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    <p><code>\S\s\S\s\S</code> will indeed match three non-space characters separated by space characters. However, <code>^.*?[\S\s\S\s\S].{5}</code> does something completely different:</p> <ul> <li><code>^</code> matches the beginning of the line.</li> <li><code>.*?</code> matches characters until the next match can start (<em>not</em> as many as it can). Since you specify <code>/s</code>, <code>.</code> will match newline as well.</li> <li><code>[\S\s\S\s\S]</code> is a character class, and so is the same as <code>[\S\s]</code>—match either <code>\S</code> or <code>\s</code>, which is to say anything.</li> <li><code>.{5}</code> will match five characters.</li> </ul> <p>Since <code>[\S\s]</code> and <code>.</code> with <code>/s</code> match the same things, the <code>.*?</code> will never match any characters as it wants to match as little as possible. Thus, this is the same as <code>s/^.{6}//s</code>—delete the first six characters from the string. As you can see, that's not what you wanted!</p> <p>One way to keep the first three numbers is to explicitly match them: <code>s/^(\d \d \d).*/$1/s</code>. Here, <code>\d</code> matches a single digit (<code>0</code>–<code>9</code>) with literal spaces in between them. We match the first three followed by anything at all, and then replace the whole match—since it ends in <code>.*</code>, that's the whole string—with just the bit in between parentheses, <em>i.e.</em> the first three numbers. If your numbers can be more than one digit long, then <code>s/^(\d+ \d+ \d+).*/$1/s</code> will do what you want; if you can have arbitrary space-like characters (space, tab, newline) separating them, then <code>s/^(\d\s\d\s\d\s).*/$1/s</code> is what you want (or <code>\s+</code> if you can have multiple spaces). If you want to catch lines which have things other than digits, you can use <code>\S</code> or <code>\S+</code>, just as you were.</p> <p>Another approach, using lookbehind, would be <code>s/(?&lt;=^\d \d \d).*//s</code>. In other words, delete any characters which are preceded by <code>^\d \d \d</code>—the beginning of the string followed by three space-separated numbers. There's no real advantage to this approach—I'd probably do it the other way—but since you mentioned lookbehind, here's how you can do it. (Again, things like <code>s/(?&lt;=^\S\s\S\s\S).*//s</code> are more general.)</p>
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    1. USAntal Spector-Zabusky
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    1. COThank you so much. Your solution was what I ultimately used, just substituting S for d so I could have decimal and E formed answers. `s/^(\S+ \S+ \S+).*/$1/s`
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    2. COGlad I could help! If this answer solved your problem, you could "accept" it by clicking the green check mark next to it. ([The FAQ entry on how to ask](http://stackoverflow.com/faq/#howtoask) explains what this is for.)
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