高效获取 pandas 区间类别的右边缘

Question

如何有效地得到 pandas 区间类别的右边缘？在下面的示例中，如何有效地创建 z？

import pandas as pd, numpy as np
bins = pd.interval_range(start=0, end=4, freq=1, closed='left')
x = pd.Series(np.linspace(0.0,3.8,num=20))
y = pd.cut(x, bins)

# How can one create z efficiently?
z = pd.Series(y.iat[n].right for n in range(len(y)))

感谢您的帮助！

Answer 1

对于高性能方法，您可以使用 np.bincount:

np.digitize(x, range(0,4))
# array([1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4])

对于 pd.Series:

pd.Series(np.digitize(x, range(0,4)), index=x.index)

0     1
1     1
2     1
3     1
4     1
5     1
6     2
7     2
8     2
9     2
10    2
11    3
...

更大数据帧上的计时 -

bins = pd.interval_range(start=0, end=400, freq=1, closed='left')
x = pd.Series(np.linspace(0.0,380,num=20_000))

%timeit pd.Series(np.digitize(x, range(0,400)))
# 567 µs ± 28.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

def op(x):
    y = pd.cut(x, bins)
    z = pd.Series(y.iat[n].right for n in range(len(y)))

%timeit op(x)
# 682 ms ± 49.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

两者给出相同的地方：

np.allclose(op(x), pd.Series(np.digitize(x, range(0,400))))
# True

因此，对于包含 20000 行的更大数据帧，我们得到 1200x 加速

高效获取 pandas 区间类别的右边缘

Get right edges of pandas interval Categories efficiently

python

pandas

indexing

categorical-data