我有一个看起来像这样的DataFrame
Unit ID Shipping to:
90 With x
91 With y
92 With z
116 Shipped to x 01/04/16. / Shipped to y - 09/08/18.
233 Shipped to z 03/01/17
265 Shipped to x 03/01/17 returned shipped to x 02/05/17
280 Shipped to x 06/01/17 Shipped to y 03/05/17 Shipped to z 12/12/17
我希望能够提取x,y或z的所有出现以及出现的日期(如果有的话)。我无法确定将会出现z,y或z的次数,但我希望最终结果看起来像这样:
Unit ID Occurrence 1 Occurrence 2 Occurrence 3 Shipping to:
90 x With x
91 y With y
92 z With z
116 x 01/04/16 y 09/08/18 Shipped to x 01/04/16. / Shipped to y - 09/08/18.
233 z 03/01/17 Shipped to z 03/01/17
265 x 03/01/17 Shipped to x 03/01/17 returned shipped to x 02/05/17
280 x 06/01/17 y 03/05/17 z 12/12/17 Shipped to x 06/01/17 Shipped to y 03/05/17 Shipped to z 12/12/17
到目前为止,我只能使用此方法来提取出现在每列中的第一个日期
date_col = []
for row in df['Shipping to:']:
match = re.search('\d{2}/\d{2}/\d{2}',str(row),re.IGNORECASE)
date_col.append(match)
df['dates'] = date_col
答案 0 :(得分:1)
数据框本身具有非常好的功能:
df['Shipping to:'].str.extractall(r'(\d{1,2}/\d{1,2}/\d{2})').unstack()
请注意,我将您的正则表达式更改为包括一个组(带有()
),并且我也匹配了月份和日期的个位数。
测试以下DataFrame(我知道这是胡说八道,但这只是一个测试):
df = pd.DataFrame([['1/22/33'], ['2/33/44 aaa 22/112/3 gook'], ['22/4/55'], [''], [None], ['aaa 22/5/66 aa 11/22/33']], columns=['Shipping to:'])
我得到以下输出:
match 0 1
0 1/22/33 NaN
1 2/33/44 NaN
2 22/4/55 NaN
5 22/5/66 11/22/33
要在开头包含x / y / z,请将正则表达式更改为r'([xyz] \d{1,2}/\d{1,2}/\d{2})'
。最后,如果要将这些匹配项添加为原始数据框的新列,则可以使用join
。然后,代码变为:
df.join(df['Shipping to:'].str.extractall(r'([xyz] \d{1,2}/\d{1,2}/\d{2})')\
.unstack()[0])
请注意,我在调用unstack
后得到了第0列-这实际上删除了1级的多索引并防止join
抱怨。现在只是因为我很开心地玩这个,所以我添加了一些代码来修复列名,以便它们与您的示例匹配:
df.join(df['Shipping to:'].str.extractall(r'([xyz] \d{1,2}/\d{1,2}/\d{2})')\
.unstack()[0]\
.rename(columns=lambda x: "Occurence " + str(x)))