Question

我的输入数据集如下-

INPUT = [
'ABCD , D.O.B: - Jun/14/1999.',
'EFGH , DOB; - Jan/10/1998,',
'IJKL , D-O-B - Jul/15/1985..',
'MNOP , (DOB)* - Dec/21/1999,',
'QRST , *DOB* - Apr/01/2000.',
'UVWX , D O B, - Feb/11/2001 '
]

我希望它采用以下格式的输出格式-

OUTPUT = [
('ABCD, Jun/14/1999'),
('EFGH, Jan/10/1998'),
('IJKL, Jul/15/1985'),
('MNOP, Dec/21/1999'),
('QRST, Apr/1/2000'),
('UVWX, Feb/11/2001')
]

我尝试了下面的部分起作用的代码，但无法以所需的OUTPUT格式进行格式化-

import re

INPUT = [
'ABCD , D.O.B: - Jun/14/1999.',
'EFGH , DOB; - Jan/10/1998,',
'IJKL , D-O-B - Jul/15/1985..',
'MNOP , (DOB)* - Dec/21/1999,',
'QRST , *DOB* - Apr/01/2000.',
'UVWX , D O B, - Feb/11/2001 '
]


def formatted_def(input):
    for n in input:
        t = re.sub('[^a-zA-Z0-9 ]+','',n).split('DOB')
        print(t)


formatted_def(INPUT)

输出-

['ABCD  ', '  Jun141999']
['EFGH  ', '  Jan101998']
['IJKL  ', '  Jul151985']
['MNOP  ', '  Dec211999']
['QRST  ', '  Apr012000']
['UVWX  D O B  Feb112001 ']

任何指针都将非常有帮助。预先感谢！

Answer 1

您可以使用select t.*, (case when seqnum = 1 then 'ADD' else 'CHANGE' end) as audit, (case when seqnum = 1 then 'NEW' when seqnum_day = 1 then 'CURRENT' else 'BEFORE' end) as history from (select t.*, row_number() over (partition by custname order by recordedtime) as seqnum, row_number() over (partition by custname, cast(recordedtime as date) order by recordedtime desc) as seqnum_day from t ) t;：

re.findall

输出：

import re
l = ['ABCD , D.O.B: - Jun/14/1999.', 'EFGH , DOB; - Jan/10/1998,', 'IJKL , D-O-B - Jul/15/1985..', 'MNOP , (DOB)* - Dec/21/1999,', 'QRST , *DOB* - Apr/01/2000.', 'UVWX , D O B, - Feb/11/2001 ']
final_data = [', '.join(re.findall('^\w+|[a-zA-Z]+/\d+/\d+(?=\W)', i)) for i in l]

Answer 2

除了其他答案，您还可以使用re.sub：

INPUT = [
    'ABCD , D.O.B: - Jun/14/1999.',
    'EFGH , DOB; - Jan/10/1998,',
    'IJKL , D-O-B - Jul/15/1985..',
    'MNOP , (DOB)* - Dec/21/1999,',
    'QRST , *DOB* - Apr/01/2000.',
    'UVWX , D O B, - Feb/11/2001 '
]

pattern = r'(?i)^([a-z]+).*([a-z]{3}/\d{2}/\d{4}).*$'

OUTPUT = [re.sub(pattern, r'\1, \2', x) for x in INPUT]

# OUTPUT:

[
    'ABCD, Jun/14/1999',
    'EFGH, Jan/10/1998',
    'IJKL, Jul/15/1985',
    'MNOP, Dec/21/1999',
    'QRST, Apr/01/2000',
    'UVWX, Feb/11/2001'
]

Answer 3

import re
re.findall(r'(\w+)\s+,.*?-\s+([^., ]*)', ' '.join(INPUT))
# [('ABCD', 'Jun/14/1999'), ('EFGH', 'Jan/10/1998'), ('IJKL', 'Jul/15/1985'), ('MNOP', 'Dec/21/1999'), ('QRST', 'Apr/01/2000'), ('UVWX', 'Feb/11/2001')]

Answer 4

主要困难是获取('ABCD, Jun/14/1999'),内容。

它不能是单元素元组，因为它会被打印为('ABCD, Jun/14/1999',),（在,之前注意额外的)）。

因此要准确获得您想要的结果，我使用一系列print语句。

整个脚本（在Python 3中）可以如下：

import re
input = [
  'ABCD , D.O.B: - Jun/14/1999.',
  'EFGH , DOB; - Jan/10/1998,',
  'IJKL , D-O-B - Jul/15/1985..',
  'MNOP , (DOB)* - Dec/21/1999,',
  'QRST , *DOB* - Apr/01/2000.',
  'UVWX , D O B, - Feb/11/2001 '
]
result = [ re.sub(r'^([a-z]+).*? - ([a-z]{3}/\d{2}/\d{4}).*',
                  r'\1, \2', txt, flags = re.IGNORECASE) for txt in input ]
print('OUTPUT = [')
for txt in result:
    print(" ('{}')".format(txt))
print(']')

Python中的正则表达式和格式

4 个答案: