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python - 使用有时包含较少项目 (< x) 的 tolist() 函数填充固定数量 (x) 的新列

问题描述

我正在使用 tolist() 将 1 列（'modelGreeks'）中的 8 项列表拆分为同一数据框中的 8 个新列：

pd.DataFrame(df['modelGreeks'].tolist(), index=df.index)
df[['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice']] = pd.DataFrame(df['modelGreeks'].tolist(), index=df.index)

这是我通常在“modelGreeks”列中获得的列表：

(0.2953686167703842, -1.9317880628477724e-14, 1.4648640549124297e-15, 0.0, 6.240571011994176e-13, 1.1840837166645831e-15, -1.4648640549124297e-15, 10.444000244140625)

10 次中有 9 次完美运行。但有时我通过 API 检索的数据并不完美/完整。而不是在“modelGreeks”列中包含 8 个项目的预期列表，它在该字段中提供了一个“无”值，并且我在第二个代码行的代码执行中收到以下错误消息（从逻辑上讲，因为它尝试用只有 1 个可用值：

ValueError: Columns must be same length as key

我正在寻找一个解决方案，无论如何都会创建和填充 8 个新列，例如使用 0 或 NaN 或 None。

希望有人能帮忙。提前感谢您的努力。

以下代码有效：

df1 = pd.DataFrame(columns=['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice','modelGreeks'])
df1['modelGreeks'] = [[None, None, None, None, None, None, None, None], None, None, None, None]
df1[['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice']] = df1['modelGreeks'].apply(pd.Series)

它返回：

   IV_model  59  Price_model  61  62  63  64  undPrice  modelGreeks
0  NaN       NaN NaN          NaN NaN NaN NaN NaN       [None, None, None, None, None, None, None, None]
1  NaN       NaN NaN          NaN NaN NaN NaN NaN       None
2  NaN       NaN NaN          NaN NaN NaN NaN NaN       None
3  NaN       NaN NaN          NaN NaN NaN NaN NaN       None
4  NaN       NaN NaN          NaN NaN NaN NaN NaN       None

这很好。唯一的问题是，在某些时刻，我通过 API 从 Interactive Brokers 收到的数据集只会在列 modelGreeks 的所有行中提供标量 None 值。如果我将其应用于测试用例，那么我会再次收到错误消息（“ValueError：列必须与键长度相同”）：

df1 = pd.DataFrame(columns=['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice','modelGreeks'])
df1['modelGreeks'] = [None, None, None, None, None]
df1[['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice']] = df1['modelGreeks'].apply(pd.Series)

Traceback (most recent call last):
File "/Users/floris/PycharmProjects/ib_insync/test1.py", line 9, in <module>
df1[['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice']] = df1['modelGreeks'].apply(pd.Series)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/pandas/core/frame.py", line 3367, in __setitem__
self._setitem_array(key, value)
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/pandas/core/frame.py", line 3389, in _setitem_array
raise ValueError('Columns must be same length as key')
ValueError: Columns must be same length as key

在这种情况下，我也希望在 8 列中只看到 NaN 值。

标签： pythonpandasmultiple-columnstolist

解决方案

不要创建新的 DataFrame，而是将列表列转换为系列：

df[['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice']] = df['modelGreeks'].apply(pd.Series)

测试：

df = pd.DataFrame(columns=['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice','modelGreeks'])
df['modelGreeks'] = [[1,2,3,4,5,6,7,8], [1,2,None,4,5,6,7,8], [1,2,3,4,5,6,7], [None], None, [None,None,None,None,None]]
df[['IV_model', 59, 'Price_model', 61, 62, 63, 64, 'undPrice']] = df['modelGreeks'].apply(pd.Series)

输出：

   IV_model   59  Price_model  ...   64  undPrice                     modelGreeks
0       1.0  2.0          3.0  ...  7.0       8.0        [1, 2, 3, 4, 5, 6, 7, 8]
1       1.0  2.0          NaN  ...  7.0       8.0     [1, 2, None, 4, 5, 6, 7, 8]
2       1.0  2.0          3.0  ...  7.0       NaN           [1, 2, 3, 4, 5, 6, 7]
3       NaN  NaN          NaN  ...  NaN       NaN                          [None]
4       NaN  NaN          NaN  ...  NaN       NaN                            None
5       NaN  NaN          NaN  ...  NaN       NaN  [None, None, None, None, None]

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