我正在尝试创建一个屏蔽数组(或至少填充NaN),该数组仅在第n个(示例中为第8个)位置提供值。该数组应与原始数组的长度相同。
这样做的方法不那么荒谬吗?
b = np.array([[i for i in 7*[np.nan] + [val]] for val in a[::8]]).flatten()[7:]
答案 0 :(得分:2)
一种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])
答案 1 :(得分:0)
您可以使用生成器表达式来创建"列表"更优雅。
def val_only_on_nth(n, limit):
for v in xrange(limit):
if v%n == 0:
yield v
else:
yield np.NaN
并使用np.from_iter将其转换为np数组
b = np.fromiter(val_only_on_nth(8,64))
答案 2 :(得分:0)
这适用于一般情况:
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])