Pandas Grouper按频率和完整性要求

时间:2015-11-10 18:49:33

标签: python pandas python-datetime

我有月度时间序列数据,这些数据都缺少某些条目,并且由于其他原因而分散了NaN值。我需要将数据汇总到季度和年度系列,但我不想报告缺少数据的季度/年数据。例如,在下面的数据中,我不想报告2014年第一季度的数据,因为我错过了那年的1月份。

import pandas as pd, numpy as np

df = pd.DataFrame([
  ('Monthly','2014-02-1', 529.1),
  ('Monthly','2014-03-1',  67.1),
  ('Monthly','2014-04-1', np.nan), 
  ('Monthly','2014-05-1', 146.8),
  ('Monthly','2014-06-1', 469.7),
  ('Monthly','2014-07-1',  82.9),
  ('Monthly','2014-08-1', 636.9),
  ('Monthly','2014-09-1', 520.9),
  ('Monthly','2014-10-1', 217.4),
  ('Monthly','2014-11-1', 776.6),
  ('Monthly','2014-12-1',  18.4),
  ('Monthly','2015-01-1', 376.7),
  ('Monthly','2015-02-1', 266.5),
  ('Monthly','2015-03-1', np.nan),
  ('Monthly','2015-04-1', 144.1), 
  ('Monthly','2015-05-1', 385.0),
  ('Monthly','2015-06-1', 527.1),
  ('Monthly','2015-07-1', 748.5),
  ('Monthly','2015-08-1', 518.2)],
  columns=['Frequency','Date','Value'])

df['Date'] = pd.to_datetime(df['Date'])
df.set_index(['Frequency','Date'],inplace=True)
df

                      Value
Frequency Date
          2014-02-01  529.1
          2014-03-01   67.1
          2014-04-01    NaN
          2014-05-01  146.8
          2014-06-01  469.7
          2014-07-01   82.9
          2014-08-01  636.9
          2014-09-01  520.9
          2014-10-01  217.4
          2014-11-01  776.6
          2014-12-01   18.4
          2015-01-01  376.7
          2015-02-01  266.5
          2015-03-01    NaN
          2015-04-01  144.1
          2015-05-01  385.0
          2015-06-01  527.1
          2015-07-01  748.5
          2015-08-01  518.2

我尝试过使用Grouper函数,但是groupby忽略了NaN值,而Grouper实用程序并没有强制执行时间序列完整性,据我所知:

df.groupby(pd.Grouper(level='Date', freq='Q')).sum()

             Value
Date
2014-03-31  1571.2
2014-06-30   616.5
2014-09-30  1240.7
2014-12-31  1012.4
2015-03-31   643.2
2015-06-30  1056.2
2015-09-30  1266.7

我希望看到的是:

             Value
Date
2014-03-31     NaN  # Because of missing 2014-01-01
2014-06-30     NaN  # Because of NaN in 2014-04-01
2014-09-30  1240.7
2014-12-31  1012.4
2015-03-31     NaN  # Because of NaN in 2015-03-01
2015-06-30  1056.2
2015-09-30     NaN  # Because of missing 2015-09-01

这是一个很好的方法吗?

2 个答案:

答案 0 :(得分:2)

您可能想要编写自己的aggergate函数,1,如果有nan,则返回nan; 2,如果期限太短,也返回nan; 3,否则,返还总和:

In [43]:

gpy = df.groupby(pd.Grouper(level='Date', freq='Q'))

print gpy.agg(lambda x: np.nan if (np.isnan(x).any() or len(x)<3) else x.sum())

             Value
Date              
2014-03-31     NaN
2014-06-30     NaN
2014-09-30  1240.7
2014-12-31  1012.4
2015-03-31     NaN
2015-06-30  1056.2
2015-09-30     NaN

答案 1 :(得分:1)

您可以为每个具有3个元素的组创建一个布尔掩码,该掩码为True:

mask = (df.groupby(pd.Grouper(level='Date', freq='Q'))['Value'].count() != 3).values

然后只需将相应的行设置为NaN。

grouped = df.groupby(pd.Grouper(level='Date', freq='Q'))
result = grouped.sum()
mask = (grouped['Value'].count() != 3).values
result.loc[mask, 'Value'] = np.nan

产量

             Value
Date              
2014-03-31     NaN
2014-06-30     NaN
2014-09-30  1240.7
2014-12-31  1012.4
2015-03-31     NaN
2015-06-30  1056.2
2015-09-30     NaN