Find and Capture RegEx Matches In Python

Working on a new thing :

python

import re

text = "the quick brown fox"

text = re.sub(r'(q\S+) (b\S+)', r'\2 \1', text)


print(text)

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This is the basic way to find and captures regular expression matches in python.

python

import re

  string = 'Test Date: 2017-04-05'
  matches = re.search(r"\d\d\d\d-(\d\d)-\d\d", string)

  if matches:
      print("Entire group: ", end="")
      print(matches.group())
      print("Subgroup by index: ", end="")
      print(matches.group(1))

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This is how to do it with a compiled regular expression

python

import re

  string = 'Test Date: 2017-04-05'

  pattern = re.compile('\d\d\d\d-(\d\d)-\d\d')

  matches = pattern.search(string)

  if matches:
      print("Entire match: ", end="")
      print(matches.group())
      print("Subgroup by index: ", end="")
      print(matches.group(1))

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Old Notes :

python

with open('/tmp/output-2.csv', 'r') as input2:
    with open('/tmp/output-3.csv', 'w') as output3:
      csv_reader = csv.reader(input2)
      csv_writer = csv.writer(output3, quoting=csv.QUOTE_ALL)
      pattern = re.compile("(\d+)/(\d+)/(\d+)")
      def replacement(m):
        return f"20{m.group(3).zfill(2)}-{m.group(1).zfill(2)}-{m.group(2).zfill(2)}"
      for row in csv_reader:
        row[7] = re.sub(pattern, replacement, row[7])
        row[8] = re.sub(pattern, replacement, row[8])
        row[12] = row[12].replace('-', '')
        row[12] = row[12].replace('%', '')
        row[15] = row[15].replace('%', '')
        csv_writer.writerow(row)