Code Sample, a copy-pastable example if possible

x = datetime.datetime.now()
df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'])
print(df.apply(lambda row: {'result': row['b']}, axis=1))
print()
x = datetime.datetime.now()
df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'], dtype='int64')
print(df.apply(lambda row: {'result': row['b']}, axis=1))
print()
x = 101
df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'])
print(df.apply(lambda row: {'result': row['b']}, axis=1))
print()
x = datetime.date.today()
df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'])
print(df.apply(lambda row: {'result': row['b']}, axis=1))
print()
x = {'x': 2}
df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'])
print(df.apply(lambda row: {'result': row['b']}, axis=1))

Problem description

The latter 4 cases produce the results that I can expect -- Series containing dicts. The first one is somehow strange as it produces a DataFrame: a b 0 NaN NaN 1 NaN NaN

Also, when I change the first case's print line to print(df.apply(lambda row: [10,20], axis=1)) I am also receiving a DataFrame: a b 0 10 20 1 10 20

Expected Output

0 {'result': 1} 1 {'result': 2} dtype: object

0 {'result': 1} 1 {'result': 2} dtype: object

0 {'result': 1} 1 {'result': 2} dtype: object

0 {'result': 1} 1 {'result': 2} dtype: object

Output of pd.show_versions()

[paste the output of ``pd.show_versions()`` here below this line] INSTALLED VERSIONS ------------------ commit: None python: 3.6.3.final.0 python-bits: 64 OS: Linux OS-release: 4.4.0-96-generic machine: x86_64 processor: x86_64 byteorder: little LC_ALL: None LANG: en_US.UTF-8 LOCALE: en_US.UTF-8 pandas: 0.20.3 pytest: None pip: 9.0.1 setuptools: 36.5.0 Cython: None numpy: 1.13.3 scipy: 0.19.1 xarray: None IPython: 6.2.1 sphinx: None patsy: None dateutil: 2.6.1 pytz: 2017.2 blosc: None bottleneck: None tables: None numexpr: None feather: None matplotlib: 2.1.0 openpyxl: None xlrd: None xlwt: None xlsxwriter: None lxml: None bs4: None html5lib: 1.0b10 sqlalchemy: None pymysql: None psycopg2: 2.7.3.1 (dt dec pq3 ext lo64) jinja2: 2.9.6 s3fs: None pandas_gbq: None pandas_datareader: None

Comment From: sinhrks

Basically, .apply returning a dict results in DataFrame. Thus, I think your expected output is caused by a bug.

x = datetime.datetime.now()
df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'])
print(df.apply(lambda row: {'a': row['b'], 'b': row['a']}, axis=1))
#    a                          b
# 0  1 2017-10-23 19:40:42.481066
# 1  2 2017-10-23 19:40:42.481066

df = pd.DataFrame([[x, 1], [x, 2]], columns=['a', 'b'])
print(df.apply(lambda row: {'a': row['b']}, axis=1))
#      a   b
# 0  1.0 NaN
# 1  2.0 NaN

Comment From: jreback

we already note this bug multiple times: https://github.com/pandas-dev/pandas/issues?q=is%3Aissue+label%3AApply+is%3Aclosed

Comment From: jreback

though it’s not actually a bug see he related issues

Comment From: jreback

master is here: https://github.com/pandas-dev/pandas/issues/15628