Terry and MRAB,
thanks for yours suggestions,
in the end i found this solution
mask=( a != 0 ) & ( b != 0 )
a_mask=a[mask]
b_mask=b[mask]
array2D = np.array(zip(a_mask,b_mask))
unique=dict()
for row in array2D :
row = tuple(row)
if row in unique:
unique[row] += 1
else:
unique[row] = 1
print unique
{(4, 5): 1, (5, 4): 1, (4, 4): 2, (2, 3): 1, (4, 3): 2}
I choose this solution because i could not install "from collections import
Counter".
Anyway how i can print to a file the unique results without the brackets and
obtain something like this?
4 5 1
5 4 1
4 4 2
2 3 1
4 3 2
Thanks in advance
Best regards.
Giuseppe
On Tuesday, 24 July 2012 13:27:05 UTC-5, giuseppe...@gmail.com wrote:
> Hi,
>
> would like to take eliminate a specific number in an array and its
> correspondent in an other array, and vice-versa.
>
>
>
> given
>
>
>
> a=np.array([1,2,4,4,5,4,1,4,1,1,2,4])
>
> b=np.array([1,2,3,5,4,4,1,3,2,1,3,4])
>
>
>
> no_data_a=1
>
> no_data_b=2
>
>
>
> a_clean=array([4,4,5,4,4,4])
>
> b_clean=array([3,5,4,4,3,4])
>
>
>
> after i need to calculate unique combination in pairs to count the
> observations
>
> and obtain
>
> (4,3,2)
>
> (4,5,1)
>
> (5,4,1)
>
> (4,4,2)
>
>
>
> For the fist task i did
>
>
>
> a_No_data_a = a[a != no_data_a]
>
> b_No_data_a = b[a != no_data_a]
>
>
>
> b_clean = b_No_data_a[b_No_data_a != no_data_b]
>
> a_clean = a_No_data_a[a_No_data_a != no_data_b]
>
>
>
> but the results are not really stable.
>
>
>
> For the second task
>
> The np.unique would solve the problem if it can be apply to a two arrays.
>
>
>
> Any idea?
>
> thanks in advance
>
> Giuseppe
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