如何用正则表达式匹配带千分位的6数字符串中的第4个数字？

云上老周

2026-5-13

Fixing Regex to Match the 4th Number in Your Space-Separated String

Let's break down why your current regex is grabbing multiple results, then tweak it to target exactly the 4th number you need.

The Issue with Your Current Regex

Your regex (?<= )(\d*,*\d*\.\d*)(?= \d*,*\d*\.) is too loose:

The positive lookbehind (?<= ) only checks for a single preceding space, which applies to every number except the first one.
The positive lookahead (?= \d*,*\d*\.) looks for a space followed by any number with a decimal, which works for every number except the last one.
That’s why it ends up matching the 2nd, 3rd, and 4th numbers instead of just the 4th.

Solution 1: Capture the 4th Number with a Non-Capturing Prefix

This approach explicitly matches the first three numbers (plus their trailing spaces) as a throwaway non-capturing group, then captures the 4th number cleanly:

^(?:\d+(?:,\d+)*\.\d+ ){3}(\d+(?:,\d+)*\.\d+)

Let’s unpack this:

^: Anchors the match to the start of the string (avoids accidental partial matches later on).
(?:...): Non-capturing group (we don’t need to save the first three numbers).
- \d+(?:,\d+)*: Matches the integer part, including optional thousand separators (like 35,362 or 0).
- \.\d+: Matches the mandatory decimal part (like .843 or .000).
- : Matches the trailing space after each of the first three numbers.
{3}: Repeats the non-capturing group exactly 3 times (covers the first three numbers).
(\d+(?:,\d+)*\.\d+): Capturing group that grabs the 4th number using the same valid number pattern.

Solution 2: Use a Precise Positive Lookbehind

If you don’t want to include the prefix in your match at all, use a positive lookbehind to confirm exactly three valid numbers come before the 4th:

(?<=^(?:\d+(?:,\d+)*\.\d+ ){3})\d+(?:,\d+)*\.\d+

This works like the first solution, but uses a lookbehind to validate the prefix instead of including it. The result is just the 4th number, no extra capturing group required.