Pandas-如何获取另一列中每个对应值的行数

Question

I have quite a complicated problem that I need help figuring out.

To begin, I have a dataframe:

 one       two     three     four      Date
comedy      a       asad      123      2013-01-18 10:00:00  
romantic    b       fas       563      2015-01-28 12:00:00
comedy      c       ewf       134      2014-01-22 09:00:00
action      a       qef       561      2013-02-18 18:00:00
action      z       adwq      1323     2016-01-23 16:00:00
...

I am trying to find the best way to count the number of occurrences(frequency) for each unique value in column 'one', for each week in the data column. I then want to be able to somehow compare does a higher frequency of each occurrences in each week, result in a higher or lower number for column 'four'.

My desired output is something like this, but I am open to better solutions:

 ones       2013-01-00  2013-01-07  2013-01-14.....    Total_frequency
 comedy         4          5           6                15
 romantic       1          2           0                3 
 action         0          0           0                0 
 ....

Each unique value from column 'one' is under 'ones', and their total number of occurrences for each week is under each week column. (The week columns will begin at a specified week (eg in the above case -> 2013-01-00).

Although, I am having trouble trying to think of the best way to relate the total frequency to column four across the dataframe.

If anyone has any idea of the best way I could go about doing this, it'd be very much appreciated.

If you need anymore information please let me know.

Edit:

  ones       2013-01-00  2013-01-07  2013-01-14.....    Total_frequency
 comedy         4          5           6                15
 romantic       1          2           0                3 
 action       NaN          1           0                1 

Thanks.

Answer 1

Use:

#changed data sample for better verify output
print (df)
        one two three  four                 Date
0    comedy   a  asad   123  2013-01-18 10:00:00
1  romantic   b   fas   563  2013-01-28 12:00:00
2    comedy   c   ewf   134  2013-01-22 09:00:00
3    action   a   qef   561  2013-02-18 18:00:00
4    action   z  adwq  1323  2013-01-23 16:00:00

Use Grouper with DataFrameGroupBy.size and unstack :

df['Date'] = pd.to_datetime(df['Date'])
df = (df.groupby(['one',pd.Grouper(freq='W-MON', key='Date')])
        .size()
        .unstack(fill_value=0)
        .sort_index(axis=1))

df.columns = df.columns.date
df['Total_frequency'] = df.sum(axis=1)
print (df)
          2013-01-21  2013-01-28  2013-02-18  Total_frequency
one                                                          
action             0           1           1                2
comedy             1           1           0                2
romantic           0           1           0                1

EDIT: Create boolean mask for values after first NaN s and replace missing values only by this mask:

print (df)
          2013-01-00  2013-01-07  2013-01-14
ones                                        
comedy           4.0           5         6.0
romantic         1.0           2         NaN
action           NaN           1         NaN

mask = df.notnull().cumsum(axis=1).ne(0)
#another solution
#mask = df.ffill(axis=1).notnull()

df = df.mask(mask, df.fillna(0))
print (df)
          2013-01-00  2013-01-07  2013-01-14
ones                                        
comedy           4.0           5         6.0
romantic         1.0           2         0.0
action           NaN           1         0.0

Answer 2

You could try this:

df = pd.DataFrame({'one': [random.choice(['comedy', 'action', 'romantic']) for i in range(1000)],
                  'Date': pd.date_range(start = '2013-01-01', periods = 1000)})
df.head()

      one        Date
0   romantic    2013-01-01
1   romantic    2013-01-02
2   romantic    2013-01-03
3   action       2013-01-04
4   romantic    2013-01-05

df.groupby([pd.Grouper(key = 'Date', freq = 'W'), 'one'])['one'].count().unstack(level = 0)

Date          2013-01-06  2013-01-13  2013-01-20.....
one           
comedy         2         2           2              
romantic       NaN       2           2               
action         4         3           3