Roy*_*ith 32 python filtering percentile pandas
我有一个DataFrame叫做data列的大熊猫ms.我想消除data.ms高于95%百分位数的所有行.现在,我这样做:
limit = data.ms.describe(90)['95%']
valid_data = data[data['ms'] < limit]
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哪个有效,但我想把它推广到任何百分位数.最好的方法是什么?
Phi*_*oud 67
使用Series.quantile()方法:
In [48]: cols = list('abc')
In [49]: df = DataFrame(randn(10, len(cols)), columns=cols)
In [50]: df.a.quantile(0.95)
Out[50]: 1.5776961953820687
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过滤掉的行df,其中df.a大于或等于第95百分位的更多信息:
In [72]: df[df.a < df.a.quantile(.95)]
Out[72]:
a b c
0 -1.044 -0.247 -1.149
2 0.395 0.591 0.764
3 -0.564 -2.059 0.232
4 -0.707 -0.736 -1.345
5 0.978 -0.099 0.521
6 -0.974 0.272 -0.649
7 1.228 0.619 -0.849
8 -0.170 0.458 -0.515
9 1.465 1.019 0.966
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2di*_*com 25
对于这类事情,numpy比熊猫快得多:
numpy.percentile(df.a,95) # attention : the percentile is given in percent (5 = 5%)
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相当于但比以下快3倍:
df.a.quantile(.95) # as you already noticed here it is ".95" not "95"
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所以对于你的代码,它给出:
df[df.a < np.percentile(df.a,95)]
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