Converting Pandas groupby.groups result into dataframe, using index tuple value as row and columns name of dataframe

Question

the resuts of my groupby.groups return :

{(2014, 36): [2], (2014, 41): [3, 4], (2013, 10): [0], (2014, 48): [5], (2014, 37): [1]}

i want to convert it into a dataframe that will looks like :

      2013 2014
10    1    0
36    0    1
37    0    1
41    0    2
48    0    1 

-- adds -- here is my workflow to the groupby.groups results :

def tr_epoch(epoch):
       y,wn,dn = epoch.isocalendar()
       return y, wn

d = [1362826800, 1410260400, 1409828400, 1412766000, 1412769600, 1417262400 ] 
l = map(lambda x:  tr_epoch(datetime.datetime.fromtimestamp(x)), d)
df = pd.DataFrame(l, columns=['year','week_idx'])
res = df.groupby(['year','week_idx']).groups

-- adds -- in pythonic way, using iteration, i will do :

def  to_dict(k,v):
    yr, wk = k
    return {'week': wk, yr: len(v)}

data =  map(lambda(k,v): to_dict(k,v), res.iteritems())
df = pd.DataFrame.from_records(data, index='week').fillna(0).sort()

But i am sure, there is a pandas way to do.

Answer 1

So you want to calculate the size of each group? Then you can do:

In [31]: df.groupby(['year','week_idx']).size()
Out[31]:
year  week_idx
2013  10          1
2014  36          1
      37          1
      41          2
      48          1
dtype: int64

To reshape this to the expected output, we can now use unstack to move the 'year' index level from the rows to the columns (and the use fillna to get 0's):

In [33]: df.groupby(['year','week_idx']).size().unstack(0).fillna(0)
Out[33]:
year      2013  2014
week_idx
10           1     0
36           0     1
37           0     1
41           0     2
48           0     1