Issue
I've recently been trying my hand at numpy, and I'm trying to find a solution to delete the elements inside the matrix at column 2 equal to the value stored in the variable element. Since I am a large amount of data I would need to know if there was a more efficient method which takes less time to execute than the classic for. I enclose an example:
element = [ 85., 222., 166., 238.]
matrix = [[228., 1., 222.],
[140., 0., 85.],
[140., 0., 104.],
[230., 0., 217.],
[115., 1., 250.],
[12., 1., 166.],
[181., 1., 238.]]
the output:
matrix = [[140., 0., 104.],
[230., 0., 217.],
[115., 1., 250.]]
The method I used is the following:
for y in element:
matrix = matrix[(matrix[:,2]!= y)]
When running it for a large amount of data it takes a long time. Is there anything more efficient, so that you can save on execution?
Solution
Since you tagged numpy
, I'd assume matrix
is a numpy array. With that, you can use np.isin
for your purpose:
matrix = np.array(matrix)
matrix[~np.isin(np.array(matrix)[:,2], element)]
Output:
array([[140., 0., 104.],
[230., 0., 217.],
[115., 1., 250.]])
Answered By - Quang Hoang
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