在一个列中基于值屏蔽一个2D numpy数组。

Suppose I have the following numpy array:

假设我有下面的numpy数组:

a = [[1, 5, 6],
     [2, 4, 1],
     [3, 1, 5]]

I want to mask all the rows which have 1 in the first column. That is, I want

我想掩盖第一列中有1的所有行。这是我想要的

   [[--, --, --],
     [2, 4, 1],
     [3, 1, 5]]

Is this possible to do using numpy masked array operations? How can one do it?

是否可以使用numpy掩蔽数组操作?怎么做呢?

Thanks.

谢谢。

3 个解决方案

#1

import numpy as np

a = np.array([[1, 5, 6],
              [2, 4, 1],
              [3, 1, 5]])

np.ma.MaskedArray(a, mask=(np.ones_like(a)*(a[:,0]==1)).T)

# Returns: 
masked_array(data =
 [[-- -- --]
 [2 4 1]
 [3 1 5]],
             mask =
 [[ True  True  True]
 [False False False]
 [False False False]])

#2

You can create the desired mask by

您可以通过以下方法创建所需的掩码

mask = numpy.repeat(a[:,0]==1, a.shape[1])

and the masked array by

还有蒙面阵列

masked_a = numpy.ma.array(a, mask=numpy.repeat(a[:,0]==1, a.shape[1]))

#3

You could simply create an empty mask and then use numpy-broadcasting (like @eumiro showed) but using the element- and bitwise "or" operator |:

您可以简单地创建一个空掩码，然后使用numpy-broadcast(如@eumiro所示)，但是使用元素-和位方式”或“操作符|:

>>> a = np.array([[1, 5, 6], [2, 4, 1], [3, 1, 5]])

>>> mask = np.zeros(a.shape, bool) | (a[:, 0] == 1)[:, None]

>>> np.ma.array(a, mask=mask)
masked_array(data =
 [[-- -- --]
 [2 4 1]
 [3 1 5]],
             mask =
 [[ True  True  True]
 [False False False]
 [False False False]],
       fill_value = 999999)

A bit further explanation:

进一步的解释:

>>> # select first column
>>> a[:, 0]  
array([1, 2, 3])

>>> # where the first column is 1
>>> a[:, 0] == 1  
array([ True, False, False], dtype=bool)

>>> # added dimension so that it correctly broadcasts to the empty mask
>>> (a[:, 0] == 1)[:, None]  
array([[ True],
       [False],
       [False]], dtype=bool)

>>> # create the final mask
>>> np.zeros(a.shape, bool) | (a[:, 0] == 1)[:, None]  
array([[ True,  True,  True],
       [False, False, False],
       [False, False, False]], dtype=bool)

One further advantage of this approach is that it doesn't need to use potentially expensive multiplications or np.repeat so it should be quite fast.

这种方法的另一个优点是，它不需要使用可能昂贵的乘法或np。重复，这样应该很快。

#1

import numpy as np

a = np.array([[1, 5, 6],
              [2, 4, 1],
              [3, 1, 5]])

np.ma.MaskedArray(a, mask=(np.ones_like(a)*(a[:,0]==1)).T)

# Returns: 
masked_array(data =
 [[-- -- --]
 [2 4 1]
 [3 1 5]],
             mask =
 [[ True  True  True]
 [False False False]
 [False False False]])

#2