Numpy: Subtract array element by element

Eg: x - any element in x < a Should return something like

[True, True, False, True]

If element 0,1 and 3 in x meets the condition. I notice numpy operations are far faster than if tests and this has helped me speed up things ALOT.

Yes, it's just m < a. For example:

>>> m = np.array((1, 3, 10, 5))
>>> a = 6
>>> m2 = m < a
>>> m2
array([ True,  True, False,  True], dtype=bool)

Now, to the question:

Q: Is it in numpy possible to take the difference of an array with each of its own element based on a bool condition? I want to avoid looping.

I'm not sure what you're asking for here, but it doesn't seem to match the example directly below it. Are you trying to, e.g., subtract 1 from each element that satisfies the predicate? In that case, you can rely on the fact that False==0 and True==1 and just subtract the boolean array:

>>> m3 = m - m2
>>> m3
>>> array([ 0,  2, 10,  4])

From your clarification, you want the equivalent of this pseudocode loop:

for i in len(x):
  for j in in len(x):
    #!= not so important
    ##earlier question I asked lets me figure that one out
    if i!=j: 
      if x[j] - x[i] < a:
        True

I think the confusion here is that this is the exact opposite of what you said: you don't want "the difference of an array with each of its own element based on a bool condition", but "a bool condition based on the difference of an array with each of its own elements". And even that only really gets you to a square matrix of len(m)*len(m) bools, but I think the part left over is that the "any".

At any rate, you're asking for an implicit cartesian product, comparing each element of m to each element of m.

You can easily reduce this from two loops to one (or, rather, implicitly vectorize one of them, gaining the usual numpy performance benefits). For each value, create a new array by subtracting that value from each element and comparing the result with a, and then join those up:

>>> a = -2
>>> comparisons = np.array([m - x < a for x in m])
>>> flattened = np.any(comparisons, 0)
>>> flattened
array([ True,  True, False,  True], dtype=bool)

But you can also turn this into a simple matrix operation pretty easily. Subtracting every element of m from every other element of m is just m - m.T. (You can make the product more explicit, but the way numpy handles adding row and column vectors, it isn't necessary.) And then you just compare every element of that to the scalar a, and reduce with any, and you're done:

>>> a = -2
>>> m = np.matrix((1, 3, 10, 5))
>>> subtractions = m - m.T
>>> subtractions
matrix([[ 0,  2,  9,  4],
        [-2,  0,  7,  2],
        [-9, -7,  0, -5],
        [-4, -2,  5,  0]])
>>> comparisons = subtractions < a
>>> comparisons
matrix([[False, False, False, False],
        [False, False, False, False],
        [ True,  True, False,  True],
        [ True, False, False, False]], dtype=bool)
>>> np.any(comparisons, 0)
matrix([[ True,  True, False,  True]], dtype=bool)

Or, putting it all together in one line:

>>> np.any((m - m.T) < a, 0)
matrix([[ True,  True,  True,  True]], dtype=bool)

If you need m to be an array rather than a matrix, you can replace the subtraction line with m - np.matrix(m).T.

For higher dimensions, you actually do need to work in arrays, because you're trying to cartesian-product a 2D array with itself to get a 4D array, and numpy doesn't do 4D matrices. So, you can't use the simple "row vector - column vector = matrix" trick. But you can do it manually:

>>> m = np.array([[1,2], [3,4]]) # 2x2
>>> m4d = m.reshape(1, 1, 2, 2)  # 1x1x2x2
>>> m4d
array([[[[1, 2],
         [3, 4]]]])
>>> mt4d = m4d.T # 2x2x1x1
>>> mt4d
array([[[[1]],
        [[3]]],
       [[[2]],
        [[4]]]])
>>> subtractions = m - mt4d # 2x2x2x2
>>> subtractions
array([[[[ 0,  1],
         [ 2,  3]],
        [[-2, -1],
         [ 0,  1]]],
       [[[-1,  0],
         [ 1,  2]],
        [[-3, -2],
         [-1,  0]]]])

And from there, the remainder is the same as before. Putting it together into one line:

>>> np.any((m - m.reshape(1, 1, 2, 2).T) < a, 0)

(If you remember my original answer, I'd somehow blanked on reshape and was doing the same thing by multiplying m by a column vector of 1s, which obviously is a much stupider way to proceed.)

One last quick thought: If your algorithm really is "the bool result of (for any element y of m, x - y < a) for each element x of m", you don't actually need "for any element y", you can just use "for the maximal element y". So you can simplify from O(N^2) to O(N):

>>> (m - m.max()) < a

Or, if a is positive, that's always false, so you can simplify to O(1):

>>> np.zeros(m.shape, dtype=bool)

But I'm guessing your real algorithm is actually using abs(x - y), or something more complicated, which can't be simplified in this way.