pandas 将不同长度的列表作为新列添加到数据框中
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Adding list with different length as a new column to a dataframe
提问by Jaffer Wilson
I am willing to add or insert the list values in the dataframe. The dataframe len is 49, whereas the length of list id 47. I am getting the following error while implementing the code.
我愿意在数据框中添加或插入列表值。数据帧 len 是49，而列表 id 的长度47。执行代码时出现以下错误。
print("Lenght of dataframe: ",datasetTest.open.count())
print("Lenght of array: ",len(test_pred_list))
datasetTest['predict_close'] = test_pred_list
The error is:
错误是：
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-105-68114a4e9a82> in <module>()
5 # datasetTest = datasetTest.dropna()
6 # print(datasetTest.count())
----> 7 datasetTest['predict_close'] = test_pred_list
8 # test_shifted['color_predicted'] = test_shifted.apply(determinePredictedcolor, axis=1)
9 # test_shifted['color_original'] =
c:\python35\lib\site-packages\pandas\core\frame.py in __setitem__(self, key, value)
2517 else:
2518 # set column
-> 2519 self._set_item(key, value)
2520
2521 def _setitem_slice(self, key, value):
c:\python35\lib\site-packages\pandas\core\frame.py in _set_item(self, key, value)
2583
2584 self._ensure_valid_index(value)
-> 2585 value = self._sanitize_column(key, value)
2586 NDFrame._set_item(self, key, value)
2587
c:\python35\lib\site-packages\pandas\core\frame.py in _sanitize_column(self, key, value, broadcast)
2758
2759 # turn me into an ndarray
-> 2760 value = _sanitize_index(value, self.index, copy=False)
2761 if not isinstance(value, (np.ndarray, Index)):
2762 if isinstance(value, list) and len(value) > 0:
c:\python35\lib\site-packages\pandas\core\series.py in _sanitize_index(data, index, copy)
3119
3120 if len(data) != len(index):
-> 3121 raise ValueError('Length of values does not match length of ' 'index')
3122
3123 if isinstance(data, PeriodIndex):
ValueError: Length of values does not match length of index
How I can get rid of this error. Please help me.
我怎样才能摆脱这个错误。请帮我。
采纳答案by EdChum
If you convert the list to a Series then it will just work:
如果您将列表转换为系列，那么它将起作用：
datasetTest.loc[:,'predict_close'] = pd.Series(test_pred_list)
example:
例子：
In[121]:
df = pd.DataFrame({'a':np.arange(3)})
df
Out[121]:
a
0 0
1 1
2 2
In[122]:
df.loc[:,'b'] = pd.Series(['a','b'])
df
Out[122]:
a b
0 0 a
1 1 b
2 2 NaN
The docs refer to this as setting with enlargementwhich talks about adding or expanding but it also works where the length is less than the pre-existing index.
文档将此称为带有放大的设置，它谈论添加或扩展，但它也适用于长度小于预先存在的索引的情况。
To handle where the index doesn't start at 0or in fact is not an int:
要处理索引不在0或实际上不是 int 的位置：
In[126]:
df = pd.DataFrame({'a':np.arange(3)}, index=np.arange(3,6))
df
Out[126]:
a
3 0
4 1
5 2
In[127]:
s = pd.Series(['a','b'])
s.index = df.index[:len(s)]
s
Out[127]:
3 a
4 b
dtype: object
In[128]:
df.loc[:,'b'] = s
df
Out[128]:
a b
3 0 a
4 1 b
5 2 NaN
You can optionally replace the NaNif you wish calling fillna
NaN如果您愿意，您可以选择替换fillna
回答by jpp
You can add items to your list with an arbitrary fillerscalar.
您可以使用任意filler标量将项目添加到列表中。
Data from @EdChum.
来自@EdChum 的数据。
filler = 0
lst = ['a', 'b']
df.loc[:, 'b'] = lst + [filler]*(len(df.index) - len(lst))
print(df)
a b
0 0 a
1 1 b
2 2 0
回答by YOBEN_S
You still can assign it by using locdata from Ed
您仍然可以使用loc来自 Ed 的数据来分配它
l = ['a','b']
df.loc[range(len(l)),'b'] = l
df
Out[546]:
a b
0 0 a
1 1 b
2 2 NaN
