从一系列 Numpy 数组的字典创建 MultiIndex DataFrame

Question

给定一个包含pandas.Series 的字典，每个单元格中都有numpy.array，

import pandas as pd
import numpy as np

N = 5
foo = [x for x in np.random.randint(10, size=(N,8))]        # list of ndarray
bar = [x for x in np.random.randint(10, size=(N,8))]        # list of ndarray
baz = [x for x in np.random.randint(10, size=(N,8))]        # list of ndarray
input = {
    'foo': pd.Series(foo, index=pd.date_range('2020-01-01', periods=N, freq='D')),
    'bar': pd.Series(bar, index=pd.date_range('2020-01-01', periods=N, freq='D')),
    'baz': pd.Series(baz, index=pd.date_range('2020-01-01', periods=N, freq='D')),
}
print(input)

# {'foo': 
# 2020-01-01    [4, 1, 3, 3, 4, 6, 0, 2]
# 2020-01-02    [7, 7, 1, 2, 1, 2, 8, 6]
# 2020-01-03    [1, 0, 6, 8, 1, 8, 2, 3]
# 2020-01-04    [1, 5, 6, 0, 1, 8, 8, 4]
# 2020-01-05    [4, 7, 9, 3, 5, 3, 0, 1]
# Freq: D, dtype: object, 
# 'bar': 
# 2020-01-01    [0, 2, 2, 5, 4, 9, 7, 9]
# 2020-01-02    [7, 0, 8, 0, 7, 8, 8, 9]
# 2020-01-03    [6, 7, 2, 7, 2, 9, 8, 7]
# 2020-01-04    [1, 8, 8, 9, 6, 1, 4, 6]
# 2020-01-05    [9, 4, 4, 2, 6, 2, 7, 7]
# Freq: D, dtype: object, 
# 'baz': 
# 2020-01-01    [9, 2, 9, 2, 5, 3, 5, 3]
# 2020-01-02    [6, 5, 3, 3, 9, 7, 7, 9]
# 2020-01-03    [5, 7, 0, 6, 1, 5, 6, 7]
# 2020-01-04    [3, 9, 2, 6, 1, 5, 9, 9]
# 2020-01-05    [2, 7, 6, 4, 1, 2, 9, 2]
# Freq: D, dtype: object}

将其转换为 MultiIndex pandas DataFrame 的最有效方法是什么，其中字典键位于第一个多索引级别，而系列的 DateTimeIndex 位于第二个多索引级别？

使用上面给出的示例，所需的 pandas DataFrame 将有 15 行和 8 列

Answer 1

使用随机时，请使用seed，这样您的数据是可重现的。

你可以使用 pandas concat，结合 numpy 的 vstack 来获得你想要的输出：

np.random.seed(5)
N = 5
foo = [x for x in np.random.randint(10, size=(N, 8))]  # list of ndarray
bar = [x for x in np.random.randint(10, size=(N, 8))]  # list of ndarray
baz = [x for x in np.random.randint(10, size=(N, 8))]  # list of ndarray
data = {
    "foo": pd.Series(foo, index=pd.date_range("2020-01-01", periods=N, freq="D")),
    "bar": pd.Series(bar, index=pd.date_range("2020-01-01", periods=N, freq="D")),
    "baz": pd.Series(baz, index=pd.date_range("2020-01-01", periods=N, freq="D")),
}

box = pd.concat(data)
pd.DataFrame(np.vstack(box), index=box.index)


                0   1   2   3   4   5   6   7
foo 2020-01-01  3   6   6   0   9   8   4   7
    2020-01-02  0   0   7   1   5   7   0   1
    2020-01-03  4   6   2   9   9   9   9   1
    2020-01-04  2   7   0   5   0   0   4   4
    2020-01-05  9   3   2   4   6   9   3   3
bar 2020-01-01  2   1   5   7   4   3   1   7
    2020-01-02  3   1   9   5   7   0   9   6
    2020-01-03  0   5   2   8   6   8   0   5
    2020-01-04  2   0   7   7   6   0   0   8
    2020-01-05  5   5   9   6   4   5   2   8
baz 2020-01-01  8   1   6   3   4   1   8   0
    2020-01-02  2   2   4   1   6   3   4   3
    2020-01-03  1   4   2   3   4   9   4   0
    2020-01-04  6   6   9   2   9   3   0   8
    2020-01-05  8   9   7   4   8   6   8   0

Answer 2

一个简单的方法是充分利用 pandas：stack()、to_frame() 和 swaplevel() 的魔力

df = pd.DataFrame(inputs).stack().to_frame().swaplevel()
df.iloc[:,0].apply(lambda x: pd.Series({idx: value for idx, value in enumerate(x)}))

产生：

                0   1   2   3   4   5   6   7
foo 2020-01-01  2   3   5   1   7   0   8   2
bar 2020-01-01  8   1   4   6   1   7   3   1
baz 2020-01-01  7   3   4   3   9   0   5   0
foo 2020-01-02  8   3   8   1   6   5   5   4
bar 2020-01-02  2   1   9   5   6   6   1   4
baz 2020-01-02  4   3   3   8   7   4   2   4
foo 2020-01-03  8   8   5   2   9   4   1   1
bar 2020-01-03  0   0   0   8   8   5   8   5
baz 2020-01-03  1   5   5   9   5   2   2   7
foo 2020-01-04  2   7   6   3   0   8   2   5
bar 2020-01-04  1   8   0   3   1   5   1   3
baz 2020-01-04  5   0   7   6   1   7   7   9
foo 2020-01-05  9   0   8   5   9   9   6   8
bar 2020-01-05  0   3   1   6   4   1   9   6
baz 2020-01-05  4   6   6   7   9   3   0   5