我想尝试以下内容：
DF1
price side timestamp
timestamp
2016-01-04 00:01:15.631331072 0.7286 2 1451865675631331
2016-01-04 00:01:15.631399936 0.7286 2 1451865675631400
2016-01-04 00:01:15.631860992 0.7286 2 1451865675631861
2016-01-04 00:01:15.631866112 0.7286 2 1451865675631866
和
DF2
bid bid_size offer offer_size
timestamp
2016-01-04 00:00:31.331441920 0.7284 4000000 0.7285 1000000
2016-01-04 00:00:53.631324928 0.7284 4000000 0.7290 4000000
2016-01-04 00:01:03.131234048 0.7284 5000000 0.7286 4000000
2016-01-04 00:01:12.131444992 0.7285 1000000 0.7286 4000000
2016-01-04 00:01:15.631364096 0.7285 4000000 0.7290 4000000
用
data = pd.concat([df1,df2], axis=1)
但是我得到了以下输出：
InvalidIndexError Traceback (most recent call last)
<ipython-input-38-2e88458f01d7> in <module>()
----> 1 data = pd.concat([df1,df2], axis=1)
2 data = data.fillna(method='pad')
3 data = data.fillna(method='bfill')
4 data['timestamp'] = data.index.values#converting to datetime
5 data['timestamp'] = pd.to_datetime(data['timestamp'])#converting to datetime
/usr/local/lib/python2.7/site-packages/pandas/tools/merge.pyc in concat(objs, axis, join, join_axes, ignore_index, keys, levels, names, verify_integrity, copy)
810 keys=keys, levels=levels, names=names,
811 verify_integrity=verify_integrity,
--> 812 copy=copy)
813 return op.get_result()
814
/usr/local/lib/python2.7/site-packages/pandas/tools/merge.pyc in __init__(self, objs, axis, join, join_axes, keys, levels, names, ignore_index, verify_integrity, copy)
947 self.copy = copy
948
--> 949 self.new_axes = self._get_new_axes()
950
951 def get_result(self):
/usr/local/lib/python2.7/site-packages/pandas/tools/merge.pyc in _get_new_axes(self)
1013 if i == self.axis:
1014 continue
-> 1015 new_axes[i] = self._get_comb_axis(i)
1016 else:
1017 if len(self.join_axes) != ndim - 1:
/usr/local/lib/python2.7/site-packages/pandas/tools/merge.pyc in _get_comb_axis(self, i)
1039 raise TypeError("Cannot concatenate list of %s" % types)
1040
-> 1041 return _get_combined_index(all_indexes, intersect=self.intersect)
1042
1043 def _get_concat_axis(self):
/usr/local/lib/python2.7/site-packages/pandas/core/index.pyc in _get_combined_index(indexes, intersect)
6120 index = index.intersection(other)
6121 return index
-> 6122 union = _union_indexes(indexes)
6123 return _ensure_index(union)
6124
/usr/local/lib/python2.7/site-packages/pandas/core/index.pyc in _union_indexes(indexes)
6149
6150 if hasattr(result, 'union_many'):
-> 6151 return result.union_many(indexes[1:])
6152 else:
6153 for other in indexes[1:]:
/usr/local/lib/python2.7/site-packages/pandas/tseries/index.pyc in union_many(self, others)
959 else:
960 tz = this.tz
--> 961 this = Index.union(this, other)
962 if isinstance(this, DatetimeIndex):
963 this.tz = tz
/usr/local/lib/python2.7/site-packages/pandas/core/index.pyc in union(self, other)
1553 result.extend([x for x in other._values if x not in value_set])
1554 else:
-> 1555 indexer = self.get_indexer(other)
1556 indexer, = (indexer == -1).nonzero()
1557
/usr/local/lib/python2.7/site-packages/pandas/core/index.pyc in get_indexer(self, target, method, limit, tolerance)
1890
1891 if not self.is_unique:
-> 1892 raise InvalidIndexError('Reindexing only valid with uniquely'
1893 ' valued Index objects')
1894
InvalidIndexError: Reindexing only valid with uniquely valued Index objects
我删除了其他列并删除了重复项和NA可能存在冲突 - 但我根本不知道什么是错的。
请帮忙 感谢
答案 0 :(得分：26)
pd.concat要求 indices 是唯一的。要remove rows with duplicate indices，请使用
df = df.loc[~df.index.duplicated(keep='first')]
import pandas as pd
from pandas import Timestamp
df1 = pd.DataFrame(
{'price': [0.7286, 0.7286, 0.7286, 0.7286],
'side': [2, 2, 2, 2],
'timestamp': [1451865675631331, 1451865675631400,
1451865675631861, 1451865675631866]},
index=pd.DatetimeIndex(['2000-1-1', '2000-1-1', '2001-1-1', '2002-1-1']))
df2 = pd.DataFrame(
{'bid': [0.7284, 0.7284, 0.7284, 0.7285, 0.7285],
'bid_size': [4000000, 4000000, 5000000, 1000000, 4000000],
'offer': [0.7285, 0.729, 0.7286, 0.7286, 0.729],
'offer_size': [1000000, 4000000, 4000000, 4000000, 4000000]},
index=pd.DatetimeIndex(['2000-1-1', '2001-1-1', '2002-1-1', '2003-1-1', '2004-1-1']))
df1 = df1.loc[~df1.index.duplicated(keep='first')]
# price side timestamp
# 2000-01-01 0.7286 2 1451865675631331
# 2001-01-01 0.7286 2 1451865675631861
# 2002-01-01 0.7286 2 1451865675631866
df2 = df2.loc[~df2.index.duplicated(keep='first')]
# bid bid_size offer offer_size
# 2000-01-01 0.7284 4000000 0.7285 1000000
# 2001-01-01 0.7284 4000000 0.7290 4000000
# 2002-01-01 0.7284 5000000 0.7286 4000000
# 2003-01-01 0.7285 1000000 0.7286 4000000
# 2004-01-01 0.7285 4000000 0.7290 4000000
result = pd.concat([df1, df2], axis=0)
print(result)
bid bid_size offer offer_size price side timestamp
2000-01-01 NaN NaN NaN NaN 0.7286 2 1.451866e+15
2001-01-01 NaN NaN NaN NaN 0.7286 2 1.451866e+15
2002-01-01 NaN NaN NaN NaN 0.7286 2 1.451866e+15
2000-01-01 0.7284 4000000 0.7285 1000000 NaN NaN NaN
2001-01-01 0.7284 4000000 0.7290 4000000 NaN NaN NaN
2002-01-01 0.7284 5000000 0.7286 4000000 NaN NaN NaN
2003-01-01 0.7285 1000000 0.7286 4000000 NaN NaN NaN
2004-01-01 0.7285 4000000 0.7290 4000000 NaN NaN NaN
注意还有pd.join，它可以根据索引加入DataFrames，
并根据how参数处理非唯一索引。行重复
索引未被删除。
In [94]: df1.join(df2)
Out[94]:
price side timestamp bid bid_size offer \
2000-01-01 0.7286 2 1451865675631331 0.7284 4000000 0.7285
2000-01-01 0.7286 2 1451865675631400 0.7284 4000000 0.7285
2001-01-01 0.7286 2 1451865675631861 0.7284 4000000 0.7290
2002-01-01 0.7286 2 1451865675631866 0.7284 5000000 0.7286
offer_size
2000-01-01 1000000
2000-01-01 1000000
2001-01-01 4000000
2002-01-01 4000000
In [95]: df1.join(df2, how='outer')
Out[95]:
price side timestamp bid bid_size offer offer_size
2000-01-01 0.7286 2 1.451866e+15 0.7284 4000000 0.7285 1000000
2000-01-01 0.7286 2 1.451866e+15 0.7284 4000000 0.7285 1000000
2001-01-01 0.7286 2 1.451866e+15 0.7284 4000000 0.7290 4000000
2002-01-01 0.7286 2 1.451866e+15 0.7284 5000000 0.7286 4000000
2003-01-01 NaN NaN NaN 0.7285 1000000 0.7286 4000000
2004-01-01 NaN NaN NaN 0.7285 4000000 0.7290 4000000
答案 1 :(得分：15)
您可以缓解此错误，而无需更改数据或删除重复项。只需使用DataFrame.reset_index创建一个新索引：
df = df.reset_index()
旧索引在数据框中保留为一列，但如果您不需要，则可以执行以下操作：
df = df.reset_index(drop=True)
有些人更喜欢：
df.reset_index(inplace=True, drop=True)
答案 2 :(得分：7)
当您搜索错误但答案不完整时，这篇文章会出现在最前面，所以让我添加我的。发生此错误的另一个原因是：如果您的数据框中有重复的列，您将无法连接并引发此错误。事实上，即使在原始问题中，也有两列名为 timestamp。因此，对于您尝试连接的所有数据帧，检查 len(df.columns) = len(set(df.columns)) 是否是值得的。
答案 3 :(得分：1)
作为对 Nicholas Morley 的回答的补充，当您发现即使这样也行不通时：
df = df.reset_index(drop=True)
您应该检查列是否唯一。如果不是，即使重置索引也不起作用。应首先删除重复的列以使其正常工作。
答案 4 :(得分：0)
另一种可能引发这种类型的错误的情况是，当您的列中具有唯一值（熵为0）时。 在这种情况下，您可以在该列中注入小的高斯噪声，或者将其完全消除。
答案 5 :(得分：0)
此页面上的最佳解决方案：
https://pandas.pydata.org/pandas-docs/version/0.20/merging.html
df = pd.concat([df1, df2], axis=1, join_axes=[df1.index])
答案 6 :(得分：0)
就我而言，问题是因为我有重复的列名。