“Piping” output from one function to another using Python infix syntax

瘦欲@ 提交于 2019-11-28 20:24:21

It is hard to implement this using the bitwise or operator because pandas.DataFrame implements it. If you don't mind replacing | with >>, you can try this:

import pandas as pd

def select(df, *args):
    cols = [x for x in args]
    return df[cols]


def rename(df, **kwargs):
    for name, value in kwargs.items():
        df = df.rename(columns={'%s' % name: '%s' % value})
    return df


class SinkInto(object):
    def __init__(self, function, *args, **kwargs):
        self.args = args
        self.kwargs = kwargs
        self.function = function

    def __rrshift__(self, other):
        return self.function(other, *self.args, **self.kwargs)

    def __repr__(self):
        return "<SinkInto {} args={} kwargs={}>".format(
            self.function, 
            self.args, 
            self.kwargs
        )

df = pd.DataFrame({'one' : [1., 2., 3., 4., 4.],
                   'two' : [4., 3., 2., 1., 3.]})

Then you can do:

>>> df
   one  two
0    1    4
1    2    3
2    3    2
3    4    1
4    4    3

>>> df = df >> SinkInto(select, 'one') \
            >> SinkInto(rename, one='new_one')
>>> df
   new_one
0        1
1        2
2        3
3        4
4        4

In Python 3 you can abuse unicode:

>>> print('\u01c1')
ǁ
>>> ǁ = SinkInto
>>> df >> ǁ(select, 'one') >> ǁ(rename, one='new_one')
   new_one
0        1
1        2
2        3
3        4
4        4

[update]

Thanks for your response. Would it be possible to make a separate class (like SinkInto) for each function to avoid having to pass the functions as an argument?

How about a decorator?

def pipe(original):
    class PipeInto(object):
        data = {'function': original}

        def __init__(self, *args, **kwargs):
            self.data['args'] = args
            self.data['kwargs'] = kwargs

        def __rrshift__(self, other):
            return self.data['function'](
                other, 
                *self.data['args'], 
                **self.data['kwargs']
            )

    return PipeInto


@pipe
def select(df, *args):
    cols = [x for x in args]
    return df[cols]


@pipe
def rename(df, **kwargs):
    for name, value in kwargs.items():
        df = df.rename(columns={'%s' % name: '%s' % value})
    return df

Now you can decorate any function that takes a DataFrame as the first argument:

>>> df >> select('one') >> rename(one='first')
   first
0      1
1      2
2      3
3      4
4      4

Python is awesome!

I know that languages like Ruby are "so expressive" that it encourages people to write every program as new DSL, but this is kind of frowned upon in Python. Many Pythonists consider operator overloading for a different purpose as a sinful blasphemy.

[update]

User OHLÁLÁ is not impressed:

The problem with this solution is when you are trying to call the function instead of piping. – OHLÁLÁ

You can implement the dunder-call method:

def __call__(self, df):
    return df >> self

And then:

>>> select('one')(df)
   one
0  1.0
1  2.0
2  3.0
3  4.0
4  4.0

Looks like it is not easy to please OHLÁLÁ:

In that case you need to call the object explicitly:
select('one')(df) Is there a way to avoid that? – OHLÁLÁ

Well, I can think of a solution but there is a caveat: your original function must not take a second positional argument that is a pandas dataframe (keyword arguments are ok). Lets add a __new__ method to our PipeInto class inside the docorator that tests if the first argument is a dataframe, and if it is then we just call the original function with the arguments:

def __new__(cls, *args, **kwargs):
    if args and isinstance(args[0], pd.DataFrame):
        return cls.data['function'](*args, **kwargs)
    return super().__new__(cls)

It seems to work but probably there is some downside I was unable to spot.

>>> select(df, 'one')
   one
0  1.0
1  2.0
2  3.0
3  4.0
4  4.0

>>> df >> select('one')
   one
0  1.0
1  2.0
2  3.0
3  4.0
4  4.0

While I can't help mentioning that using dplyr in Python might the closest thing to having in dplyr in Python (it has the rshift operator, but as a gimmick), I'd like to also point out that the pipe operator might only be necessary in R because of its use of generic functions rather than methods as object attributes. Method chaining gives you essentially the same without having to override operators:

dataf = (DataFrame(mtcars).
         filter('gear>=3').
         mutate(powertoweight='hp*36/wt').
         group_by('gear').
         summarize(mean_ptw='mean(powertoweight)'))

Note wrapping the chain between a pair of parenthesis lets you break it into multiple lines without the need for a trailing \ on each line.

You can use sspipe library, and use the following syntax:

from sspipe import p
df = df | p(select, 'one') \
        | p(rename, one = 'new_one')

I couldn't find a built-in way of doing this, so I created a class that uses the __call__ operator because it supports *args/**kwargs:

class Pipe:
    def __init__(self, value):
        """
        Creates a new pipe with a given value.
        """
        self.value = value
    def __call__(self, func, *args, **kwargs):
        """
        Creates a new pipe with the value returned from `func` called with
        `args` and `kwargs` and it's easy to save your intermedi.
        """
        value = func(self.value, *args, **kwargs)
        return Pipe(value)

The syntax takes some getting used to, but it allows for piping.

def get(dictionary, key):
    assert isinstance(dictionary, dict)
    assert isinstance(key, str)
    return dictionary.get(key)

def keys(dictionary):
    assert isinstance(dictionary, dict)
    return dictionary.keys()

def filter_by(iterable, check):
    assert hasattr(iterable, '__iter__')
    assert callable(check)
    return [item for item in iterable if check(item)]

def update(dictionary, **kwargs):
    assert isinstance(dictionary, dict)
    dictionary.update(kwargs)
    return dictionary


x = Pipe({'a': 3, 'b': 4})(update, a=5, c=7, d=8, e=1)
y = (x
    (keys)
    (filter_by, lambda key: key in ('a', 'c', 'e', 'g'))
    (set)
    ).value
z = x(lambda dictionary: dictionary['a']).value

assert x.value == {'a': 5, 'b': 4, 'c': 7, 'd': 8, 'e': 1}
assert y == {'a', 'c', 'e'}
assert z == 5
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!