我有以下熊猫交易数据,关于一家零售店:
print(df)
product Date Assistant_name
product_1 2017-01-02 11:45:00 John
product_2 2017-01-02 11:45:00 John
product_3 2017-01-02 11:55:00 Mark
...我希望为市场篮子分析创建以下dataset:
product Date Assistant_name Invoice_number
product_1 2017-01-02 11:45:00 John 1
product_2 2017-01-02 11:45:00 John 1
product_3 2017-01-02 11:55:00 Mark 2
...简单地说,如果一个事务具有相同的 Assistant_name和Date,那么我假设它确实生成了一个新的发票。
发布于 2018-02-18 13:13:37
最简单的是将列连接在一起的factorize:
df['Invoice'] = pd.factorize(df['Date'].astype(str) + df['Assistant_name'])[0] + 1
print (df)
product Date Assistant_name Invoice
0 product_1 2017-01-02 11:45:00 John 1
1 product_2 2017-01-02 11:45:00 John 1
2 product_3 2017-01-02 11:55:00 Mark 2如果性能很重要,请使用pd.lib.fast_zip
df['Invoice']=pd.factorize(pd.lib.fast_zip([df.Date.values, df.Assistant_name.values]))[0]+1时间
#[30000 rows x 3 columns]
df = pd.concat([df] * 10000, ignore_index=True)
In [178]: %%timeit
...: df['Invoice'] = list(zip(df['Date'], df['Assistant_name']))
...: df['Invoice'] = df['Invoice'].astype('category').cat.codes + 1
...:
9.16 ms ± 54.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [179]: %%timeit
...: df['Invoice'] = pd.factorize(df['Date'].astype(str) + df['Assistant_name'])[0] + 1
...:
11.2 ms ± 395 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [180]: %%timeit
...: df['Invoice'] = pd.factorize(pd.lib.fast_zip([df.Date.values, df.Assistant_name.values]))[0] + 1
...:
6.27 ms ± 93.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)发布于 2018-02-18 12:51:45
使用pandas类别是一种方法:
df['Invoice'] = list(zip(df['Date'], df['Assistant_name']))
df['Invoice'] = df['Invoice'].astype('category').cat.codes + 1
# product Date Assistant_name Invoice
# product_1 2017-01-02 11:45:00 John 1
# product_2 2017-01-02 11:45:00 John 1
# product_3 2017-01-02 11:55:00 Mark 2此方法的优点是可以轻松检索映射字典:
dict(enumerate(df['Invoice'].astype('category').cat.categories, 1))
# {1: ('11:45:00', 'John'), 2: ('11:55:00', 'Mark')}https://stackoverflow.com/questions/48851820
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