Python Pandas Groupby在try 聚合列时引发ValueError：LEN(索引)！=len(标签)

我有一些列是浮点数的数据，我想将它们按四舍五入为的整数进行聚合.在下面的MWE中，预期yields 应为

   912
0  2.5
1  1.5

因为所有列元素都四舍五入为912.

MWe:

import pandas as pd
temp = pd.DataFrame({911.7: {0: 0, 1: 1}, 911.9: {0: 2.0, 1: 0.0}, 912.0: {0: 0.5, 1: 0.5}})
round_to = 1
price_digits=1
rounded = [round(round(x / round_to) * round_to, price_digits) for x in temp.columns]
temp.groupby(by=rounded, axis=1).sum()

当实际运行时，traceback将是:

Traceback (most recent call last):
  File "D:\Anaconda3\lib\site-packages\IPython\core\interactiveshell.py", line 3331, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)
  File "<ipython-input-17-983fbc3f7113>", line 1, in <module>
    temp.groupby(by=rounded, axis=1).sum()
  File "D:\Anaconda3\lib\site-packages\pandas\core\groupby\groupby.py", line 1378, in f
    return self._cython_agg_general(alias, alt=npfunc, **kwargs)
  File "D:\Anaconda3\lib\site-packages\pandas\core\groupby\generic.py", line 1004, in _cython_agg_general
    how, alt=alt, numeric_only=numeric_only, min_count=min_count
  File "D:\Anaconda3\lib\site-packages\pandas\core\groupby\generic.py", line 1033, in _cython_agg_blocks
    block.values, how, axis=1, min_count=min_count
  File "D:\Anaconda3\lib\site-packages\pandas\core\groupby\ops.py", line 587, in aggregate
    "aggregate", values, how, axis, min_count=min_count
  File "D:\Anaconda3\lib\site-packages\pandas\core\groupby\ops.py", line 530, in _cython_operation
    result, counts, values, codes, func, is_datetimelike, min_count
  File "D:\Anaconda3\lib\site-packages\pandas\core\groupby\ops.py", line 608, in _aggregate
    agg_func(result, counts, values, comp_ids, min_count)
  File "pandas\_libs\groupby.pyx", line 464, in pandas._libs.groupby._group_add
ValueError: len(index) != len(labels)

这很令人费解，因为len(rounded)==len(temp.columns)==3.似乎没有长度不匹配的情况.

怎样才是达到我的目的的合适方式？提前谢谢！

Pandas 版:'1.0.1'.Python版本:Python 3.7.6 (default, Jan 8 2020, 16:21:45) [MSC v.1916 32 bit (Intel)].

在大多数情况下，MWE确实有效.例如，当我们将第三个Column元素从912.0更改为912.3时:

import pandas as pd
round_to = 1
price_digits=1
temp = pd.DataFrame({911.7: {0: 0, 1: 1}, 911.9: {0: 2.0, 1: 0.0}, 912.3: {0: 0.5, 1: 0.5}})
rounded = [round(round(x / round_to) * round_to, price_digits) for x in temp.columns]
temp.groupby(by=rounded, axis=1).sum()

输出将为

Out[14]: 
   912
0  2.5
1  1.5