python - 更高效的时间增量计算 python 3

Question

抱歉，我是Python新手。

我有一个实体数据框，每月记录一次值。对于数据框中的每个唯一实体，我找到最大值，然后找到最大值对应的月份。使用最大值月份，可以以天为单位计算每个其他唯一实体的月份与最大月份之间的时间增量。这适用于小型数据帧。

我知道我的循环性能不佳，无法扩展到更大的数据帧(例如，3M 行(+156MB))。经过几周的研究，我发现我的循环已经退化，并且感觉有一个 numpy 解决方案或更Pythonic 的解决方案。有人能找到一种更高效的方法来计算以天为单位的时间增量吗？

我在 lambda 函数中尝试了各种 value.shift(x) 计算，但峰值不一致。我还尝试计算更多列以最小化循环计算。

import pandas as pd

df = pd.DataFrame({'entity':['A','A','A','A','B','B','B','C','C','C','C','C'], 'month': ['10/31/2018','11/30/2018','12/31/2018','1/31/2019','1/31/2009','2/28/2009','3/31/2009','8/31/2011','9/30/2011','10/31/2011','11/30/2011','12/31/2011'], 'value':['80','600','500','400','150','300','100','200','250','300','200','175'], 'month_number': ['1','2','3','4','1','2','3','1','2','3','4','5']})

df['month'] = df['month'].apply(pd.to_datetime)

for entity in set(df['entity']):
    # set peak value
    peak_value = df.loc[df['entity'] == entity, 'value'].max()
    # set peak value date
    peak_date = df.loc[(df['entity'] == entity) & (df['value'] == peak_value), 'month'].min()
    # subtract peak date from current date
    delta = df.loc[df['entity'] == entity, 'month'] - peak_date
    # update days_delta with delta in days
    df.loc[df['entity'] == entity, 'days_delta'] = delta

结果:

entity   month   value   month_number   days_delta
A       2018-10-31   80    1    0 days
A       2018-11-30    600    2  30 days
A       2018-12-31  500 3   61 days
A       2019-01-31  400 4   92 days
B       2009-01-31  150 1   -28 days
B       2009-02-28  300 2   0 days
B       2009-03-31  100 3   31 days
C       2011-08-31  200 1   -61 days
C       2011-09-30  250 2   -31 days
C       2011-10-31  300 3   0 days
C       2011-11-30  200 4   30 days
C       2011-12-31  175 5   61 days

Answer 1

设置

首先我们还要确保value 是数字

df = pd.DataFrame({
    'entity':['A','A','A','A','B','B','B','C','C','C','C','C'],
    'month': ['10/31/2018','11/30/2018','12/31/2018','1/31/2019',
              '1/31/2009','2/28/2009','3/31/2009','8/31/2011',
              '9/30/2011','10/31/2011','11/30/2011','12/31/2011'],
    'value':['80','600','500','400','150','300','100','200','250','300','200','175'],
    'month_number': ['1','2','3','4','1','2','3','1','2','3','4','5']
})

df['month'] = df['month'].apply(pd.to_datetime)
df['value'] = pd.to_numeric(df['value'])

<小时/>

变换和idxmax

max_months = df.groupby('entity').value.transform('idxmax').map(df.month)
df.assign(days_delta=df.month - max_months)

   entity      month  value month_number days_delta
0       A 2018-10-31     80            1   -30 days
1       A 2018-11-30    600            2     0 days
2       A 2018-12-31    500            3    31 days
3       A 2019-01-31    400            4    62 days
4       B 2009-01-31    150            1   -28 days
5       B 2009-02-28    300            2     0 days
6       B 2009-03-31    100            3    31 days
7       C 2011-08-31    200            1   -61 days
8       C 2011-09-30    250            2   -31 days
9       C 2011-10-31    300            3     0 days
10      C 2011-11-30    200            4    30 days
11      C 2011-12-31    175            5    61 days