如何用Python正则表达式在无空格连续字符串中查找数字/IP/整数？

First, let's break down why your initial loop approach didn't work:

• You're iterating over individual characters, so you can only capture single digits instead of continuous sequences like 123 or 1.2.3.4.
• The check if a in range(9) is invalid: range(9) produces integers 0-8, but a is a string character (like '9'), so they can't be compared directly. Even if you used a.isdigit(), you'd still only get isolated digits, not the grouped sequences you need.

Working Regex Approach #

Regular expressions are perfect here because they can match patterns of continuous digits and optional dot-separated number segments. Here's the code that will extract exactly what you need:

import re

# Your input text
input_text = "randomblahblah123iwanttosearchthis666number1.2.3.4withpython"

# Regex pattern to match:
# - \b: Word boundary (ensures we don't capture partial sequences attached to letters)
# - \d+: One or more digits
# - (?:\.\d+)*: Non-capturing group for zero or more ".digit" segments (handles IP-like strings)
pattern = r'\b\d+(?:\.\d+)*\b'

# Extract all matches
matches = re.findall(pattern, input_text)

# Format the output as requested
print(', '.join(matches))  # Output: 123, 666, 1.2.3.4

How the Regex Works #

Let's break down the pattern piece by piece:

• \b: Acts as a marker between alphanumeric characters and non-alphanumeric characters (or start/end of string). This ensures we don't accidentally capture something like blah123 as part of a longer word—we only get the standalone numeric sequences.
• \d+: Matches one or more consecutive digits (covers 123 and 666).
• (?:\.\d+)*: The (?:...) is a non-capturing group (we don't need to extract the individual dot segments separately), and \.\d+ matches a dot followed by one or more digits. The * means this can repeat zero or more times—so it handles both plain numbers (zero repeats) and IP-like strings (three repeats, like 1.2.3.4).

Optional: Strict IP Validation #

If you ever need to ensure the dot-separated sequence is a valid IP address (each segment 0-255), you can use a more precise pattern:

strict_ip_pattern = r'\b(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b'
strict_matches = re.findall(strict_ip_pattern, input_text)

But for your current use case, the simpler pattern is sufficient.

内容的提问来源于stack exchange，提问作者lola