How to turn a boolean array into index array in numpy

Is there an efficient Numpy mechanism to retrieve the integer indexes of locations in an array based on a condition is true as opposed to the Boolean mask array?

For example:

x=np.array([range(100,1,-1)])
#generate a mask to find all values that are a power of 2
mask=x&(x-1)==0
#This will tell me those values
print x[mask]

In this case, I'd like to know the indexes i of mask where mask[i]==True. Is it possible to generate these without looping?

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You should be able to use numpy.nonzero() to find this information.

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最佳答案

Another option:

In [13]: numpy.where(mask)
Out[13]: (array([36, 68, 84, 92, 96, 98]),)

which is the same thing as numpy.where(mask==True).

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np.arange(100,1,-1)
array([100,  99,  98,  97,  96,  95,  94,  93,  92,  91,  90,  89,  88,
87,  86,  85,  84,  83,  82,  81,  80,  79,  78,  77,  76,  75,
74,  73,  72,  71,  70,  69,  68,  67,  66,  65,  64,  63,  62,
61,  60,  59,  58,  57,  56,  55,  54,  53,  52,  51,  50,  49,
48,  47,  46,  45,  44,  43,  42,  41,  40,  39,  38,  37,  36,
35,  34,  33,  32,  31,  30,  29,  28,  27,  26,  25,  24,  23,
22,  21,  20,  19,  18,  17,  16,  15,  14,  13,  12,  11,  10,
9,   8,   7,   6,   5,   4,   3,   2])


x=np.arange(100,1,-1)


np.where(x&(x-1) == 0)
(array([36, 68, 84, 92, 96, 98]),)

Now rephrase this like :

x[x&(x-1) == 0]
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If you prefer the indexer way, you can convert your boolean list to numpy array:

print x[nd.array(mask)]

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