Question

我正在尝试计算某个公司在其收益日期后一年内出现在新闻上的次数，并将该次数与其他人在同一时间段内进行比较。我有两个pandas数据帧，一个有收入日期，另一个有新闻。我的方法很慢。是否有更好的熊猫/ numpy方式？

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

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

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

我该如何重写？这有效，但是因为每个数据帧都是非常慢的。 500k行。

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

好的，这就是。

要获得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，您可以使用file a JIRA，这非常简洁：

12MonOtherNewsCount

其效果大约是当前实施的两倍（并且会更好地扩展）

对于companies['12MonOtherNewsCount'] = companies.apply( lambda x: len( news[ (news.NewsDate.between(x.EarningsDate-pd.Timedelta('365D'), x.EarningsDate, inclusive=False)) &(news.CompanyName!=x.CompanyName) ] ), axis=1 )，我无法在没有循环的情况下找到一种方法。我想这有点简洁：

{{1}}

看起来确实快一点。

Answer 2

我找不到一种不迭代companies行的方法。但是，您可以为companies设置开始日期列，迭代companies行，并为符合您标准的news的日期和公司名称创建布尔索引。然后只需执行布尔and操作并对得到的布尔数组求和。

我发誓当你看到代码时会更有意义。

# 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

Pandas为每行加入两个具有不同时间帧的数据帧

2 个答案: