Question

我有一个看起来像这样的DataFrame（请参阅下面的代码以重现它）：

         date      id_  val
0  2017-01-08     a; b  9.3
1  2017-01-07  a; b; c  7.9
2  2017-01-07        a  7.3
3  2017-01-06        b  9.0
4  2017-01-06        c  8.1
5  2017-01-05        c  7.4
6  2017-01-05        d  7.1
7  2017-01-05        a  7.0
8  2017-01-04     b; a  7.7
9  2017-01-04     c; a  5.3
10 2017-01-04     a; c  8.0

我希望将id_的各个（分号分隔）元素分组，并计算val 的累积平均值，但不包括每个日期。这应该为任何id_的第一次出现提供NaN，然后我填充一些任意值（这里为6.0）。

输出：

id_    
a    0     6.0000
     1     9.3000
     2     8.6000
     7     8.1667
     8     7.8750
     9     7.8400
     10    7.4167
b    0     6.0000
     1     9.3000
     3     8.6000
     8     8.7333
c    1     6.0000  # fill value
     4     7.9000  # first actual occurrence of id_='c'
     5     8.0000  # cumulative mean of the first two 'c'  
     9     7.8000
     10    7.1750
d    6     6.0000
Name: val, dtype: float64

这是我目前的流程，速度很慢 - 可以改进吗？其次，我可以在最终结果中保留date col吗？

# seems like `pd.melt` might be more direct here
df.sort_values('date', inplace=True)
stacked = df.id_.str.split('; ', expand=True).stack()
stacked.index = stacked.index.droplevel(1)
stacked = stacked.to_frame()\
    .merge(df, left_index=True, right_index=True)\
    .drop('id_', axis=1)\
    .rename({0: 'id_'}, axis=1)


def trend_scorer(s: pd.Series, fillvalue=6.):
    return s['val'].expanding().mean().shift(1).fillna(fillvalue)


stacked.groupby('id_').apply(trend_scorer)

DataFrame创建：

import pandas as pd

data = \
{'id_': {0: 'a; b',
             1: 'a; b; c',
             2: 'a',
             3: 'b',
             4: 'c',
             5: 'c',
             6: 'd',
             7: 'a',
             8: 'b; a',
             9: 'c; a',
             10: 'a; c'},
 'date': {0: '1/8/17',
          1: '1/7/17',
          2: '1/7/17',
          3: '1/6/17',
          4: '1/6/17',
          5: '1/5/17',
          6: '1/5/17',
          7: '1/5/17',
          8: '1/4/17',
          9: '1/4/17',
          10: '1/4/17'},
 'val': {0: 9.3,
           1: 7.9,
           2: 7.3,
           3: 9.0,
           4: 8.1,
           5: 7.4,
           6: 7.1,
           7: 7.0,
           8: 7.7,
           9: 5.3,
           10: 8.0}}

df = pd.DataFrame(data)
df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)

Answer 1

groupby/apply通常是一个相对较慢的操作（与Pandas的Cythonized或NumPy的矢量化操作相比），因为它需要为每个组调用一次Python函数。尽可能避免使用它。在这种情况下，您可以使用groupby/expanding来获得适度的优势：

result = stacked.groupby('id_').expanding()['val'].mean()
result = result.groupby(level='id_').shift(1).fillna(fillvalue)

要使用result重新加入此stacked，您可以使用DataFrame.join - 这里的主要问题是DataFrames必须共享相同的索引级别在你加入之前：

result = stacked.set_index('id_', append=True).swaplevel().join(result)

在您的小示例DataFrame上，alt比orig快约1.3倍：

In [500]: %timeit orig(df)
100 loops, best of 3: 12.5 ms per loop

In [501]: %timeit alt(df)
100 loops, best of 3: 9.49 ms per loop

在一个包含10K行和1000组的大型DataFrame上，alt的速度优势大致相同：

In [504]: %timeit orig(df)
1 loop, best of 3: 2.34 s per loop

In [505]: %timeit alt(df)
1 loop, best of 3: 1.95 s per loop

（固定费用，例如stacked.set_index('id_', append=True).swaplevel().join(result)会影响使用groupby/expanding代替groupby/apply的相对较小的好处。

以下是用于制作上述基准的代码：

import pandas as pd
import numpy as np

def trend_scorer(s: pd.Series, fillvalue=6.):
    return s['val'].expanding().mean().shift(1).fillna(fillvalue)

def orig(df):
    stacked = df.id_.str.split('; ', expand=True).stack()
    stacked.index = stacked.index.droplevel(1)
    stacked = (stacked.to_frame()
               .merge(df, left_index=True, right_index=True)
               .drop('id_', axis=1)
               .rename(columns={0: 'id_'}))                   
    result = stacked.groupby('id_').apply(trend_scorer)
    result = result.rename('expanding mean')
    result = stacked.set_index('id_', append=True).swaplevel().join(result)
    return result

def alt(df, fillvalue=6.0):
    stacked = df['id_'].str.split('; ', expand=True).stack()
    stacked.index = stacked.index.droplevel(1)
    stacked = (df.drop('id_', axis=1)
               .join(stacked.rename('id_')))
    result = stacked.groupby('id_').expanding()['val'].mean()
    result = result.groupby(level='id_').shift(1).fillna(fillvalue)
    result = result.rename('expanding mean')
    result = stacked.set_index('id_', append=True).swaplevel().join(result)
    return result

data = {'id_': {0: 'a; b', 1: 'a; b; c', 2: 'a', 3: 'b', 4: 'c', 5: 'c', 6: 'd', 7: 'a', 8: 'b; a', 9: 'c; a', 10: 'a; c'}, 'date': {0: '1/8/17', 1: '1/7/17', 2: '1/7/17', 3: '1/6/17', 4: '1/6/17', 5: '1/5/17', 6: '1/5/17', 7: '1/5/17', 8: '1/4/17', 9: '1/4/17', 10: '1/4/17'}, 'val': {0: 9.3, 1: 7.9, 2: 7.3, 3: 9.0, 4: 8.1, 5: 7.4, 6: 7.1, 7: 7.0, 8: 7.7, 9: 5.3, 10: 8.0}}
df = pd.DataFrame(data)
df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)
df = df.sort_values('date')

assert alt(df).equals(orig(df))

这就是我为上述基准创建更大的测试数据框架的方法：

import numpy as np
def make_df(N=10000, seed=2018):
    np.random.seed(seed)
    data = []
    for date in pd.date_range('2017-1-1', periods=N):
        for i in range(np.random.randint(1, 10)):
            ids = '; '.join(np.random.choice(1000, size=np.random.randint(1, 10)).astype(str))
            data.append((date, ids))

    df = pd.DataFrame(data, columns=['date', 'id_'])
    df['val'] = np.random.uniform(1, 10, size=len(df))
    return df

df = make_df()

Groupby累积平均值，具有宽/长旋转

1 个答案: