inconsistency in concat behavior in pandas 0.21.0

[dsspiegel]

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

[jreback]

this was just fixed in #18191 in master

[jreback]

duplicated #18178 and #18187