Check if values in a set are in a numpy array in python

Question

I want to check if a NumPyArray has values in it that are in a set, and if so set that area in an array = 1. If not set a keepRaster = 2.

numpyArray = #some imported array
repeatSet= ([3, 5, 6, 8])

confusedRaster = numpyArray[numpy.where(numpyArray in repeatSet)]= 1

Yields:

<type 'exceptions.TypeError'>: unhashable type: 'numpy.ndarray'

Is there a way to loop through it?

 for numpyArray
      if numpyArray in repeatSet
           confusedRaster = 1
      else
           keepRaster = 2

To clarify and ask for a bit further help:

What I am trying to get at, and am currently doing, is putting a raster input into an array. I need to read values in the 2-d array and create another array based on those values. If the array value is in a set then the value will be 1. If it is not in a set then the value will be derived from another input, but I'll say 77 for now. This is what I'm currently using. My test input has about 1500 rows and 3500 columns. It always freezes at around row 350.

for rowd in range(0, width):
    for cold in range (0, height):
        if numpyarray.item(rowd,cold) in repeatSet:
            confusedArray[rowd][cold] = 1
        else:
            if numpyarray.item(rowd,cold) == 0:
                confusedArray[rowd][cold] = 0
            else:
                confusedArray[rowd][cold] = 2

Answer 1

In versions 1.4 and higher, numpy provides the in1d function.

>>> test = np.array([0, 1, 2, 5, 0])
>>> states = [0, 2]
>>> np.in1d(test, states)
array([ True, False,  True, False,  True], dtype=bool)

You can use that as a mask for assignment.

>>> test[np.in1d(test, states)] = 1
>>> test
array([1, 1, 1, 5, 1])

Here are some more sophisticated uses of numpy's indexing and assignment syntax that I think will apply to your problem. Note the use of bitwise operators to replace if-based logic:

>>> numpy_array = numpy.arange(9).reshape((3, 3))
>>> confused_array = numpy.arange(9).reshape((3, 3)) % 2
>>> mask = numpy.in1d(numpy_array, repeat_set).reshape(numpy_array.shape)
>>> mask
array([[False, False, False],
       [ True, False,  True],
       [ True, False,  True]], dtype=bool)
>>> ~mask
array([[ True,  True,  True],
       [False,  True, False],
       [False,  True, False]], dtype=bool)
>>> numpy_array == 0
array([[ True, False, False],
       [False, False, False],
       [False, False, False]], dtype=bool)
>>> numpy_array != 0
array([[False,  True,  True],
       [ True,  True,  True],
       [ True,  True,  True]], dtype=bool)
>>> confused_array[mask] = 1
>>> confused_array[~mask & (numpy_array == 0)] = 0
>>> confused_array[~mask & (numpy_array != 0)] = 2
>>> confused_array
array([[0, 2, 2],
       [1, 2, 1],
       [1, 2, 1]])

Another approach would be to use numpy.where, which creates a brand new array, using values from the second argument where mask is true, and values from the third argument where mask is false. (As with assignment, the argument can be a scalar or an array of the same shape as mask.) This might be a bit more efficient than the above, and it's certainly more terse:

>>> numpy.where(mask, 1, numpy.where(numpy_array == 0, 0, 2))
array([[0, 2, 2],
       [1, 2, 1],
       [1, 2, 1]])

Answer 2

Here is one possible way of doing what you whant:

numpyArray = np.array([1, 8, 35, 343, 23, 3, 8]) # could be n-Dimensional array
repeatSet = np.array([3, 5, 6, 8])
mask = (numpyArray[...,None] == repeatSet[None,...]).any(axis=-1) 
print mask
>>> [False  True False False False  True  True]