我正在解释我实际上希望做什么，以防有更高级别的建议完全避免这个问题。
我有存储在三个数组中的科学数据:wave、flux、error。这些代表波长、通量和误差值。
阵列长约4000个元素(阵列的索引号对应探测器的像素数)。
我做了各种测试，但对于这个例子，我们只说我做了 2 个测试，我需要在这些测试中有效地屏蔽掉关联的数组。
masks = []
masks.append(wave > 5500.35)
masks.append(flux / wave > 8.5)
子问题: 我可以很容易地做 2-mask 案例:
fullmask = [x[0] and x[1] for x in zip(masks[0], masks[1])]
但是对于任意数量的掩码，该怎么做呢？
真题:
有没有办法将所有掩码应用于每个数组(wave、flux、error)并保留原始索引号？ “保留原始索引号”是指原则上我可以采用掩蔽波阵列的平均像素数(原始索引号)？也就是说:如果 wave[98:99] 是唯一未被屏蔽的部分，则平均像素将为 98.5。
元问题:这是做这些事情的最佳方式吗？
编辑
所以这里有一些示例数据可供使用。
wave = array([5000, 5001, 5002, 5003, 5004, 5005, 5006, 5007, 5008, 5009, 5010,
5011, 5012, 5013, 5014, 5015, 5016, 5017, 5018, 5019, 5020, 5021,
5022, 5023, 5024, 5025, 5026, 5027, 5028, 5029, 5030, 5031, 5032,
5033, 5034, 5035, 5036, 5037, 5038, 5039, 5040, 5041, 5042, 5043,
5044, 5045, 5046, 5047, 5048, 5049, 5050, 5051, 5052, 5053, 5054,
5055, 5056, 5057, 5058, 5059, 5060, 5061, 5062, 5063, 5064, 5065,
5066, 5067, 5068, 5069, 5070, 5071, 5072, 5073, 5074, 5075, 5076,
5077, 5078, 5079, 5080, 5081, 5082, 5083, 5084, 5085, 5086, 5087,
5088, 5089, 5090, 5091, 5092, 5093, 5094, 5095, 5096, 5097, 5098,
5099])
flux = array([ 112.65878609, 109.2008992 , 113.30629929, 117.17002715,
103.19663878, 110.42131523, 106.00841123, 100.27882741,
103.89160905, 102.29402469, 105.58894696, 103.21314852,
96.97242814, 106.70130478, 108.83891225, 110.60598803,
95.10361887, 109.39734257, 103.08289878, 104.97258911,
96.46606257, 106.75993458, 99.25386914, 105.91429417,
105.83752232, 100.53312657, 99.74871394, 107.12735837,
108.81187473, 96.51418895, 99.71311101, 94.08702553,
98.81198643, 93.84567201, 103.21444519, 94.7027134 ,
99.61842203, 103.71336458, 100.8697998 , 92.1564786 ,
96.56711985, 94.7728761 , 82.65194671, 83.52280884,
86.57960844, 73.6700194 , 66.11794666, 61.01624627,
63.19944529, 55.50283247, 62.09172307, 59.55436092,
75.66399466, 70.69397378, 64.27899192, 73.80248662,
89.17119606, 78.97024327, 82.3334254 , 100.82581489,
102.77937201, 99.37717696, 96.2215563 , 104.52291339,
93.7581944 , 93.32154346, 103.57018896, 108.08682518,
105.2711359 , 100.00242988, 100.86934866, 103.20764384,
104.19274473, 101.3314802 , 102.75057114, 94.02347591,
95.48758551, 106.0099397 , 99.50733501, 97.88110415,
107.54266965, 107.76126331, 98.14882302, 101.55654606,
101.02418212, 106.82324958, 95.52086925, 102.65957133,
104.93806492, 103.22762427, 108.02087993, 106.71911141,
97.24396195, 103.3450277 , 113.99870588, 106.4145751 ,
110.08294674, 109.40908288, 118.61518086, 114.37341062])
error = array([ 11.72799338, 22.33423611, 16.89347382, 12.80063102,
23.99242356, 25.15863754, 20.44765811, 14.84358628,
19.16343785, 19.5703491 , 18.44427035, 19.08648083,
19.09116433, 12.22098884, 14.81280352, 11.35010222,
18.59850136, 15.78855734, 21.85877638, 20.12179042,
22.04894395, 21.986731 , 13.26738352, 16.10987762,
24.28528627, 30.11866128, 25.30220842, 25.02100014,
29.38560916, 16.8192307 , 29.15097205, 23.56805267,
15.17285709, 18.27495747, 18.63750452, 18.61618504,
11.45940025, 21.95805701, 24.22923951, 11.76824052,
19.75465065, 14.72979889, 15.45936176, 14.73227474,
28.91683627, 22.90534472, 16.82376093, 21.47830226,
20.05012214, 16.74393817, 17.79456361, 20.80008233,
19.32059989, 23.23471888, 13.77434964, 17.56121752,
15.96716163, 18.5294016 , 28.31005939, 13.66340359,
10.38160267, 16.09621015, 18.25125683, 20.95954331,
21.31996941, 24.51998489, 16.58831953, 15.25427142,
23.93065281, 30.4552266 , 16.94527367, 16.92730802,
17.79659417, 18.85080572, 18.0839428 , 23.93949481,
26.60243553, 13.68320208, 16.74669921, 20.30238694,
12.74773905, 19.20810456, 20.7189417 , 20.73402554,
17.12106905, 25.06475175, 13.0947528 , 28.16437938,
22.4803386 , 13.71143627, 6.60617725, 20.41186825,
23.54924934, 22.25930658, 20.09337438, 24.94705884,
18.58056249, 5.58653271, 18.71242702, 17.83578444])
# How I created masks, or just jump to next comment if it's too painful to look at...
masks = []
masks.append(flux/error > 4.0) # high error
absorptionMask1 = (wave < 5060)
absorptionMask2 = (wave > 5040)
bob = [all(x) for x in zip(absorptionMask1, absorptionMask2)]
absorptionMask = ~np.array(bob)
masks.append(absorptionMask)
# The resulting mask
masks = [array([ True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, False, False,
True, False, True, False, False, True, True, True, True,
True, True, True, True, True, True, True, True, False,
False, False, False, False, False, False, False, False, False,
True, True, True, True, False, True, True, True, True,
True, True, False, True, True, True, False, True, True,
True, True, True, False, False, True, True, True, True,
True, True, True, True, True, True, False, True, True,
True, True, True, True, True, True, True, True, True, True], dtype=bool),
array([ True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True, True], dtype=bool)]
# More in a bit, should get you a feel for what I'm looking at.
最佳答案
否则你可以使用 bool 运算符，让我们定义一个例子:
d=np.arange(10)
masks = [d>5, d % 2 == 0, d<8]
你可以使用 reduce 来组合它们:
from functools import reduce
total_mask = reduce(np.logical_and, masks)
如果您需要手动选择掩码，您也可以明确使用 bool 运算符:
total_mask = masks[0] & masks[1] & masks[2]
关于python - 将多个掩码应用于数组，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11536367/