sel_pattern,:]
>>
>>
>> Cheers
>> --
>> Oleksandr (Sasha) Huziy
>>
>>
>>
>>
>> 2012/11/13 Andreas Hilboll <lists@hilboll.de>
>> Am Di 13 Nov 2012 17:07:19 CET schrieb Juan Luis Cano Rodríguez:
>> >
>> >   alt    temp    press    dens
>> >   10.0    223.3    26500    0.414
>> >   10.5    220.0    24540    0.389
>> >   11.0    216.8    22700    0.365
>> >   11.5    216.7    20985    0.337
>> >   12.0    216.7    19399    0.312
>> >   12.5    216.7    17934    0.288
>> >   13.0    216.7    16579    0.267
>> >   13.5    216.7    15328    0.246
>> >   14.0    216.7    14170    0.228
>> >
>> > into an ordinary NumPy array using np.loadtxt. I would like though to
>> > select the rows according to the altitude level, that is:
>> >
>> >     >>> data = np.loadtxt('data.txt', skiprows=1)
>> >     >>> altitudes = [10.5, 11.5, 14.0]
>> >     >>> d = ...  # some simple syntax involving data and altitudes
>> >     >>> d
>> >     10.5    220.0    24540    0.389
>> >     11.5    216.7    20985    0.337
>> >     14.0    216.7    14170    0.228
>> >
>> > I have tried a cumbersome expression which traverses all the array,
>> > then uses a list comprehension, converts to an array... but I'm sure
>> > there must be a simpler way. I've also looked at argwhere. Or maybe I
>> > should use pandas?
>> >
>> > Thank you in advance.
>> >
>> >
>> > _______________________________________________
>> > SciPy-User mailing list
>> > SciPy-User@scipy.org
>> > http://mail.scipy.org/mailman/listinfo/scipy-user
>>
>> +1 for using pandas
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Use isin:

In [5]: df[df.alt.isin(altitudes)]
Out[5]:
alt   temp  press   dens
1  10.5  220.0  24540  0.389
3  11.5  216.7  20985  0.337
8  14.0  216.7  14170  0.228

- Wes
```