Python 消除给定百分位数上的所有数据
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Eliminating all data over a given percentile
提问by Roy Smith
I have a pandas DataFramecalled datawith a column called ms. I want to eliminate all the rows where data.msis above the 95% percentile. For now, I'm doing this:
我有一个DataFrame名为.pandasdata的列ms。我想消除data.ms95% 以上的所有行。现在，我正在这样做：
limit = data.ms.describe(90)['95%']
valid_data = data[data['ms'] < limit]
which works, but I want to generalize that to any percentile. What's the best way to do that?
哪个有效，但我想将其推广到任何百分位数。这样做的最佳方法是什么？
采纳答案by Phillip Cloud
Use the Series.quantile()method:
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
To filter out rows of dfwhere df.ais greater than or equal to the 95th percentile do:
过滤掉的行df，其中df.a大于或等于第95百分位做：
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
回答by 2diabolos.com
numpy is much faster than Pandas for this kind of things :
在这种情况下，numpy 比 Pandas 快得多：
numpy.percentile(df.a,95) # attention : the percentile is given in percent (5 = 5%)
is equivalent but 3 times faster than :
等效但比 快 3 倍：
df.a.quantile(.95) # as you already noticed here it is ".95" not "95"
so for your code, it gives :
所以对于你的代码，它给出：
df[df.a < np.percentile(df.a,95)]
