pandas-refract 1.2.1

Split a dataframe by boolean array

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``pandas-refract``: Convenient partitioning by Truthy/Falsey array
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**pandas-refract** is an MIT licensed Python package with a simple function that allows users to divide their
dataframes by the 'Truthy' and 'Falseyness' of a provided array.

Eventually, the goal of this package is an additional feature to the Pandas library that allows users to .pop rows
from a dataframe where a condition is met. As far as I can tell this is not possible like the below example.

Ideal case would be::

target_df = df.pop(df['target_column'] == 'target_value')
non_target_df = df

What is required now is::

target_df = df[df['target_column'] == 'target_value']
non_target_df = df[df['target_column'] != 'targe_value']


Obviously, this package is not providing anything not currently possible in the current Pandas library. It does,
however, add a layer of convenience for more complex slicing where you need to separate, not remove, rows by conditions.


Examples
========

Simplest example of current Pandas requires::

df1 = df[df.column.notnull()].reset_index(drop=True)
df2 = df[df.column.isnull()].reset_index(drop=True)

or::

df1 = df[df.column == 'test_string'].reset_index(drop=True)
df2 = df[df.column != 'test_string'].reset_index(drop=True)


With pandas-refract this becomes::

df1, df2 = refract(df, df.column.notnull(), True]

and::

df1, df2 = refract(df, df.column == test_string', True]


But you don't have to pass it explicit boolean arrays::

data = {'a': ['', 'truthy', '', 'truthy'],
'b': [0, 1, 2, 3]
}

df = pd.DataFrame(data)

truthy_df, falsey_df = refract(df, df.a)


More complex examples:
*(where 'a' is Falsey and 'b' is an odd number)*
::

df1, df2 = refract(df, ((~df.a) & (df.b % 2 == 1)))