I don't understand why there isn't a B2
in your dict. I'm also not sure what you want to happen in the case of repeated column values (every one except the last, I mean.) Assuming the first is an oversight, we could use recursion:
def recur_dictify(frame):
if len(frame.columns) == 1:
if frame.values.size == 1: return frame.values[0][0]
return frame.values.squeeze()
grouped = frame.groupby(frame.columns[0])
d = {k: recur_dictify(g.ix[:,1:]) for k,g in grouped}
return d
which produces
>>> df
name v1 v2 v3
0 A A1 A11 1
1 A A2 A12 2
2 B B1 B12 3
3 C C1 C11 4
4 B B2 B21 5
5 A A2 A21 6
>>> pprint.pprint(recur_dictify(df))
{'A': {'A1': {'A11': 1}, 'A2': {'A12': 2, 'A21': 6}},
'B': {'B1': {'B12': 3}, 'B2': {'B21': 5}},
'C': {'C1': {'C11': 4}}}
It might be simpler to use a non-pandas approach, though:
def retro_dictify(frame):
d = {}
for row in frame.values:
here = d
for elem in row[:-2]:
if elem not in here:
here[elem] = {}
here = here[elem]
here[row[-2]] = row[-1]
return d
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