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NumPy: Find first n columns according to mask

Say I have an array arr in shape (m, n) and a boolean array mask in the same shape as arr. I would like to obtain the first N columns from arr that are True in mask as well.
An example:
arr = np.array([[1,2,3,4,5],
[6,7,8,9,10],
[11,12,13,14,15]])
mask = np.array([[False, True, True, True, True],
[True, False, False, True, False],
[True, True, False, False, False]])
N = 2
Given the above, I would like to write a (vectorized) function that outputs the following:
output = maskify_n_columns(arr, mask, N)
output = np.array(([2,3],[6,9],[11,12]))

You can use broadcasting, numpy.cumsum() and numpy.argmax().
def maskify_n_columns(arr, mask, N):
m = (mask.cumsum(axis=1)[..., None] == np.arange(1,N+1)).argmax(axis=1)
r = arr[np.arange(arr.shape[0])[:, None], m]
return r
maskify_n_columns(arr, mask, 2)
Output:
[[ 2 3]
[ 6 9]
[11 12]]

How can i find the intersection of two multidimensional arrays faster?

there are two multidimensional boolean arrays with a different number of rows. I want to quickly find indexes of True values in common rows. I wrote the following code but it is too slow.
Is there a faster way to do this?
a=np.random.choice(a=[False, True], size=(100,100))
b=np.random.choice(a=[False, True], size=(1000,100))
for i in a:
for j in b:
if np.array_equal(i, j):
print(np.where(i))

Let's start with an edition to the question that makes sense and usually prints something:
a = np.random.choice(a=[False, True], size=(2, 2))
b = np.random.choice(a=[False, True], size=(4, 2))
print(f"a: \n {a}")
print(f"b: \n {b}")
matches = []
for i, x in enumerate(a):
for j, y in enumerate(b):
if np.array_equal(x, y):
matches.append((i, j))
And the solution using scipy.cdist which compares all rows in a against all rows in b, using hamming distance for Boolean vector comparison:
import numpy as np
import scipy
from scipy import spatial
d = scipy.spatial.distance.cdist(a, b, metric='hamming')
cdist_matches = np.where(d == 0)
mathces_values = [(a[i], b[j]) for (i, j) in matches]
cdist_values = a[cdist_matches[0]], b[cdist_matches[1]]
print(f"matches_inds = \n{matches}")
print(f"matches = \n{mathces_values}")
print(f"cdist_inds = \n{cdist_matches}")
print(f"cdist_matches =\n {cdist_values}")
out:
a:
[[ True False]
[False False]]
b:
[[ True True]
[ True False]
[False False]
[False True]]
matches_inds =
[(0, 1), (1, 2)]
matches =
[(array([ True, False]), array([ True, False])), (array([False, False]), array([False, False]))]
cdist_inds =
(array([0, 1], dtype=int64), array([1, 2], dtype=int64))
cdist_matches =
(array([[ True, False],
[False, False]]), array([[ True, False],
[False, False]]))
See this for a pure numpy implementation if you don't want to import scipy

The comparision of each row of a to each row of b can be made by making the shape of a broadcastable to the shape of b with the use of np.newaxis and np.tile
import numpy as np
a=np.random.choice(a=[True, False], size=(2,5))
b=np.random.choice(a=[True, False], size=(10,5))
broadcastable_a = np.tile(a[:, np.newaxis, :], (1, b.shape[0], 1))
a_equal_b = np.equal(b, broadcastable_a)
indexes = np.where(a_equal_b)
indexes = np.stack(np.array(indexes[1:]), axis=1)

Add together two numpy masked arrays

Is there a convenient way to add another array with actual values to masked positions in another array?
import numpy as np
arr1 = np.ma.array([0,1,0], mask=[True, False, True])
arr2 = np.ma.array([2,3,0], mask=[False, False, True])
arr1+arr2
Out[4]:
masked_array(data = [-- 4 --],
mask = [ True False True],
fill_value = 999999)
Note: in arr2 the value 2 is not masked -> should be in the resulting array
The result should be [2, 4, --]. I'd think there must be an easy solution for this?

Try this (choosing the logical operator that you want to use for your masks from http://docs.python.org/3/library/operator.html)
>>> from operator import and_
>>> np.ma.array(arr1.data+arr2.data,mask=map(and_,arr1.mask,arr2.mask))
masked_array(data = [2 4 --],
mask = [False False True],
fill_value = 999999)
In Python 3, map() returns an iterator and not a list, so it is necessary to add list():
>>> np.ma.array(arr1.data+arr2.data,mask=list(map(and_,arr1.mask,arr2.mask)))

Getting a grid of a matrix via logical indexing in Numpy

I'm trying to rewrite a function using numpy which is originally in MATLAB. There's a logical indexing part which is as follows in MATLAB:
X = reshape(1:16, 4, 4).';
idx = [true, false, false, true];
X(idx, idx)
ans =
1 4
13 16
When I try to make it in numpy, I can't get the correct indexing:
X = np.arange(1, 17).reshape(4, 4)
idx = [True, False, False, True]
X[idx, idx]
# Output: array([6, 1, 1, 6])
What's the proper way of getting a grid from the matrix via logical indexing?

You could also write:
>>> X[np.ix_(idx,idx)]
array([[ 1, 4],
[13, 16]])

In [1]: X = np.arange(1, 17).reshape(4, 4)
In [2]: idx = np.array([True, False, False, True]) # note that here idx has to
# be an array (not a list)
# or boolean values will be
# interpreted as integers
In [3]: X[idx][:,idx]
Out[3]:
array([[ 1, 4],
[13, 16]])

In numpy this is called fancy indexing. To get the items you want you should use a 2D array of indices.
You can use an outer to make from your 1D idx a proper 2D array of indices. The outers, when applied to two 1D sequences, compare each element of one sequence to each element of the other. Recalling that True*True=True and False*True=False, the np.multiply.outer(), which is the same as np.outer(), can give you the 2D indices:
idx_2D = np.outer(idx,idx)
#array([[ True, False, False, True],
# [False, False, False, False],
# [False, False, False, False],
# [ True, False, False, True]], dtype=bool)
Which you can use:
x[ idx_2D ]
array([ 1, 4, 13, 16])
In your real code you can use x=[np.outer(idx,idx)] but it does not save memory, working the same as if you included a del idx_2D after doing the slice.

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

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]])

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]

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]])