Cleanest way to replace np.array value with np.nan by user defined index
One question about mask 2-d np.array data.
- one 2-d np.array value in the shape of 20 x 20.
- An index t = [(1,2),(3,4),(5,7),(12,13)]
How to mask the 2-d array value by the (y,x) in index?
Usually, replacing with np.nan are based on the specific value like y[y==7] = np.nan
On my example, I want to replace the value specific location with np.nan.
For now, I can do it by:
- Creating a new array value_mask in the shape of 20 x 20
- Loop the value and testify the location by (i,j) == t[k]
- If True, value_mask[i,j] = value[i,j] ; In verse, value_mask[i,j] = np.nan
My method was too bulky especially for hugh data(3 levels of loops). Are there some efficiency method to achieve that? Any advice would be appreciate.
You are nearly there.
You can pass arrays of indices to arrays. You probably know this with 1D-arrays.
With a 2D-array you need to pass the array a tuple of lists (one tuple for each axis; one element in the lists (which have to be of equal length) for each array-element you want to chose). You have a list of tuples. So you have just to "transpose" it.
t1 = zip(*t)
gives you the right shape of your index array; which you can now use as index for any assignment, for example: value[t1] = np.NaN
(There are lots of nice explanation of this trick (with zip and *) in python tutorials, if you don't know it yet.)
You can use np.logical_and
arr = np.zeros((20,20))
You can select by location, this is just an example location.
arr[4:8,4:8] = 1
You can create a mask the same shape as arr
mask = np.ones((20,20)).astype(bool)
Then you can use the np.logical_and.
mask = np.logical_and(mask, arr == 1)
And finally, you can replace the 1s with the np.nan
arr[mask] = np.nan