Armin Burger wrote:
maybe someone on this list has some experience using numpy in
combination with GDAL/Python and could give me some advice.
For further questions about numpy, the numpy list is very helpful.
difference can be easily calculated after reading both images
into an array and substract one from the other.
The problem is that both images can contain clouds or snow that have
predefined pixel values (252,253).
Does anybody know if this is possible and how to perform it?
yep.
Looking
through the numpy docs I was not able to identify required methods or
functions for this. There is something like 'masked arrays' but I have
not understood if this could be used for my purpose.
This is exactly the kind of thing masked arrays are for. Yu can create a
masked array out of your data with something like:
>>> a
array([ 1, 2, 3, 4, 5, 6, 3, 4, 67, 4, 3, 5, 6, 7])
#a regular array
>>> import numpy.ma as ma
# create a masked array with the mask set at all elements with a value of 3:
>>> a = ma.masked_values(a, 3)
>>> a
masked_array(data = [1 2 -- 4 5 6 -- 4 67 4 -- 5 6 7],
mask = [False False True False False False True False False
False True False
False False],
fill_value=3)
#another one:
>>> b = np.array((1,3,4,2,4,7,4,5,23,5,7,3,8,5))
>>> b = ma.masked_values(b, 3)
>>> b
masked_array(data = [1 -- 4 2 4 7 4 5 23 5 7 -- 8 5],
mask = [False True False False False False False False False
False False True
False False],
fill_value=3)
add them together:
>>> c = a+b
>>> c
masked_array(data = [2 -- -- 6 9 13 -- 9 90 9 -- -- 14 12],
mask = [False True True False False False True False False
False True True
False False],
fill_value=3)
If your values are Floating Point, then Another option would be to
replace all the "cloud" values with NaN:
>>> a = np.array((1,2,3,4,5,6,3,4,67,4,3,5,6,7), dtype=np.float)
>>> a[a==3] = np.nan
>>> a
array([ 1., 2., NaN, 4., 5., 6., NaN, 4., 67., 4., NaN,
5., 6., 7.])
>>> b = np.array((1,3,4,2,4,7,4,5,23,5,7,3,8,5), dtype=np.float)
>>> b[b==3] = np.nan
>>> b
array([ 1., NaN, 4., 2., 4., 7., 4., 5., 23., 5., 7.,
NaN, 8., 5.])
>>> a+b
array([ 2., NaN, NaN, 6., 9., 13., NaN, 9., 90., 9., NaN,
NaN, 14., 12.])
-Chris
--
Christopher Barker, Ph.D.
Oceanographer
Emergency Response Division
NOAA/NOS/OR&R (206) 526-6959 voice
7600 Sand Point Way NE (206) 526-6329 fax
Seattle, WA 98115 (206) 526-6317 main reception
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