python - 去除 Python 中数组列表中的空数组和零数组

Question

我正在考虑一些 Python 数据，它们是以下形式的数组列表:

LA=
[array([  99.08322813,  253.42371683,  300.792029  ])
array([  51.55274095,  106.29707418,  0])
array([0, 0 ,0 , 0, 0])
array([ 149.07283952,  191.45513754,  251.19610503,  393.50806493, 453.56783459])
array([ 105.61643877,  442.76668729,  450.37335607])
array([ 348.84179544])
array([], dtype=float64)]
array([0, 0 , 0])
array([ 295.05603151,  0,  451.77083268,  500.81771919])
array([ 295.05603151,  307.37232315,  451.77083268,  500.81771919])
array([  91.86758237,  148.70156948,  488.70648486,  507.31389766])
array([ 353.68691095])
array([ 208.21919198,  246.57665959,  0,  251.33820305, 394.34266882])
array([], dtype=float64)]

在我的数据中，我得到了一些空数组:

array([], dtype=float64)] 

和用零填充的数组:

array([0, 0, 0])

我怎样才能以自动化的简单方式摆脱这两种数组

LA=
[array([  99.08322813,  253.42371683,  300.792029  ])
array([  51.55274095,  106.29707418,  0])
array([ 149.07283952,  191.45513754,  251.19610503,  393.50806493, 453.56783459])
array([ 105.61643877,  442.76668729,  450.37335607])
array([ 348.84179544])
array([ 295.05603151,  0,  451.77083268,  500.81771919])
array([ 295.05603151,  307.37232315,  451.77083268,  500.81771919])
array([  91.86758237,  148.70156948,  488.70648486,  507.31389766])
array([ 353.68691095])
array([ 208.21919198,  246.57665959,  0,  251.33820305, 394.34266882])

最后我想删除零并保持数组列表格式

LA=
[array([  99.08322813,  253.42371683,  300.792029  ])
array([  51.55274095,  106.29707418])
array([ 149.07283952,  191.45513754,  251.19610503,  393.50806493, 453.56783459])
array([ 105.61643877,  442.76668729,  450.37335607])
array([ 348.84179544])
array([ 295.05603151,  451.77083268,  500.81771919])
array([ 295.05603151,  307.37232315,  451.77083268,  500.81771919])
array([  91.86758237,  148.70156948,  488.70648486,  507.31389766])
array([ 353.68691095])
array([ 208.21919198,  246.57665959,  251.33820305, 394.34266882])

提前致谢

Answer 1

列表理解应该做第一部分

[x for x in LA if x.any()]

你可以用compress做第二部分

[x.compress(x) for x in LA if x.any()]

基于 Ashwini 想法的更快版本

[x.compress(x) for x in LA if count_nonzero(x)]

时间:

In [89]: %timeit [x.compress(x) for x in LA if count_nonzero(x)]  #clear winner                                
10000 loops, best of 3: 20.2 µs per loop