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Martin Hope
amance
Asked: 2024-12-17 00:55:33 +0800 CST
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我正在尝试寻找一种更好/更快的方法来做到这一点。我有一个相当大的数据集（约 2 亿行），每行都有单独的日期。我想找出 2018 年至 2019 年每组的平均年利率。我知道我可以用结果创建一个小的 df 并将其合并回去，但我试图找到一种使用转换的方法。不确定合并是否会更快。一行代码可以加分。

示例数据

rng = np.random.default_rng(seed=123)
df = pd.DataFrame({'group':rng.choice(list('ABCD'), 100),
                   'date':[(pd.to_datetime('2018')+pd.Timedelta(days=x)).normalize() for x in rng.integers(0, 365*5, 100)],
                   'foo':rng.integers(1, 100, 100),
                   'bar':rng.integers(50, 200, 100)})
df['year'] = df['date'].dt.year

这有效

#find average 2018 and 2019 'foo' and 'bar'
for col in ['foo', 'bar']:
    for y in [2018, 2019]:
        df[col+'_'+str(y)+'_total'] = df.groupby('group')['year'].transform(lambda x: df.loc[x.where(x==y).dropna().index, col].sum())

#find 2018 and 2019 rates
for y in [2018, 2019]:
    df['rate_'+str(y)] =  df['foo_'+str(y)+'_total'].div(df['bar_'+str(y)+'_total'])

#find average rate
df['2018_2019_avg_rate'] = df[['rate_2018', 'rate_2019']].mean(axis=1)

我尝试过但不太管用的东西（我apply在切换到之前用它来测试它是否有效transform）

#gives yearly totals for each year and each column, but further 'apply'ing to find rates then averaging isn't working after I switch to transform
df.groupby(['group', 'year'])['year'].apply(lambda x: df.loc[x.where(x.between(2018, 2019)).dropna().index, ['foo', 'bar']].sum())

#close but is averaging too early
df.groupby(['group', 'year'])['year'].apply(lambda x: df.loc[i, 'foo'].sum()/denom if (denom:=df.loc[i:=x.where(x.between(2018, 2019)).dropna().index, 'bar'].sum())>0 else np.nan)
python