I have a nested dictionary, whereby the sub-dictionary use lists:
nested_dict = {\'string1\': {69: [1231, 232], 67:[682, 12], 65: [1, 1]},
`string2` :{2
Here's a method which uses a recursive generator to unroll the nested dictionaries. It won't assume that you have exactly two levels, but continues unrolling each dict
until it hits a list
.
nested_dict = {
'string1': {69: [1231, 232], 67:[682, 12], 65: [1, 1]},
'string2' :{28672: [82, 23], 22736:[82, 93, 1102, 102], 19423: [64, 23]},
'string3': [101, 102]}
def unroll(data):
if isinstance(data, dict):
for key, value in data.items():
# Recursively unroll the next level and prepend the key to each row.
for row in unroll(value):
yield [key] + row
if isinstance(data, list):
# This is the bottom of the structure (defines exactly one row).
yield data
df = pd.DataFrame(list(unroll(nested_dict)))
Because unroll
produces a list of lists rather than dicts, the columns will be named numerically (from 0 to 5 in this case). So you need to use rename
to get the column labels you want:
df.rename(columns=lambda i: 'col{}'.format(i+1))
This returns the following result (note that the additional string3
entry is also unrolled).
col1 col2 col3 col4 col5 col6
0 string1 69 1231 232.0 NaN NaN
1 string1 67 682 12.0 NaN NaN
2 string1 65 1 1.0 NaN NaN
3 string2 28672 82 23.0 NaN NaN
4 string2 22736 82 93.0 1102.0 102.0
5 string2 19423 64 23.0 NaN NaN
6 string3 101 102 NaN NaN NaN