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Suppose I have the following numpy array:
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
[[--, --, --],
[2, 4, 1],
[3, 1, 5]]
Is this possible to do using numpy masked array operations? How can one do it?
Thanks.
579
To mask rows and/or columns of a 2D array that contain masked values, use the np. ma. mask_rowcols() method in Numpy. The function returns a modified version of the input array, masked depending on the value of the axis parameter.
For splitting the 2d array,you can use two specific functions which helps in splitting the NumPy arrays row wise and column wise which are split and hsplit respectively . 1. split function is used for Row wise splitting. 2.
The [:, :] stands for everything from the beginning to the end just like for lists. The difference is that the first : stands for first and the second : for the second dimension. a = numpy. zeros((3, 3)) In [132]: a Out[132]: array([[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]])
In NumPy, you filter an array using a boolean index list. A boolean index list is a list of booleans corresponding to indexes in the array. If the value at an index is True that element is contained in the filtered array, if the value at that index is False that element is excluded from the filtered array.
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]])
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