将多个蒙版应用于数组 [英] Applying multiple masks to arrays
问题描述
我要解释的是我有什么更高水平的建议可以完全消除该问题时的实际操作。
I'm explaining what I'm actually hoping to do in case there's a higher level suggestion that obviates the question entirely.
我有存储的科学数据分为三个数组：波，通量，错误。这些代表波长，通量和误差值。
数组长约4000个元素（数组的索引号对应于检测器的像素数）。
I have scientific data that I store in three arrays: wave, flux, error. These stand for wavelength, flux, and error values.
The arrays are about 4000 elements long (and the index number of the arrays corresponds to the pixel number of the detector).
我进行了各种测试可以，但是对于这个示例，我们仅做2个测试，这些测试需要有效地屏蔽关联的数组。
There are various tests that I do, but for this example let's just say I do 2 tests where I need to effectively mask out the associated arrays.
masks = []
masks.append(wave > 5500.35)
masks.append(flux / wave > 8.5)
Subquestion ：
我可以轻松进行2个掩码的情况，例如：
Subquestion: I can easily do the 2-mask case like:
fullmask = [x[0] and x[1] for x in zip(masks[0], masks[1])]
但是对任意数字怎么做
实际问题：
是否可以将所有掩码应用于每个阵列（ wave ， flux ，错误），并保留原始索引号？所谓保持原始索引号，是指我原则上可以获取掩蔽波阵列的平均像素数（原始索引号）吗？那就是：如果 wave [98:99] 是唯一未被遮盖的部分，则平均像素为98.5。
Real question:
Is there a way to apply all masks to each of the arrays (wave, flux, error), and keep the original index numbers? By "keep the original index numbers" I mean that I could, in principle, take the average pixel number of the masked wave array (the original index numbers)? That is: if wave[98:99] were the only parts not masked, the average pixel would be 98.5.
元问题：这是做任何此类事情的最佳方法吗？
Meta question: is this the best way to be doing any of this stuff?
编辑
因此，这里是一些示例数据。
So here's some sample data to play around with.
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.
推荐答案
否则，您可以使用布尔运算符，让我们定义一个例子：
otherwise you can use boolean operators, let's define en example:
d=np.arange(10)
masks = [d>5, d % 2 == 0, d<8]
您可以使用reduce组合所有这些内容：
you can use reduce to combine all of them:
total_mask = reduce(np.logical_and, masks)
如果需要手动选择掩码，也可以显式使用布尔运算符：
you can also use boolean operators explicitely if you need to manually choose the masks:
total_mask = masks[0] & masks[1] & masks[2]
这篇关于将多个蒙版应用于数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
