在Python中将系列重塑为Dataframe矩阵

3 回答

• 0

  s
  
  1992-01-31     1.123077
  1992-02-28    -2.174845
  1992-03-31    -3.884848
  1992-04-30     8.682919
  1992-05-29     1.312976
  1992-06-30     7.851080
  1992-07-31    -3.192788
  1992-08-31    -7.351976
  1992-09-30    -6.782217
  1992-10-30   -17.182738
  1992-11-30     3.898782
  1992-12-31   -26.190414
  1993-01-29     2.233359
  1993-02-26     6.709006
  Name: 1, dtype: float64
  
  type(s)
  pandas.core.series.Series
  

  如有必要，将索引转换为 datetime -

  s.index = pd.to_datetime(s.index, errors='coerce')
  

  现在，使用 pd.pivot -

  x = pd.Series(s.index.strftime('%Y %B')).str.split()
  y, m = x.str[0], x.str[1]
  
  pd.pivot(y, m, s)
  
           April    August   December  February   January      July     June  \
  1992  8.682919 -7.351976 -26.190414 -2.174845  1.123077 -3.192788  7.85108   
  1993       NaN       NaN        NaN  6.709006  2.233359       NaN      NaN   
  
           March       May  November    October  September  
  1992 -3.884848  1.312976  3.898782 -17.182738  -6.782217  
  1993       NaN       NaN       NaN        NaN        NaN
  

  回复于 2024-05-05T19:03:45+08:00
• 2

  尝试类似的东西

  #Give your series index a name so that we can reset index and have a new column
  your_series.index = your_series.index.rename('Time')
  df = your_series.toframe('Values').reset_index()
  
  #Create variables for month and year
  df['Month'] = df.Time.dt.month
  df['Year'] = df.Time.dt.Year
  
  #Assuming they are unique, create a pivot table
  df.pivot('Year','Month','Values')
  

  月份将是数字 . 如果你想要你必须要做的月份名称

  import datetime as dt
  
  df['Month'] = df.Time.date.apply(lambda x: dt.datetime.strftime(x,'%B'))
  

  如果您的月/年对不是唯一的，那么请执行类似的操作

  df.groupby(['Year','Month']).Values.sum().unstack()
  

  回复于 2024-05-05T19:03:45+08:00
• 2

  来源DF：

  In [159]: df
  Out[159]:
                    val
  date
  1992-01-31   1.123077
  1992-02-28  -2.174845
  1992-03-31  -3.884848
  1992-04-30   8.682919
  1992-05-29   1.312976
  1992-06-30   7.851080
  1992-07-31  -3.192788
  1992-08-31  -7.351976
  1992-09-30  -6.782217
  1992-10-30 -17.182738
  1992-11-30   3.898782
  1992-12-31 -26.190414
  1993-01-29   2.233359
  1993-02-26   6.709006
  

  解：

  import calendar
  
  In [158]: (df.assign(year=df.index.year, mon=df.index.month)
               .pivot(index='year', columns='mon', values='val')
               .rename(columns=dict(zip(range(13), calendar.month_name))))
  Out[158]:
  mon    January  February     March     April       May     June      July    August  September    October  November   December
  year
  1992  1.123077 -2.174845 -3.884848  8.682919  1.312976  7.85108 -3.192788 -7.351976  -6.782217 -17.182738  3.898782 -26.190414
  1993  2.233359  6.709006       NaN       NaN       NaN      NaN       NaN       NaN        NaN        NaN       NaN        NaN
  

  UPDATE: 或者更好更短version from @COLDSPEED：

  In [164]: pd.pivot(df.index.year, df.index.month, df['val']) \
              .rename(columns=calendar.month_name.__getitem__)
  Out[164]:
  date   January  February     March     April       May     June      July    August  September    October  November   December
  date
  1992  1.123077 -2.174845 -3.884848  8.682919  1.312976  7.85108 -3.192788 -7.351976  -6.782217 -17.182738  3.898782 -26.190414
  1993  2.233359  6.709006       NaN       NaN       NaN      NaN       NaN       NaN        NaN        NaN       NaN        NaN
  

  回复于 2024-05-05T19:03:45+08:00