CODEWITHSUNDEEP
pythonpandasdata-representation
spd
spd

Reputation: 354

List to a Readable Representation using Python

I have data as

[{'name': 'A', 'subsets': ['X_1', 'X_A', 'X_B'], 'cluster': 0},
 {'name': 'B', 'subsets': ['B_1', 'B_A'], 'cluster': 2},
 {'name': 'C', 'subsets': ['X_1', 'X_A', 'X_B'], 'cluster': 0},
 {'name': 'D', 'subsets': ['D_1', 'D_2', 'D_3', 'D_4'], 'cluster': 1}]

I need to represent it as

Cluster Number    Subset                         Name
0                 ['X_1', 'X_A', 'X_B']          A, C
1                 ['D_1', 'D_2', 'D_3', 'D_4']   D
2                 ['B_1', 'B_A']                 B

Upvotes: 1

Views: 63

Answers (2)

AloneTogether
AloneTogether

Reputation: 26708

For the sake of completeness, I think it is fair to mention that you can actually create a dataframe without json_normalize in your case and apply groupby as originally shown here:

import pandas as pd

data = [{'name': 'A', 'subsets': ['X_1', 'X_A', 'X_B'], 'cluster': 0},
 {'name': 'B', 'subsets': ['B_1', 'B_A'], 'cluster': 2},
 {'name': 'C', 'subsets': ['X_1', 'X_A', 'X_B'], 'cluster': 0},
 {'name': 'D', 'subsets': ['D_1', 'D_2', 'D_3', 'D_4'], 'cluster': 1}]

df = pd.DataFrame(data).groupby('cluster')
                       .agg({'subsets':'first','name':', '.join})
                       .reset_index()
                       .set_index('cluster')
                       .rename_axis('Cluster Number')

                             subsets  name
Cluster Number                            
0                    [X_1, X_A, X_B]  A, C
1               [D_1, D_2, D_3, D_4]     D
2                         [B_1, B_A]     B

Upvotes: 1

user7864386
user7864386

Reputation:

You can use json_normalize + groupby "cluster" and apply join to "name" and first to "subsets":

df = pd.json_normalize(data).groupby('cluster').agg({'subsets':'first','name':', '.join}).reset_index()

Output:

   cluster               subsets  name
0        0       [X_1, X_A, X_B]  A, C
1        1  [D_1, D_2, D_3, D_4]     D
2        2            [B_1, B_A]     B

Upvotes: 1

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