如何在迭代器中掩蓋除每個第n個元素之外的每個元素？

Question

我正在嘗試創建一個只在第n位（例如第8位）提供值的掩碼數組（至少是NaN）。數組的長度應與原始長度相同。

有沒有一個不太可笑的方式來做到這一點？

b = np.array([[i for i in 7*[np.nan] + [val]] for val in a[::8]]).flatten()[7:] 

Answer 1

一種方法是使用切片分配：

>>> a 
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 
     17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 
     34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 
     51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63]) 
>>> b = numpy.array([numpy.NaN] * len(a)) 
>>> b[::8] = a[::8] 
>>> b 
array([ 0., nan, nan, nan, nan, nan, nan, nan, 8., nan, nan, 
     nan, nan, nan, nan, nan, 16., nan, nan, nan, nan, nan, 
     nan, nan, 24., nan, nan, nan, nan, nan, nan, nan, 32., 
     nan, nan, nan, nan, nan, nan, nan, 40., nan, nan, nan, 
     nan, nan, nan, nan, 48., nan, nan, nan, nan, nan, nan, 
     nan, 56., nan, nan, nan, nan, nan, nan, nan]) 

Answer 2

你可以使用一個生成器表達式來創建「清單」更優雅。

def val_only_on_nth(n, limit): 
    for v in xrange(limit): 
     if v%n == 0: 
      yield v 
     else: 
      yield np.NaN 

，並使用np.from_iter將其轉換成一個NP陣列

b = np.fromiter(val_only_on_nth(8,64)) 

Answer 3

這在一般情況下工作：

n=8 
a=np.random.random((64,)) # example random array to mask 
a[np.arange(0,len(a))%n!=0]=np.nan 


array([ 0.68756737,   nan,   nan,   nan,   nan, 
       nan,   nan,   nan, 0.68577462,   nan, 
       nan,   nan,   nan,   nan,   nan, 
       nan, 0.89002182,   nan,   nan,   nan, 
       nan,   nan,   nan,   nan, 0.26135927, 
       nan,   nan,   nan,   nan,   nan, 
       nan,   nan, 0.66857456,   nan,   nan, 
       nan,   nan,   nan,   nan,   nan, 
     0.39230499,   nan,   nan,   nan,   nan, 
       nan,   nan,   nan, 0.85367809,   nan, 
       nan,   nan,   nan,   nan,   nan, 
       nan, 0.48642591,   nan,   nan,   nan, 
       nan,   nan,   nan,   nan])