我正在嘗試轉換這些資料（在資料框中）：
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
放入col.0用作鍵的字典和匹配col.1資訊串列（見下文）
IE
temp_dict = {
CC363292 : [HT01, RL01, TN01]
CC363293 : [BN02]
}
我嘗試使用 for 回圈將串列附加到鍵上，但沒有成功。
有人可以幫忙嗎？
uj5u.com熱心網友回復：
zip_data = zip(df['col1'], df['col2'])
result = {}
for i in zip_data:
result.setdefault(i[1], []).append(i[0])
這可能會奏效。
uj5u.com熱心網友回復：
這些問題需要按列對行進行分組，并根據組列出另一列。一個快速的解決方案可以是：
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)
輸出應類似于：
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]
要轉換為字典，請dict(groups)改用。輸出應該是：
{
'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']
}
uj5u.com熱心網友回復：
我對 Pandas 有點熟悉，這是我想出的解決方案：
# 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
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