在数据框pyspark中创建新的列和行

Question

我有一个这样的数据框

id_1    id_desc    cat_1    cat_2
111      ask        ele     phone
222      ask hr     ele     phone
333      ask hr dk  ele     phone
444      askh       ele     phone

如果cat_1 ， cat_2对于多个id_1相同，则需要将该关联捕获为新列。

需要这样的输出，

id_1    id_desc        cat_1    cat_2   id_2
111      ask             ele    phone   222
111      ask             ele    phone   333
111      ask             ele    phone   444
222      ask hr          ele    phone   111
222      ask hr          ele    phone   333
222      ask hr          ele    phone   444
333      ask hr dk       ele    phone   111
333      ask hr dk       ele    phone   222
333      ask hr dk       ele    phone   444

如何在python中做到这一点？

Answer 1

我无法提出任何特别优雅的方法，但这应该可以完成工作：

import pandas as pd
import numpy as np

df = pd.DataFrame([[111, 'ask', 'ele', 'phone'], 
                   [222, 'ask_hr', 'ele', 'phone'], 
                   [333, 'ask_hr_dk', 'ele', 'phone'], 
                   [444, 'askh', 'ele', 'phone']], 
                   columns=['id_1', 'id_desc', 'cat_1', 'cat_2'])

grouped = df.groupby(by=['cat_1', 'cat_2'])  # group by the columns you want to be identical

data = []  # a list to store all unique groups

# In your example, this loop is not needed, but this generalizes to more than 1 pair
# of cat_1 and cat_2 values
for group in grouped.groups:  
    n_rows = grouped.get_group(group).shape[0]  # how many unique id's in a group
    all_data = np.tile(grouped.get_group(group).values, (n_rows, 1))  # tile the data n_row times
    ids = np.repeat(grouped.get_group(group)['id_1'].values, n_rows)  # repeat the ids n_row times
    data += [np.c_[all_data, ids]]  # concat the two sets of data and add to list

df_2 = pd.DataFrame(np.concatenate(data), columns=['id_1', 'id_desc', 'cat_1', 'cat_2', 'id_2'])

基本思想是按cat_1和cat_2列对数据进行cat_1 （使用groupby ），使用np.tile创建每个组的副本的次数要id_1该组中id_1唯一值，然后将结果与唯一id_1值（每组数据一个值）。

如果您不希望id_1与id_2相同，则只需选择不匹配的行：

df_2 = df_2[df_2['id_1'] != df_2['id_2']] 

如果希望它们在id_1上id_1 ：

df_2.sort_values('id_1', inplace=True)