'Two groups of alphanumeric characters (including letters, numbers, and underscores) separated by one or more whitespace characters

Fill in the code to check if the text passed has at least 2 groups of alphanumeric characters (including letters, numbers, and underscores) separated by one or more whitespace characters

import re
def check_character_groups(text):
  result = re.search(r"___", text)
  return result != None

print(check_character_groups("One")) # False
print(check_character_groups("123  Ready Set GO")) # True
print(check_character_groups("username user_01")) # True
print(check_character_groups("shopping_list: milk, bread, eggs.")) # False
regex


Solution 1:[1]

import re
def check_character_groups(text):
  result = re.search(r"\w\s+\w", text)
  return result != None

print(check_character_groups("One ")) # False
print(check_character_groups("123  Ready Set GO")) # True
print(check_character_groups("username user_01")) # True
print(check_character_groups("shopping_list: milk, bread, eggs.")) # False

This is my answer that returns what it is asking for,
the key answer is \w\s+\w

Solution 2:[2]

Use as regex pattern

\b\w+\s+\w+\b
  • \b word boundery, marks a start of a word (also works as start of string)
  • \w+ means alphanumeric characters [a-zA-Z0-9_] (one or more)
  • \s+ one or more whitespaces
  • \w+ next alphanumeric characters (1+)
  • \b word boundery (works at string end too)

So try:

import re

def check_character_groups(text):
  result = re.search(r"\b\w+\s+\w+\b", text)
  return result != None

print(check_character_groups("One "))  # False
print(check_character_groups("123  Ready Set GO"))  # True
print(check_character_groups("username user_01"))  # True
print(check_character_groups("shopping_list: milk, bread, eggs."))  # False

If there should be special characters/spaces/... left and/or right of the alphanumeric groups, you must specify it.

Test it on regex101 or debuggex.

Picture by https://jex.im/regulex/ Picture by Regulex (! The syntax there is for JavaScript !)

Solution 3:[3]

import re
def check_character_groups(text):
  result = re.search(r"\w+\s", text)
  return result != None

print(check_character_groups("One")) # False
print(check_character_groups("123  Ready Set GO")) # True
print(check_character_groups("username user_01")) # True
print(check_character_groups("shopping_list: milk, bread, eggs.")) # False

Here is your output: False True True False

Solution 4:[4]

import re
def check_character_groups(text):
  result = re.search(r"[\w]+[\s]+[\w]", text)
  return result != None

The answer is to use [\w]+[\s]+[\w] as regex.

Solution 5:[5]

This will match the condition for 2 groups of alphanumeric characters including underscore separated by one or more whitespace characters

result = re.search(r"\w+\s.*", text)

Clarification:

  1. Repeated characters will be searched by *
  2. Alphanumeric characters including underscore will be searched by \w
  3. Space characters will be searched by using \s
  4. To search character between two will be searched by a combination of a dot(.)

Solution 6:[6]

result = re.search(r"\b\w*\s\w*\b", text)

You can try this too. \b defines the boundary.

Solution 7:[7]

Use simple one

result = re.search(r"[\w] [\w]", text)

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2
Solution 3 Divya Goswami
Solution 4 Ank
Solution 5
Solution 6 Shravan Palan
Solution 7 manthan gehlot

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