How to extract the file Data in python

Question

I wanted to extract the date from the given string on the basis of tag.

My string is -

DATE: 7/25/2017 DATE OPENED: 7/25/2017 RETURN DATE: 7/26/2017 
NUMBER: 201707250008754 RATE:  10.00

I want something like this - If I give "DATE" it should return 7/25/2017 only

if I give "RETURN DATE" it should return 7/26/2017

if I give the "NUMBER" it should return 201707250008754 and so on.

How we can achieve this in Python 2.7 (Note: Dates and numbers are always random in string"

Please don't add the python 3 tag if your question is specific to python 2. — glibdud
– glibdud, Commented Jun 1, 2018 at 17:05
I'd start with writing some code. What have you attempted, so far? — SiHa
– SiHa, Commented Jun 1, 2018 at 17:08

Ajax1234 · Accepted Answer · 2018-06-01 13:58:54Z

3

You can create a dictionary from the string's contents with re:

import re
s = 'DATE: 7/25/2017 DATE OPENED: 7/25/2017 RETURN DATE: 7/26/2017 NUMBER: 201707250008754 RATE: 10.00'
results = re.findall('[a-zA-Z\s]+(?=:)|[\d/\.]+', s)
d = dict([re.sub('^\s+', '', results[i]), results[i+1]] for i in range(0, len(results), 2))
for i in ['DATE', 'RETURN DATE', 'NUMBER']:
   print(d[i])

Output:

7/25/2017
7/26/2017
201707250008754

answered Jun 1, 2018 at 13:58

Ajax1234

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Abhijith Asokan · Accepted Answer · 2018-06-01 14:24:55Z

1

Use dict to map key (eg: 'DATE' ) to its value.

import re
s = '''DATE: 7/25/2017 DATE OPENED: 7/25/2017 RETURN DATE: 7/26/2017 NUMBER: 201707250008754 RATE:  10.00'''

items = re.findall('\s*(.*?)\:\s*([0-9/.]*)',s)
#[('DATE', '7/25/2017'), ('DATE OPENED', '7/25/2017'), ('RETURN DATE', '7/26/2017'), ('NUMBER', '201707250008754'), ('RATE', '10.00')]

info = dict(items)
#{'DATE': '7/25/2017', 'DATE OPENED': '7/25/2017', 'RETURN DATE': '7/26/2017', 'NUMBER': '201707250008754', 'RATE': '10.00'}


for key in ['DATE', 'RETURN DATE', 'NUMBER']:
    print(info[key])

answered Jun 1, 2018 at 14:24

Abhijith Asokan

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