如何将一个布尔数组转换成numpy中的索引数组

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?

是否有一种有效的Numpy机制来检索数组中基于条件为真而不是布尔掩码数组的位置的整数索引?

For example:

例如:

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?

在这种情况下，我想知道mask的索引I，其中mask[I]= True。是否有可能在不循环的情况下生成它们?

4 个解决方案

#1

Another option:

另一个选择:

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

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

这和numpy.where(mask==True)是一样的。

#2

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

您应该能够使用numpy.nonzero()查找此信息。

#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 :

现在可以这样重新表达:

x[x&(x-1) == 0]

#4

If you prefer the indexer way, you can convert your boolean list to numpy array:

如果您喜欢索引器方式，可以将布尔列表转换为numpy数组:

print x[nd.array(mask)]

#1