Question

我有一个琐碎的问题。我有一个很大的df，有很多列。我试图找到最有效的方法来对所有具有不同bin大小的列进行装箱并创建新的df。这是仅对单个列进行装箱的示例：

import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,20,size=(5, 4)), columns=list('ABCD'))
newDF = pd.cut(df.A, 2, precision=0)
newDF 
0    (9.0, 18.0]
1    (-0.0, 9.0]
2    (-0.0, 9.0]
3    (-0.0, 9.0]
4    (9.0, 18.0]
Name: A, dtype: category
Categories (2, interval[float64]): [(-0.0, 9.0] < (9.0, 18.0]]

Answer 1

如果要分别处理每一列，请使用DataFrame.apply：

df = pd.DataFrame(np.random.randint(0,20,size=(5, 4)), columns=list('ABCD'))
newDF = df.apply(lambda x: pd.cut(x, 2, precision=0))
print (newDF)
            A            B             C             D
0  (2.0, 4.0]  (8.0, 15.0]   (7.0, 13.0]  (12.0, 18.0]
1  (2.0, 4.0]  (8.0, 15.0]   (7.0, 13.0]  (12.0, 18.0]
2  (4.0, 7.0]  (8.0, 15.0]  (13.0, 19.0]  (12.0, 18.0]
3  (4.0, 7.0]  (8.0, 15.0]   (7.0, 13.0]   (5.0, 12.0]
4  (4.0, 7.0]   (1.0, 8.0]   (7.0, 13.0]   (5.0, 12.0]

如果要通过相同的分箱处理所有列，请对MultiIndex Series使用DataFrame.stack，应用cut并通过Series.unstack重新整形：

newDF = pd.cut(df.stack(), 2, precision=0).unstack()
print (newDF)
              A             B             C             D
0  (10.0, 19.0]  (10.0, 19.0]  (10.0, 19.0]  (-0.0, 10.0]
1  (10.0, 19.0]  (10.0, 19.0]  (-0.0, 10.0]  (-0.0, 10.0]
2  (-0.0, 10.0]  (10.0, 19.0]  (-0.0, 10.0]  (-0.0, 10.0]
3  (-0.0, 10.0]  (-0.0, 10.0]  (10.0, 19.0]  (-0.0, 10.0]
4  (10.0, 19.0]  (10.0, 19.0]  (-0.0, 10.0]  (-0.0, 10.0]

在所有数据框列上应用具有不同bin大小的合并

1 个答案: