每天提到一个词

Question

我有以下df，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”这个词的每日提到的总数--因此按日期进行汇总。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，按列df['Date']与sum聚合

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

您可以对不同的日期进行GroupBy，使用str.count来计数Apple的出现情况，并使用sum进行聚合，以获取每个组中的计数数量：

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

您可以使用str.contains和groupby函数尝试替代解决方案，而不必一直使用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

解决办法：

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()