将原始数据框数据转换为带有列表的字典

Question

我正在尝试转换此数据（在 Dataframe 中）：

         0         1
0   HT01  CC363292
29  RL01  CC363292
50  TN01  CC363292
4   BN02  CC363293
7   MR20  CC363294
9   TN01  CC363295
10  RL01  CC363296
13  HT01  CC363297
17  HT01  CC363298
21  SU01  CC363299
22  BN02  CC363300
25  MR20  CC363301
27  MR20  CC363302
54  BN02  CC363313
57  BN02  CC363314
60  BN02  CC363315
52  SU01  EA363303
32  RL01  EA363303
35  MR20  EA363304
37  HU01  EA363305
38  HU01  EA363306
39  BN02  EA363307
63  RL01  EA363311
66  MR20  EA363312
42  HT01  SC363308
46  RL01  SC363309
51  SP01  SC363309
53  FU01  SC363309
49  SP01  SC363310

进入使用 col.0 作为键和匹配 col.1 信息列表的字典（见下文）

即

temp_dict = {

CC363292 : [HT01, RL01, TN01]
CC363293 : [BN02]

}

我曾尝试使用 for 循环将列表附加到密钥，但没有成功。

有人可以帮忙吗？

Answer 1

我对 Pandas 有点熟悉，这是我想出的解决方案：

# Create DataFrame from initial data
data = [
    ('HT01', 'CC363292'),
    ('RL01', 'CC363292'),
    ('TN01', 'CC363292'),
    ('BN02', 'CC363293'),
    ...
]

df = pandas.DataFrame(data=data, columns=['col1', 'col2'])

# This will create a such dataframe:
# col1 |     col2 |
# HT01 | CC363292 |
# RL01 | CC363292 |
# TN01 | CC363292 |
# BN02 | CC363293 |
# .... | ........ |

# The next step is to convert 'col2' to categorical
_df = pandas.get_dummies(data=df, columns=['col2'], prefix='', prefix_sep='')

# This will give us such result:
# col1 | CC363292 | CC363293 | ...
# HT01 |        1 |        0 | ...
# RL01 |        1 |        0 | ...
# TN01 |        1 |        0 | ...
# BN02 |        0 |        1 | ...
# .... | ........ | ........ | ...

# Then we'll create the simple lambda function to initialize our lists:
f = lambda col: [_df.col1[i] for i, val in enumerate(_df[col]) if val]

# And obtain the requested result using dict-comprehensions:
my_dict = {col: f(col) for col in _df.columns[1:]}

# Important: using _df.columns[1:] is not very universal, but
# will be ok for the problem you described

Answer 2

题目要求按列对行进行分组，并根据分组列出另一列。一个快速的解决方案可以是：

import pandas as pd

data = [
  [0, "HT01", "CC363292"],
  [29, "RL01", "CC363292"],
  [50, "TN01", "CC363292"],
  [4, "BN02", "CC363293"],
  [7, "MR20", "CC363294"],
  [9, "TN01", "CC363295"],
  [10, "RL01", "CC363296"],
  [13, "HT01", "CC363297"],
  [17, "HT01", "CC363298"],
  [21, "SU01", "CC363299"],
  [22, "BN02", "CC363300"],
  [25, "MR20", "CC363301"],
  [27, "MR20", "CC363302"],
  [54, "BN02", "CC363313"],
  [57, "BN02", "CC363314"],
  [60, "BN02", "CC363315"],
  [52, "SU01", "EA363303"],
  [32, "RL01", "EA363303"],
  [35, "MR20", "EA363304"],
  [37, "HU01", "EA363305"],
  [38, "HU01", "EA363306"],
  [39, "BN02", "EA363307"],
  [63, "RL01", "EA363311"],
  [66, "MR20", "EA363312"],
  [42, "HT01", "SC363308"],
  [46, "RL01", "SC363309"],
  [51, "SP01", "SC363309"],
  [53, "FU01", "SC363309"],
  [49, "SP01", "SC363310"],
]
df = pd.DataFrame(data)

# Group by the third column.
# List the second column.
groups = df.groupby(df.columns[2])[df.columns[1]].apply(list)
print(groups)

输出应类似于：

CC363292    [HT01, RL01, TN01]
CC363293                [BN02]
CC363294                [MR20]
CC363295                [TN01]
CC363296                [RL01]
CC363297                [HT01]
CC363298                [HT01]
CC363299                [SU01]
CC363300                [BN02]
CC363301                [MR20]
EA363311                [RL01]
EA363312                [MR20]
SC363308                [HT01]
SC363309    [RL01, SP01, FU01]
SC363310                [SP01]

要转换为字典，请改用 dict(groups)。输出应该是：

{
  'CC363292': ['HT01', 'RL01', 'TN01'],
  'CC363293': ['BN02'],
  'CC363294': ['MR20'],
  'CC363295': ['TN01'],
  'CC363296': ['RL01'],
  'CC363297': ['HT01'],
  'CC363298': ['HT01'],
  'CC363299': ['SU01'],
  'CC363300': ['BN02'],
  'CC363301': ['MR20'],
  'CC363302': ['MR20'], 
  'CC363313': ['BN02'], 
  'CC363314': ['BN02'],
  'CC363315': ['BN02'],
  'EA363303': ['SU01', 'RL01'],
  'EA363304': ['MR20'],
  'EA363305': ['HU01'],
  'EA363306': ['HU01'],
  'EA363307': ['BN02'],
  'EA363311': ['RL01'],
  'EA363312': ['MR20'],
  'SC363308': ['HT01'],
  'SC363309': ['RL01', 'SP01', 'FU01'],
  'SC363310': ['SP01']
}

Answer 3

zip_data = zip(df['col1'], df['col2'])

result = {}

for i in zip_data:
    result.setdefault(i[1], []).append(i[0])

这可能有效。

将原始数据框数据转换为带有列表的字典

turning a raw dataframe data into a dictionary with a list

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for-loop

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