CODEWITHSUNDEEP
pythonnumpy
mkmitchell
mkmitchell

Reputation: 701

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

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

Upvotes: 8

Views: 9839

Answers (3)

Aelius
Aelius

Reputation: 1413

In recent numpy you could use a combination of np.isin and np.where to achieve this result. The first method outputs a boolean numpy array that evaluates to True where its vlaues are equal to an array-like specified test element (see doc), while with the second you could create a new array that set some a value where the specified confition evaluates to True and another value where False.

Example

I'll make an example with a random array but using the specific values you provided.

import numpy as np

repeatSet = ([2, 5, 6, 8])

arr = np.array([[1,5,1],
                [0,1,0],
                [0,0,0],
                [2,2,2]])

out = np.where(np.isin(arr, repeatSet), 1, 77)

> out
array([[77,  1, 77],
       [77, 77, 77],
       [77, 77, 77],
       [ 1,  1,  1]])

Upvotes: 1

senderle
senderle

Reputation: 151197

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]])

Upvotes: 16

segasai
segasai

Reputation: 8548

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]

Upvotes: 1

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