I3 np.ma.masked_where(x<1, x)

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
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