Why does argwhere return column vector?
Hello list. I don't understand why argwhere returns a column vector when I apply it to a row vector:
a = arange(5) argwhere(a>1) array([[2], [3], [4]])
That seems odd and inconvenient. Any advantage that I am missing? Mark
Hi Mark 2008/5/27 mark <markbak@gmail.com>:
I don't understand why argwhere returns a column vector when I apply it to a row vector:
a = arange(5) argwhere(a>1) array([[2], [3], [4]])
Each row of the result is a coordinate into your array. So if you had a = np.arange(12).reshape((4,3)) then you'd get In [54]: np.argwhere(a>5) Out[54]: array([[2, 0], [2, 1], [2, 2], [3, 0], [3, 1], [3, 2]]) If you want to grab the elements larger than 5, just do a[a>5] Regards Stéfan
OK, I get how it works for 2D arrays. What I want to do is insert a number, say 7, before every value in the array that is larger than, for example, 1. Then I need to first find all the indices of values larger than 1, and then I can do an insert:
a = arange(5) i = argwhere( a>1 ) insert(a,i[:,0],7) array([0, 1, 7, 2, 7, 3, 7, 4])
Is there a better way to do this? So for this instance, it is inconvenient that argwhere returns a column vector. But I understand the issue for arrays with higher dimensions. Thanks for the explanation, Mark
Each row of the result is a coordinate into your array. So if you had
a = np.arange(12).reshape((4,3))
then you'd get
In [54]: np.argwhere(a>5) Out[54]: array([[2, 0], [2, 1], [2, 2], [3, 0], [3, 1], [3, 2]])
If you want to grab the elements larger than 5, just do
a[a>5]
Hi Mark 2008/5/27 mark <markbak@gmail.com>:
OK, I get how it works for 2D arrays.
What I want to do is insert a number, say 7, before every value in the array that is larger than, for example, 1.
Then I need to first find all the indices of values larger than 1, and then I can do an insert:
a = arange(5) i = argwhere( a>1 ) insert(a,i[:,0],7) array([0, 1, 7, 2, 7, 3, 7, 4])
Is there a better way to do this?
Inserting is slow, since a new array is allocated each time. In the case where you insert at multiple indexes, it may be handled better, but I haven't timed it. I would do it the following way: import numpy as np a = np.array([2,1,3,-1,2]) mask = a > 1 out = np.zeros(len(a) + sum(mask), dtype=int) out.fill(7) out[np.arange(len(a)) + mask.cumsum()] = a Regards Stéfan
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Stéfan van der Walt