Python regex: Get regex pattern in a text file and store in an array or list

Question

I have this sample data inside a text file:

09-02||||||||09-14|07:24|12:15|12:58| | |

09-03| | | | | | |09-15|||||||

I'am trying to get all the data with this kind of pattern and store it in an array or list:

\d{2,3}-\d{2,3}

the output data when printed should be like this:

['09-02','09-14','09-02','09-15']

I tried this code but it printed all the lines matching the pattern:

n_date = re.compile('\d{2,3}-\d{2,3}')
with open('sample_2.txt', 'r') as n:
    for line in n:
        re.match(n_date, line)
print(line)

Please give me an idea on how can I just get the data matching my regex pattern not the whole line. Thank you!

it printed all the lines But you only have print(line) once, shouldn't it only be printing the last line? — CertainPerformance
– CertainPerformance, Commented Oct 4, 2018 at 3:58

crow3487 · Accepted Answer · 2018-10-04 03:57:00Z

1

Try this:

import re
n_date = re.compile('\d{2,3}-\d{2,3}')
with open('sample_2.txt', 'r') as n:
    n = n.read()
    result = re.findall(n_date, n)
    print(result)

It prints out:

['09-02', '09-14', '09-03', '09-15']

Your code just prints the last line of the for loop and you're not storing or using the result of re.match. re.findall will give you what you need, a list of all the elements matching the pattern.

answered Oct 4, 2018 at 3:57

crow3487

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Comments

ipramusinto · Accepted Answer · 2018-10-04 03:57:44Z

0

You should use re.findall

n_date = re.compile('\d{2,3}-\d{2,3}')
result = []
with open(‘re.txt’, ‘r’) as n:
    for line in n:
         result += re.findall(n_date, line)
print(result)

answered Oct 4, 2018 at 3:57

ipramusinto

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