Thanks eat! It helps and it's so easy! :)
On 02.08.2012 14:59, eat wrote:
Hi,
On Thu, Aug 2, 2012 at 3:43 PM, Nicole Stoffels
<nicole.stoff...@forwind.de <mailto:nicole.stoff...@forwind.de>> wrote:
Dear all,
I have a two-dimensional array:
a = array([[1,2,3],[0,2,1],[5,7,8]])
I want to reorder it by the last column in descending order, so
that I get:
b =array([[5, 7, 8],[1, 2, 3],[0, 2, 1]])
Perhaps along the lines:
In []: a
Out[]:
array([[1, 2, 3],
[0, 2, 1],
[5, 7, 8]])
In []: ndx= a[:, 2].argsort()
In []: a[ndx[::-1], :]
Out[]:
array([[5, 7, 8],
[1, 2, 3],
[0, 2, 1]])
What I did first is the following, which reorders the array in
ascending
order (I found that method in the internet):
b = array(sorted(a, key=lambda new_entry: new_entry[2]))
b = array([[0, 2, 1],[1, 2, 3],[5, 7, 8]])
But I want it just the other way arround. So I did the following
afterwards which results in an array only containing zeros:
b_indices = b.argsort()
b_matrix = b[b_indices[::-1]]
new_b = b_matrix[len(b_matrix)-1]
Is there an easy way to reorder it? Or is there at least a complicated
way which produces the right output?
I hope you can help me! Thanks!
My 2 cents,
-eat
Best regards,
Nicole
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