如何在numpy中将布尔数组转换为索引数组

Question

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?

Answer 1

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) .

Answer 2

您应该能够使用numpy.nonzero()来查找此信息。

Answer 3

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]

Answer 4

如果您更喜欢索引器方式，则可以将布尔列表转换为numpy数组：

print x[nd.array(mask)]