如何使用熊猫从DataFrame或np.array中的列条目创建字典

Question

So I have a DataFrame , I labeled the columns a - i. I want to make a Dictionary of Dictionaries where the outer key is column "a", the inner key is column "d", and the value is "e". I know how to do this by iterating through each row, but I feel like there is a more efficient way to do this using DataFrame.to_dict() but I can't figure out how...maybe DataFrame.group_by could help but that seems to be used for grouping column or index IDs.

How can I use pandas (or numpy ) to create a Dictionary of Dictionaries efficiently without iterating through each row? I've shown an example of my current method and what the desired output should be below.

#!/usr/bin/python
import numpy as np
import pandas as pd

tmp_array = np.array([['AAA', 86880690, 86914111, '22RV1', 2, 2, 'H', '-'], ['ABA', 86880690, 86914111, 'A549', 2, 2, 'L', '-'], ['AAC', 86880690, 86914111, 'BFTC-905', 3, 3, 'H', '-'], ['AAB', 86880690, 86914111, 'BT-20', 2, 2, 'H', '-'], ['AAA', 86880690, 86914111, 'C32', 2, 2, 'H', '-']])

DF = pd.DataFrame(tmp_array,columns=["a,b,c,d,e,g,h,i".split(",")])

#print(DF)
a         b         c         d  e  g  h  i
0  AAA  86880690  86914111     22RV1  2  2  H  -
1  ABA  86880690  86914111      A549  2  2  L  -
2  AAC  86880690  86914111  BFTC-905  3  3  H  -
3  AAB  86880690  86914111     BT-20  2  2  H  -
4  AAA  86880690  86914111       C32  2  2  H  -

from collections import defaultdict
from itertools import izip

D_a_d_e = defaultdict(dict)
for a,d,e in izip(DF["a"],DF["d"],DF["e"]):
    D_a_d_e[a][d] = e

#print(D_a_d_e)
#ignore the defaultdict part

defaultdict(<type 'dict'>, {'ABA': {'A549': '2'}, 'AAA': {'22RV1': '2', 'C32': '2'}, 'AAC': {'BFTC-905': '3'}, 'AAB': {'BT-20': '2'}})

I saw this https://stackoverflow.com/questions/28820254/how-to-create-a-pandas-dataframe-using-a-dictionary-in-a-single-column but it was a little different and it also doesn't have an answer.

Answer 1

There's a to_dict method:

In [11]: DF.to_dict()
Out[11]:
{'a': {0: 'AAA', 1: 'ABA', 2: 'AAC', 3: 'AAB', 4: 'AAA'},
 'b': {0: '86880690', 1: '86880690', 2: '86880690' 3: '86880690', 4: '86880690'},
 'c': {0: '86914111', 1: '86914111', 2: '86914111', 3: '86914111', 4: '86914111'},
 'd': {0: '22RV1', 1: 'A549', 2: 'BFTC-905', 3: 'BT-20', 4: 'C32'},
 'e': {0: '2', 1: '2', 2: '3', 3: '2', 4: '2'},
 'g': {0: '2', 1: '2', 2: '3', 3: '2', 4: '2'},
 'h': {0: 'H', 1: 'L', 2: 'H', 3: 'H', 4: 'H'},
 'i': {0: '-', 1: '-', 2: '-', 3: '-', 4: '-'}}

In [12]: DF.to_dict(orient="index")
Out[12]:
{0: {'a': 'AAA', 'b': '86880690', 'c': '86914111', 'd': '22RV1', 'e': '2', 'g': '2', 'h': 'H', 'i': '-'},
 1: {'a': 'ABA', 'b': '86880690', 'c': '86914111', 'd': 'A549', 'e': '2', 'g': '2', 'h': 'L', 'i': '-'},
 2: {'a': 'AAC', 'b': '86880690', 'c': '86914111', 'd': 'BFTC-905', 'e': '3', 'g': '3', 'h': 'H', 'i': '-'},
 3: {'a': 'AAB', 'b': '86880690', 'c': '86914111', 'd': 'BT-20', 'e': '2', 'g': '2', 'h': 'H', 'i': '-'},
 4: {'a': 'AAA', 'b': '86880690', 'c': '86914111', 'd': 'C32', 'e': '2', 'g': '2', 'h': 'H', 'i': '-'}}

With that in mind you can do the groupby:

In [21]: DF.set_index("d").groupby("a")[["e"]].apply(lambda x: x["e"].to_dict())
Out[21]:
a
AAA    {'C32': '2', '22RV1': '2'}
AAB                {'BT-20': '2'}
AAC             {'BFTC-905': '3'}
ABA                 {'A549': '2'}
dtype: object

---

That said, you may be able to use a straight up MultiIndex instead of a dictionary of dictionaries:

In [31]: res = DF.set_index(["a", "d"])["e"]

In [32]: res
Out[32]:
a    d
AAA  22RV1       2
ABA  A549        2
AAC  BFTC-905    3
AAB  BT-20       2
AAA  C32         2
Name: e, dtype: object

It'll work much the same way:

In [33]: res["AAA"]
Out[33]:
d
22RV1    2
C32      2
Name: e, dtype: object

In [34]: res["AAA"]["22RV1"]
Out[34]: '2'

But will be a more space-efficient / you're still in pandas.

Answer 2

Something along these lines:

def dictmaker(df): 
    """
    wrapper for storing key, values in dict. Takes df.
    """
    dct={}  ## storage
    dct[df.d.values[0]]=df.e.values[0]
    return dct

DF[['a','d','e']].groupby('a').apply(dictmaker)

a
AAA       {u'22RV1': u'2'}
AAB       {u'BT-20': u'2'}
AAC    {u'BFTC-905': u'3'}
ABA        {u'A549': u'2'}
dtype: object