[BUG-REPORT] Trying to filter a DataFrame on a column with spaces in its name raises NameError

[vmussa]

Description Similar to #1217 and #1218. Trying to filter a DataFrame on a column with spaces in its name raises the following error: NameError: name 'column_name' is not defined.

Steps to reproduce

import vaex as vx
df = vx.open('dataset.hdf5')

boolean_mask = df['column name'] > 0 # alternatively: boolean_mask = df.column_name > 0
df[boolean_mask] # error raising bit

Software information

• Vaex version (import vaex; vaex.__version__): vaex 3.0.0 (conda-forge noarch package) vx.__version__ output: {'vaex-core': '2.0.3', 'vaex-viz': '0.4.0', 'vaex-hdf5': '0.6.0', 'vaex-server': '0.3.1', 'vaex-astro': '0.7.0', 'vaex-jupyter': '0.5.2', 'vaex-ml': '0.9.0', 'vaex-arrow': '0.5.1'}
• Vaex was installed via: conda-forge
• OS: Ubuntu 20.04.2 LTS (running on top of WSL 2, Windows 10, Build 19041)

Additional information I'm actually trying to filter this specific dataset, which came with spaced named columns, which I only saw after I had already converted the csv to hdf5. I will convert it again, but changing its columns' names first, removing the spaces. But what made me report the issue was this curious behavior: just producing the boolean mask works, while using it to filter the DataFrame produces the error, no matter which format you choose to name the column (['column name'] or .column_name).

[foooooooooooooooobar]

I see I'm not the only one https://github.com/vaexio/vaex/issues/1229

[Nicholas-Schaub]

It looks like this has to do with how vaex is using the ast library to parse expressions. It appears as though any time an expression is used, if one of the columns contains a space then it will throw an error. We are running into the same error.

[maartenbreddels]

Hi,

I cannot reproduce this, since this seems to work fine:

df = vaex.from_arrays(x=[1,2])
df['column name'] = df.x -1
df.export('spacetest.hdf5')
df = vaex.open('spacetest.hdf5')
boolean_mask = df['column name'] > 0 # alternatively: boolean_mask = df.column_name > 0
assert df[boolean_mask].sum('x') == 2

Please provide a reproducible example I can copy paste so I can fix this.

Regards,

Maarten