inconsistant result shape from dataframe.apply depending on if a date column is present

[KevinGrealish]

A small, complete example of the issue

import pandas as pd
df1 = pd.DataFrame.from_items([('A', [1,2,3]), ('B', [4, 5, 6])])
df2 = pd.DataFrame.from_items([('A', [pd.datetime(1970,1,1), pd.datetime(1970,1,1), pd.datetime(1970,1,1)]), ('B', [4, 5, 6])])
f = lambda row: [row.B + 3]
r1 = df1.apply(f, axis=1)
r2 = df2.apply(f, axis=1)
# bug: r1 and r2 are different and different in shape.
# expect: r1 and r2 to be the same values and shape.
r1
r2
# a clue to the difference:
(df1._is_mixed_type,df1._is_datelike_mixed_type)
(df2._is_mixed_type,df2._is_datelike_mixed_type)

Expected Output

0    [7]
1    [8]
2    [9]

Output of pd.show_versions()

## INSTALLED VERSIONS

commit: None
python: 3.5.1.final.0
python-bits: 64
OS: Windows
OS-release: 10
machine: AMD64
processor: Intel64 Family 6 Model 58 Stepping 9, GenuineIntel
byteorder: little
LC_ALL: None
LANG: None

pandas: 0.18.1
nose: None
pip: 8.1.2
setuptools: 20.3
Cython: None
numpy: 1.11.1
scipy: None
statsmodels: None
xarray: None
IPython: None
sphinx: None
patsy: None
dateutil: 2.5.3
pytz: 2016.6.1
blosc: None
bottleneck: None
tables: None
numexpr: None
matplotlib: None
openpyxl: None
xlrd: None
xlwt: None
xlsxwriter: None
lxml: None
bs4: None
html5lib: None
httplib2: None
apiclient: None
sqlalchemy: None
pymysql: None
psycopg2: None
jinja2: None
boto: None
pandas_datareader: None

[jreback]

this is irrespective of other dtypes. You are returning a 2 element list so naturally pandas will try to coerce to the original shape as its compatible. Returning a list is not labeled so that is the only thing pandas can do. You can do this if you want. I also suppose more/better documentation is possible.

In [8]: df1.apply(lambda row: Series([row.B + 3], ['result']), axis=1)
Out[8]: 
   result
0       7
1       8
2       9

[KevinGrealish]

Jeff, are you saying it's by design that r1 and r2 come out different, (By Design), or are you saying they should be the same (Won't Fix)?

[KevinGrealish]

In my case, f is a function that takes a string and returns a list of strings and it needs to be applied to each row. i.e. I want result cells to have lists in them, I simplified this in the example, but here is a better repro. If what I was doing is under specified to Pandas, I want it to error all the time, not just when another column happens to be a date. It cost a me a lot of time trying to figure out why this only worked some of the time.:

import pandas as pd
df1 = pd.DataFrame.from_items([('A', [1,2,3]), ('B', ["ABCD", "EFGH", "IJKL"])])
df2 = pd.DataFrame.from_items([('A', [pd.datetime(1970,1,1), pd.datetime(1970,1,1), pd.datetime(1970,1,1)]), ('B', ["ABCD", "EFGH", "IJKL"])])
f = lambda row: [char for char in row.B]
r1 = df1.apply(f, axis=1)  # works
r2 = df2.apply(f, axis=1)  # fails
# bug: apply when date present crashes.
# expect: r1 and r2 to be the same values and shape. r1 has the expected value.

The problem is that I now have code that crashes only when a date column is present. Your workaround to use a series:

f = lambda row: pd.Series([[char for char in row.B]], ["result"])
r1 = df1.apply(f, axis=1)
r2 = df2.apply(f, axis=1)