column names with leading underscores with df.itertuples cause AttributeError when accessed

Code Sample, a copy-pastable example if possible

data = [['Alex',10],['Bob',12],['Clarke',13]]
df = pd.DataFrame(data, columns=['_name','age'])
try:
    for row in df.itertuples():
        val = getattr(row, '_name')
        print(val)
except AttributeError:
    print('_name column not accessible')

Problem description

DataFrame columns named with a leading underscore are not accessible. While it is possible to rename the column to access the data, Pandas users unfamiliar with this issue might spend a bit of time developing their own workaround. The perplexing thing about this is when the column names are output for inspection you can see the column name with the leading underscore, e.g. print(df.dtypes.index)

[this should explain why the current behaviour is a problem and why the expected output is a better solution.]

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Expected Output

The expected output is the data from the column whose name begins with an underscore.

Output of pd.show_versions()

[paste the output of pd.show_versions() here below this line] INSTALLED VERSIONS

commit: None python: 3.6.5.final.0 python-bits: 64 OS: Windows OS-release: 2012ServerR2 machine: AMD64 processor: Intel64 Family 6 Model 85 Stepping 4, GenuineIntel byteorder: little LC_ALL: None LANG: None LOCALE: None.None

pandas: 0.23.4 pytest: None pip: 10.0.1 setuptools: 39.0.1 Cython: None numpy: 1.15.3 scipy: None pyarrow: None xarray: None IPython: None sphinx: None patsy: None dateutil: 2.7.5 pytz: 2018.7 blosc: None bottleneck: None tables: None numexpr: None feather: None matplotlib: None openpyxl: None xlrd: None xlwt: None xlsxwriter: None lxml: None bs4: None html5lib: None sqlalchemy: 1.2.6 pymysql: None psycopg2: None jinja2: 2.10 s3fs: None fastparquet: None pandas_gbq: None pandas_datareader: None

[TomAugspurger]

@gaetano-guerriero that’s a bug. Doing a release today with a fix.

[gaetano-guerriero]

I think it would be important to document that the silent renaming can happen elsewhere:

>>> df = pd.DataFrame({'a': [1, 2], '_b': [3, 4]})
>>> df.to_dict(orient='records')
[{'a': 1, '_1': 3}, {'a': 2, '_1': 4}]