Pandas inconsistency in concat behavior in pandas 0.21.0

Code Sample, a copy-pastable example if possible

In [1]: from collections import OrderedDict
In [2]: import pandas as pd
In [3]: pd.__version__
Out[3]: u'0.21.0'

# the following works as expected:
In [4]: df1 = pd.DataFrame([[1, 2, 3]], columns=['a', 'b', 'c'])
In [5]: df2 = pd.DataFrame(columns=['a', 'b', 'c'])
In [6]: pd.concat([df1, df2])
Out[6]: 
   a  b  c
0  1  2  3

# however, the following seems like it does an identical thing but throws an error:
In [7]: od3 = OrderedDict([('a', [1]), ('b', [2]), ('c', [3])])
In [8]: od4 = OrderedDict([('a', []), ('b', []), ('c', [])])
In [9]: df3 = pd.DataFrame(od3)
In [10]: df4 = pd.DataFrame(od4)
In [11]: pd.concat([df3, df4])
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-11-ac4fae34c928> in <module>()
----> 1 pd.concat([df3, df4])
#...
ValueError: Shape of passed values is (3, 1), indices imply (3, 0)

Rewriting it in copy/pastable form...

This works:

df1 = pd.DataFrame([[1, 2, 3]], columns=['a', 'b', 'c'])
df2 = pd.DataFrame(columns=['a', 'b', 'c'])
pd.concat([df1, df2])

And this does not:

od3 = OrderedDict([('a', [1]), ('b', [2]), ('c', [3])])
od4 = OrderedDict([('a', []), ('b', []), ('c', [])])
df3 = pd.DataFrame(od3)
df4 = pd.DataFrame(od4)
pd.concat([df3, df4])

Problem description

This is also described here.

In the above code, df1 equals df3, and df2 and df4 are both empty dataframes with the same column names (although, strangely, df2 and df4 aren't equal according to df2.equals(df4)) but pd.concat([df1, df2]) works while pd.concat([df3, df4])results in ValueError. This did not happen in previous versions of Pandas, but when I upgraded to 0.21.0, it started happening.

Oddly, as the stackoverflow link notes, using the drop_duplicates() method on either df3 or df4 (or both) results in the concat() working, even though neither of them contains any duplicates.

Expected Output

The expected output of pd.concat([df3, df4]) is

   a  b  c
0  1  2  3

Output of `pd.show_versions()`

[paste the output of ``pd.show_versions()`` here below this line] INSTALLED VERSIONS ------------------ commit: None python: 2.7.13.final.0 python-bits: 64 OS: Darwin OS-release: 16.7.0 machine: x86_64 processor: i386 byteorder: little LC_ALL: None LANG: en_US.UTF-8 LOCALE: None.None pandas: 0.21.0 pytest: 3.2.1 pip: 9.0.1 setuptools: 32.1.0 Cython: 0.23.4 numpy: 1.13.3 scipy: 0.19.1 pyarrow: None xarray: None IPython: 5.3.0 sphinx: None patsy: 0.4.1 dateutil: 2.6.1 pytz: 2017.3 blosc: None bottleneck: None tables: None numexpr: None feather: None matplotlib: 2.0.2 openpyxl: None xlrd: None xlwt: None xlsxwriter: None lxml: None bs4: 4.6.0 html5lib: 0.999999999 sqlalchemy: 1.1.14 pymysql: 0.7.11.None psycopg2: 2.7.3.1 (dt dec pq3 ext lo64) jinja2: 2.9.6 s3fs: None fastparquet: None pandas_gbq: None pandas_datareader: 0.5.0

Comment From: jreback

this was just fixed in #18191 in master