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up vote 1 down vote favorite

I'm looking for sort of the opposite of sum() I guess. Here we go:

x = array([
    [False, False, False, False, False],
    [ True, False, False, False, False],
    [ True,  True, False, False, False],
    [ True,  True,  True, False, False]])

x.sum(axis=1)
Out: array([0, 1, 2, 3])

So I want to go the opposite direction: from [0,1,2,3] to an array like x (I can specify the number of columns I want in x, of course, above it's 5).

The solution should ideally work for higher dimensions too, and I don't want to loop in Python of course, because the input could be longer than this example. That said, here's a solution using a loop:

s = np.array([0, 1, 2, 3])
y = np.zeros((len(s), 5), np.bool)
for row,col in enumerate(s):
    y[row,0:col] = True
share|improve this question
    
Looks like you are thinking about stackoverflow.com/questions/26310346/… :) –  Warren Weckesser Oct 11 '14 at 4:02
    
@WarrenWeckesser: correct! I know how to get from here to there...but got a little stuck here. –  John Zwinck Oct 11 '14 at 4:05
1  
In your loop solution, it looks like the input is x, and you create y, but earlier you say to want to go from [0, 1, 2, 3] to x (i.e. x is the output). Isn't the operation from [0,1,2,3] to the bigger boolean array what you want? –  Warren Weckesser Oct 11 '14 at 4:12
    
If so, you can do something like I did in stackoverflow.com/questions/26269893/…, but change the equality in the comparison to arange(A.max()+1) to < (as in @DSM's answer). –  Warren Weckesser Oct 11 '14 at 4:15
    
@WarrenWeckesser: yeah, the loop code I put in the question was a bit confusing so I updated it to hopefully make it more clear/self-contained/non-self-referential. –  John Zwinck Oct 11 '14 at 4:16

1 Answer 1

up vote 1 down vote accepted

IIUC -- and I'm not sure that I do -- you could use arange and a broadcasting comparison:

>>> v = np.array([0,1,3,2])
>>> np.arange(5) < v[...,None]
array([[False, False, False, False, False],
       [ True, False, False, False, False],
       [ True,  True,  True, False, False],
       [ True,  True, False, False, False]], dtype=bool)

or in 2D:

>>> v = np.array([[1,2],[0,2]])
>>> np.arange(5) < v[...,None]
array([[[ True, False, False, False, False],
        [ True,  True, False, False, False]],

       [[False, False, False, False, False],
        [ True,  True, False, False, False]]], dtype=bool)
>>> ((np.arange(5) < v[...,None]).sum(2) == v).all()
True
share|improve this answer
    
You sure did understand--even before I clarified my example. Thank you, I would not have thought of this solution. –  John Zwinck Oct 11 '14 at 4:23

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