Seleccione ciertas filas (condición cumplida), pero solo algunas columnas en Python/Numpy

I have an numpy array with 4 columns and want to select columns 1, 3 and 4, where the value of the second column meets a certain condition (i.e. a fixed value). I tried to first select only the rows, but with all 4 columns via:

I = A[A[:,1] == i]

which works. Then I further tried (similarly to matlab which I know very well):

I = A[A[:,1] == i, [0,2,3]]

which doesn't work. How to do it?

DATOS DE EJEMPLO:

 >>> A = np.array([[1,2,3,4],[6,1,3,4],[3,2,5,6]])
 >>> print A
 [[1 2 3 4]
  [6 1 3 4]
  [3 2 5 6]]
 >>> i = 2
     
 # I want to get the columns 1, 3 and 4 
 # for every row which has the value i in the second column. 
 # In this case, this would be row 1 and 3 with columns 1, 3 and 4:
 [[1 3 4]
  [3 5 6]]

I am now currently using this:

I = A[A[:,1] == i]
I = I[:, [0,2,3]]

But I thought that there had to be a nicer way of doing it... (I am used to MATLAB)

preguntado el 28 de mayo de 14 a las 13:05

A[A[:,1] == i][0,2,3] didn't work either? -

I = A[A[:,1] == i][0,2,3] --> IndexError: too many indices -

And apart from that I got to admit that I wouldn't really understand that indexing either, very different from matlab... -

@tim: Could you please post the array and what output do you expect? -

@Ankur Ankan: edited into the question. -

5 Respuestas

>>> a = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])
>>> a
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])

>>> a[a[:,0] > 3] # select rows where first column is greater than 3
array([[ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])

>>> a[a[:,0] > 3][:,np.array([True, True, False, True])] # select columns
array([[ 5,  6,  8],
       [ 9, 10, 12]])

# fancier equivalent of the previous
>>> a[np.ix_(a[:,0] > 3, np.array([True, True, False, True]))]
array([[ 5,  6,  8],
       [ 9, 10, 12]])

For an explanation of the obscure np.ix_(), consulte nuestra página, https://stackoverflow.com/a/13599843/4323

Finally, we can simplify by giving the list of column numbers instead of the tedious boolean mask:

>>> a[np.ix_(a[:,0] > 3, (0,1,3))]
array([[ 5,  6,  8],
       [ 9, 10, 12]])

contestado el 23 de mayo de 17 a las 13:05

So really two consecutive selections necessary? - Tim

If you're wishing you could do a[x][y] where x and y are boolean masks, yeah, I wish that too, but it does not work. This seems to be a known problem, and I don't know why, but it's hardly important here. - Juan Zwinck

Not only that, I wished to be able to select the rows and colums in ONE single statement like this: A[row_indices_to_select, colum_indices_to_select], mientras que row_indices_to_select would be coming from the condition I wanted to apply.. :( - Tim

I've added some more solutions--I like the last one using ix_() with a tuple. - Juan Zwinck

If you do not want to use boolean positions but the indexes, you can write it this way:

A[:, [0, 2, 3]][A[:, 1] == i]

Going back to your example:

>>> A = np.array([[1,2,3,4],[6,1,3,4],[3,2,5,6]])
>>> print A
[[1 2 3 4]
 [6 1 3 4]
 [3 2 5 6]]
>>> i = 2
>>> print A[:, [0, 2, 3]][A[:, 1] == i]
[[1 3 4]
 [3 5 6]]

Seriamente,

contestado el 28 de mayo de 14 a las 14:05

Boolean positions actually are okay for me, I just would have wanted to do the selection in ONE step and not in two consecutive selections (which your solution is doing, isn't it?) because of performance reasons. - Tim 

>>> a=np.array([[1,2,3], [1,3,4], [2,2,5]])
>>> a[a[:,0]==1][:,[0,1]]
array([[1, 2],
       [1, 3]])
>>>

respondido 15 mar '16, 04:03

Esto también funciona.

I = np.array([row[[x for x in range(A.shape[1]) if x != i-1]] for row in A if row[i-1] == i])
print I

Edit: Since indexing starts from 0, so 

i-1

debe ser usado.

contestado el 28 de mayo de 14 a las 14:05

The algorithm must be correct, but it is not very pythonic. - Taha

@Taha maybe not, bu it saves you double selection. The idea is actually simple, first choose cols then iterate over rows. - genclik27

@genclik27 I understood what you did. But lately, I am doing some numerical computation with large matrices. I always was in need of vectorized calculations. The problem of what you are proposing is that you create a new list. You cannot change the values directly in the matrix this way. It is indeed useful if you don't need to change the values of A. - Taha

I am hoping this answers your question but a piece of script I have implemented using pandas is:

df_targetrows = df.loc[df[col2filter]*somecondition*, [col1,col2,...,coln]]

Por ejemplo, 

targets = stockdf.loc[stockdf['rtns'] > .04, ['symbol','date','rtns']]

this will return a dataframe with only columns ['symbol','date','rtns'] del stockdf where the row value of rtns satisfies, stockdf['rtns'] > .04

espero que esto ayude

Respondido el 21 de diciembre de 14 a las 06:12

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