按星期几分组熊猫

Question

  I have a dataframe,df 

        Index       eventName Count   pct
     2017-08-09       ABC     24     95.00%
     2017-08-09       CDE    140     98.50%
     2017-08-10       DEF    200     50.00%
     2017-08-11       CDE    150     99.30%
     2017-08-11       CDE    150     99.30%
     2017-08-16       DEF    200     50.00%
     2017-08-17       DEF    200     50.00%

I want to group by daily weekly occurrence by counting the values in the column pct. for example, we now have:

 2017-08-09 has 2 values in pct column  and  2017-08-16 has 1 value in pct, then we have Monday:3 
  2017-08-10  has 1 value and 2017-08-17 has 1 value,then we have Tuesday:2 and so on

then the resulting dataframe should look like this:

    Index        Count   
 Monday            3
 Tuesday           2
 Wednesday         2

I have tried df2=df.groupby(pd.Grouper(freq='D')).size().sort_values(ascending=False) but its not grouping by day of the week and not transforming to the date index to words

Answer 1

By using value_counts

df.Index=pd.to_datetime(df.Index)
df.Index.dt.weekday_name.value_counts()
Out[994]: 
Wednesday    3
Thursday     2
Friday       2
Name: Index, dtype: int64

Answer 2

Wen's answer with value_counts is good, but does not account for the possibility of NaN s in the pct column.

---

Assuming Index is the index, you can call groupby + count -

df.index = pd.to_datetime(df.index)
df.groupby(df.index.weekday_name).pct.count()

Index
Friday       2
Thursday     2
Wednesday    3
Name: pct, dtype: int64

To sort on weekday, convert to pd.Categorical , as shown here .

Answer 3

You can use:

df.rename(columns={'Index': 'New_name'}, inplace=True)

df['New_name'] = pd.to_datetime(df['New_name'])

df['Day_df'] = df['New_name'].dt.weekday_name

df.groupby(['Day_df']).count()