Hi, You can try masked_array module: x = np.array([[0,1,2],[3,4,5],[6,7,8]]) I3 np.ma.masked_where(x<1, x) O3 masked_array(data = [[-- 1 2] [3 4 5] [6 7 8]], mask = [[ True False False] [False False False] [False False False]], fill_value = 999999) There might be a smarter solution than this, since ma tends to get slower when you deal with big data arrays. But you retain the 2D information instead of getting a flattened np.array. On Wed, Mar 14, 2012 at 5:35 PM, jonasr <jonas.ruebsam@web.de> wrote:
Hello,
my problem is that i want to remove some small numbers of an 2d array, for example if i want to sort out all numbers smaller then 1 of an array i get
x=[[0,1,2],[3,4,5][6,7,8]]
c=x>=1
In [213]: c Out[213]: array([[False, True, True], [ True, True, True], [ True, True, True]], dtype=bool)
In [214]: x[c] Out[214]: array([1, 2, 3, 4, 5, 6, 7, 8])
the problem ist that i now have a 1d array, is there any possibility to keep the 2d structure ?
greets jonas -- View this message in context: http://old.nabble.com/Replace-array-values-tp33506446p33506446.html Sent from the Numpy-discussion mailing list archive at Nabble.com.
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