How can I convert a Pandas DataFrame to a three level nested dictionary using column names?
The columns are not first three columns and I want it to group by column artist
then group by column album
, and I need it to be case insensitive, preferably without using defaultdict.
This is a minimal reproducible example:
from collections import defaultdict
from itertools import product
from pandas import DataFrame
tree = defaultdict(lambda: defaultdict(dict))
columns = {'a': str(), 'b': str(), 'c': str(), 'd': int(), 'e': int(), 'f': int()}
df = DataFrame(columns, index=[])
for i, j, k in product('abcd', repeat=3):
tree[i][j][k] = list(map('abcd'.index, (i, j, k)))
df.loc[len(df)] = [i, j, k, *list(map('abcd'.index, (i, j, k)))]
How can I get a nested dictionary similar to tree
from df
?
I am really sorry I can provide any actual examples because they wouldn't be minimal.
I tried to use .groupby()
but I only ever saw it being used with one column and I really don't know what to do with the pandas.core.groupby.generic.DataFrameGroupBy
object it returns, I just started using it today.
Currently I can do this:
tree1 = dict()
for index, row in df.iterrows():
if not tree1.get(row['a'].lower()):
tree1[row['a'].lower()] = dict()
if not tree1[row['a'].lower()].get(row['b'].lower()):
tree1[row['a'].lower()][row['b'].lower()] = dict()
tree1[row['a'].lower()][row['b'].lower()][row['c'].lower()] = [row['d'], row['e'], row['f']]
I actually implemented case insensitive str
and dict
but for the sake of brevity (they are very long) I wouldn't use it here.
But according to this answer https://stackoverflow.com/a/55557758/16383578 such method is bad, what is a better way?
I would probably do it like this:
cols = ['a', 'b', 'c']
for col in cols:
df[col] = df[col].str.casefold()
tree = {}
for (a, b, c), values in (df.set_index(cols).T.to_dict(orient='list')
.items()):
tree.setdefault(a, {}).setdefault(b, {})[c] = values
or
...
for (a, b, c), values in (df.set_index(cols).apply(list, axis=1)
.to_dict()).items():
tree.setdefault(a, {}).setdefault(b, {})[c] = values
This produces the same result (when the first part that casefolds is included)
def to_dict(df):
return df.set_index(df.columns[0]).iloc[:, 0].to_dict()
df['values'] = df[['d', 'e', 'f']].apply(list, axis=1)
df = df[['a', 'b', 'c', 'values']]
tree = (df.set_index(['a', 'b'])
.groupby(['a', 'b']).apply(to_dict)
.reset_index('b')
.groupby('a').apply(to_dict)
.to_dict())
but I think it's a bit too convoluted.
Results:
{'a': {'a': {'a': [0, 0, 0], 'b': [0, 0, 1], 'c': [0, 0, 2], 'd': [0, 0, 3]},
'b': {'a': [0, 1, 0], 'b': [0, 1, 1], 'c': [0, 1, 2], 'd': [0, 1, 3]},
'c': {'a': [0, 2, 0], 'b': [0, 2, 1], 'c': [0, 2, 2], 'd': [0, 2, 3]},
'd': {'a': [0, 3, 0], 'b': [0, 3, 1], 'c': [0, 3, 2], 'd': [0, 3, 3]}},
'b': {'a': {'a': [1, 0, 0], 'b': [1, 0, 1], 'c': [1, 0, 2], 'd': [1, 0, 3]},
'b': {'a': [1, 1, 0], 'b': [1, 1, 1], 'c': [1, 1, 2], 'd': [1, 1, 3]},
'c': {'a': [1, 2, 0], 'b': [1, 2, 1], 'c': [1, 2, 2], 'd': [1, 2, 3]},
'd': {'a': [1, 3, 0], 'b': [1, 3, 1], 'c': [1, 3, 2], 'd': [1, 3, 3]}},
'c': {'a': {'a': [2, 0, 0], 'b': [2, 0, 1], 'c': [2, 0, 2], 'd': [2, 0, 3]},
'b': {'a': [2, 1, 0], 'b': [2, 1, 1], 'c': [2, 1, 2], 'd': [2, 1, 3]},
'c': {'a': [2, 2, 0], 'b': [2, 2, 1], 'c': [2, 2, 2], 'd': [2, 2, 3]},
'd': {'a': [2, 3, 0], 'b': [2, 3, 1], 'c': [2, 3, 2], 'd': [2, 3, 3]}},
'd': {'a': {'a': [3, 0, 0], 'b': [3, 0, 1], 'c': [3, 0, 2], 'd': [3, 0, 3]},
'b': {'a': [3, 1, 0], 'b': [3, 1, 1], 'c': [3, 1, 2], 'd': [3, 1, 3]},
'c': {'a': [3, 2, 0], 'b': [3, 2, 1], 'c': [3, 2, 2], 'd': [3, 2, 3]},
'd': {'a': [3, 3, 0], 'b': [3, 3, 1], 'c': [3, 3, 2], 'd': [3, 3, 3]}}}
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