Pandas groupby并在列表中获得dict

Question

I'm trying to extract grouped row data to use values to plot it with label colors another file.

my dataframe is like below.

df = pd.DataFrame({'x': [1, 4, 5], 'y': [3, 2, 5], 'label': [1.0, 1.0, 2.0]})

    x   y   label
0   1   3   1.0
1   4   2   1.0
2   5   5   2.0

I want to get group of label list like

{'1.0': [{'index': 0, 'x': 1, 'y': 3}, {'index': 1, 'x': 4, 'y': 2}],
 '2.0': [{'index': 2, 'x': 5, 'y': 5}]}

How to do this?

Answer 1

df = pd.DataFrame({'x': [1, 4, 5], 'y': [3, 2, 5], 'label': [1.0, 1.0, 2.0]})
df['index'] = df.index
df
   label  x  y  index
0    1.0  1  3      0
1    1.0  4  2      1
2    2.0  5  5      2

df['dict']=df[['x','y','index']].to_dict("records")
df
   label  x  y  index                             dict
0    1.0  1  3      0  {u'y': 3, u'x': 1, u'index': 0}
1    1.0  4  2      1  {u'y': 2, u'x': 4, u'index': 1}
2    2.0  5  5      2  {u'y': 5, u'x': 5, u'index': 2}

df = df[['label','dict']]
df['label'] = df['label'].apply(str) #Converting integer column 'label' to string
df = df.groupby('label')['dict'].apply(list) 
desired_dict = df.to_dict()
desired_dict 
    {'1.0': [{'index': 0, 'x': 1, 'y': 3}, {'index': 1, 'x': 4, 'y': 2}],
     '2.0': [{'index': 2, 'x': 5, 'y': 5}]}

Answer 2

You can use collections.defaultdict with to_dict :

from collections import defaultdict

# add 'index' series
df = df.reset_index()

# initialise defaultdict
dd = defaultdict(list)

# iterate and append
for d in df.to_dict('records'):
    dd[d['label']].append(d)

Result:

print(dd)

defaultdict(list,
            {1.0: [{'index': 0.0, 'x': 1.0, 'y': 3.0, 'label': 1.0},
                   {'index': 1.0, 'x': 4.0, 'y': 2.0, 'label': 1.0}],
             2.0: [{'index': 2.0, 'x': 5.0, 'y': 5.0, 'label': 2.0}]})

In general, there's no need to convert back to a regular dict , since defaultdict is a subclass of dict .

Answer 3

You can use itertuples and defulatdict :

itertuples returns named tuples to iterate over dataframe:

for row in df.itertuples():
    print(row)
Pandas(Index=0, x=1, y=3, label=1.0)
Pandas(Index=1, x=4, y=2, label=1.0)
Pandas(Index=2, x=5, y=5, label=2.0)

So taking advantage of this:

from collections import defaultdict
dictionary = defaultdict(list)
for row in df.itertuples():
    dummy['x'] = row.x
    dummy['y'] = row.y
    dummy['index'] = row.Index
    dictionary[row.label].append(dummy)

dict(dictionary)
> {1.0: [{'x': 1, 'y': 3, 'index': 0}, {'x': 4, 'y': 2, 'index': 1}],
 2.0: [{'x': 5, 'y': 5, 'index': 2}]}

Answer 4

The quickest solution for what you want is almost along what @cph_sto offers,

>>> df.reset_index().to_dict('records')
[{'index': 0.0, 'label': 1.0, 'x': 1.0, 'y': 3.0}, {'index': 1.0, 'label': 1.0, 'x': 4.0, 'y': 2.0}, {'index': 2.0, 'label': 2.0, 'x': 5.0, 'y': 5.0}]

That is, convert the index to a regular column, then apply the records version of to_dict . Another option of interest:

>>> df.to_dict('index')
{0: {'label': 1.0, 'x': 1.0, 'y': 3.0}, 1: {'label': 1.0, 'x': 4.0, 'y': 2.0}, 2: {'label': 2.0, 'x': 5.0, 'y': 5.0}}

Check the help on to_dict for more.