Question

如何将此数据框转换为由numpy.nan行拆分的数据框字典？

import pandas
import numpy
names = ['a', 'b', 'c']
df = pandas.DataFrame([1,2,3,numpy.nan, 4,5,6,numpy.nan, 7, 8,9])
>>> df

      0
0   1.0
1   2.0
2   3.0
3   NaN
4   4.0
5   5.0
6   6.0
7   NaN
8   7.0
9   8.0
10  9.0

期望的输出：

df_dict = {'a': <df1>, 'b': <df2>, 'c': <df3>}

带

Answer 1

将groupby与d = {names[i]: x.dropna() for i, x in df.groupby(df[0].isnull().cumsum())} {'c': 0 0 7.0 1 8.0 2 9.0, 'b': 0 0 4.0 1 5.0 2 6.0, 'a': 0 0 1.0 1 2.0 2 3.0}：

一起使用

print (d['a'])
     0
0  1.0
1  2.0
2  3.0

print (d['b'])
     0
4  4.0
5  5.0
6  6.0

print (d['c'])
      0
8   7.0
9   8.0
10  9.0

SELECT b.id_match
FROM a JOIN
     b
     ON a.year = b.year
GROUP BY b.id_match
HAVING SUM(a.score <> b.score_match) = 0;

Answer 2

另一种方法是通过numpy数组拆分，即

import numpy as np
dic = {names[i]: j.dropna() for i,j in enumerate(np.array_split(df, np.where(df[0].isnull())[0]))}

%%timeit
dic = {names[i]: j.dropna() for i,j in enumerate(np.array_split(df, np.where(df[0].isnull())[0]))}
100 loops, best of 3: 2.51 ms per loop
%%timeit
d = {names[i]: x.dropna() for i, x in df.groupby(df[0].isnull().cumsum())}
100 loops, best of 3: 6.1 ms per loop

Answer 3

这是单程

最初，

In [2109]: df_dict = dict(zip(
                          names,
                          [g.dropna() for _, g in df.groupby(df[0].isnull().cumsum())]
                             ))

在编辑中意识到它与另一个答案相同。

In [2100]: df_dict = {names[i]: g.dropna() for i, g in df.groupby(df[0].isnull().cumsum())}

In [2101]: df_dict['a']
Out[2101]:
     0
0  1.0
1  2.0
2  3.0

In [2102]: df_dict['b']
Out[2102]:
     0
4  4.0
5  5.0
6  6.0

In [2103]: df_dict['c']
Out[2103]:
      0
8   7.0
9   8.0
10  9.0

如何用空格分割这个数据框？

3 个答案: