Pandas通过不同时间范围的两个数据框连接

Question

我试图计算某家公司在其盈利日后一年内出现在新闻上的次数，并与同一时间段内的其他公司进行比较。我有两个pandas数据帧，一个是收益日期，另一个是新闻。我的方法很慢。有更好的熊猫/裸体方式吗？在

import pandas as pd

companies = pd.DataFrame({'CompanyName': ['A', 'B', 'C'], 'EarningsDate': ['2013/01/15', '2015/03/25', '2017/05/03']})
companies['EarningsDate'] = pd.to_datetime(companies.EarningsDate)

news = pd.DataFrame({'CompanyName': ['A', 'A', 'A', 'B', 'B', 'C'], 
                     'NewsDate': ['2012/02/01', '2013/01/10', '2015/05/13' , '2012/05/23', '2013/01/03', '2017/05/01']})
news['NewsDate'] = pd.to_datetime(news.NewsDate)

companies看起来像

^{pr2}$

news看起来像

CompanyName NewsDate
0   A       2012-02-01
1   A       2013-01-10
2   A       2015-05-13
3   B       2012-05-23
4   B       2013-01-03
5   C       2017-05-01

我怎么能重写这个？这是可行的，但速度非常慢，因为每个数据帧的行数超过500k行。在

company_count = []
other_count = []

for _, company in companies.iterrows():
    end_date = company.EarningsDate
    start_date = end_date - pd.DateOffset(years=1)
    subset = news[(news.NewsDate > start_date) & (news.NewsDate < end_date)]

    mask = subset.CompanyName==company.CompanyName
    company_count.append(subset[mask].shape[0])
    other_count.append(subset[~mask].groupby('CompanyName').size().mean())

companies['12MonCompanyNewsCount'] = pd.Series(company_count)
companies['12MonOtherNewsCount'] = pd.Series(other_count).fillna(0)

最终结果，companies看起来像

    CompanyName EarningsDate    12MonCompanyNewsCount   12MonOtherNewsCount
0   A           2013-01-15      2                       2
1   B           2015-03-25      0                       0
2   C           2017-05-03      1                       0

Answer 1

我找不到不迭代companies行的方法。但是，您可以为companies设置一个开始日期列，遍历companies的行，并为符合条件的news的日期和公司名称创建布尔索引。然后只需执行一个布尔and操作，并对得到的布尔数组求和。在

我发誓当你看到密码的时候会更有意义。在

# create the start date column and the 12 month columns,
# fill the 12 month columns with zeros for now
companies['startdate'] = companies.EarningsDate - pd.DateOffset(years=1)
companies['12MonCompanyNewsCount'] = 0
companies['12MonOtherNewsCount'] = 0

# iterate the rows of companies and hold the index
for i, row in companies.iterrows():
    # create a boolean index when the news date is after the start date
    # and when the news date is before the end date
    # and when the company names match
    ix_start = news.NewsDate >= row.startdate
    ix_end = news.NewsDate <= row.EarningsDate
    ix_samename = news.CompanyName == row.CompanyName
    # set the news count value for the current row of `companies` using
    # boolean `and` operations on the indices.  first when the names match
    # and again when the names don't match.
    companies.loc[i,'12MonCompanyNewsCount'] = (ix_start & ix_end & ix_samename).sum()
    companies.loc[i,'12MonOtherNewsCount'] = (ix_start & ix_end & ~ix_samename).sum()

companies
#returns:

  CompanyName EarningsDate  startdate  12MonCompanyNewsCount  \
0           A   2013-01-15 2012-01-15                      1
1           B   2015-03-25 2014-03-25                      0
2           C   2017-05-03 2016-05-03                      1

   12MonOtherNewsCount
0                    2
1                    1
2                    0

Answer 2

好的，给你。在

要获得12MonCompanyNewsCount，可以使用merge_asof，这真的很好：

companies['12MonCompanyNewsCount'] = pd.merge_asof(
    news, 
    companies, 
    by='CompanyName',
    left_on='NewsDate',
    right_on='EarningsDate',
    tolerance=pd.Timedelta('365D'),
    direction='forward'
).groupby('CompanyName').count().NewsDate

它的工作速度大约是当前实现的两倍（并且可以更好地扩展）

对于12MonOtherNewsCount，我真的无法想出一种不循环的方法来完成它。我想这是一个更简洁的，虽然：

^{pr2}$

看起来确实有点快。在