每日提词

Question

我有以下df，其中包含来自不同来源的每日文章：

print(df)

Date         content

2018-11-01    Apple Inc. AAPL 1.54% reported its fourth cons...
2018-11-01    U.S. stocks climbed Thursday, Apple is a real ...
2018-11-02    GONE are the days when smartphone manufacturer...
2018-11-03    To historians of technology, the story of the ...
2018-11-03    Apple Inc. AAPL 1.54% reported its fourth cons...
2018-11-03    Apple is turning to traditional broadcasting t...

(...)

我想计算“苹果”一词的每日提及总数-因此按日期汇总。如何创建“ final_df”？

print(final_df) 

    2018-11-01    2
    2018-11-02    0
    2018-11-03    2
    (...)

Answer 1

将count用于新的Series，并按列df['Date']与sum进行汇总：

df1 = df['content'].str.count('Apple').groupby(df['Date']).sum().reset_index(name='count')
print (df1)
         Date  count
0  2018-11-01      2
1  2018-11-02      0
2  2018-11-03      2

Answer 2

您可以GroupBy个不同的日期，使用str.count对Apple的出现进行计数，并与sum进行累加以获得每个组中的计数数量：< / p>

df.groupby('Date').apply(lambda x: x.content.str.count('Apple').sum())
                  .reset_index(name='counts')

      Date     counts
0 2018-11-01       2
1 2018-11-02       0
2 2018-11-03       2

Answer 3

您可以尝试使用具有groupby功能的str.contains替代解决方案，而无需始终使用sum。

>>> df
         Date                                         content
0  2018-11-01  Apple Inc. AAPL 1.54% reported its fourth cons
1  2018-11-01   U.S. stocks climbed Thursday, Apple is a real
2  2018-11-02  GONE are the days when smartphone manufacturer
3  2018-11-03   To historians of technology, the story of the
4  2018-11-03  Apple Inc. AAPL 1.54% reported its fourth cons
5  2018-11-03  Apple is turning to traditional broadcasting t

解决方案：

df.content.str.contains("Apple").groupby(df['Date']).count().reset_index(name="count")

         Date  count
0  2018-11-01      2
1  2018-11-02      1
2  2018-11-03      3


# df["content"].str.contains('Apple',case=True,na=False).groupby(df['Date']).count()