Regular Expression - Match all but first letter in each word in sentence

Question

I've almost got the answer here, but I'm missing something and I hope someone here can help me out.

I need a regular expression that will match all but the first letter in each word in a sentence. Then I need to replace the matched letters with the correct number of asterisks. For example, if I have the following sentence:

There is an enormous apple tree in my backyard.

I need to get this result:

T**** i* a* e******* a**** t*** i* m* b*******.

I have managed to come up with an expression that almost does that:

(?<=(\b[A-Za-z]))([a-z]+)

Using the example sentence above, that expression gives me:

T* i* a* e* a* t* i* m* b*.

How do I get the right number of asterisks?

Thank you.

Chausses

Do you need to use regular expressions for any particular reason? Depending on the programming language you're writing in, you can use a substring with replacement to get the same effect – Uird

Answer 1

Try this:

\B[a-z]

\B is the opposite of \b - it matches where there is no word boundary - when we see a letter that is after another letter.

Your regex is replacing the whole tail of the word - [a-z]+, with a single asterisks. You should replace them one by one. If you want it to work, you should match a single letter, but check is has a word behind it (which is a little pointless, since you might as well check for a single letter (?<=[A-Za-z])[a-z]):

(?<=\b[A-Za-z]+)[a-z]

(note that the last regex has a variable length lookbehind, which isn't implemented in most regex flavors)

Bashful

The shortest regex here is probably \B\w, but \w adds upper case letters and underscores. – Bashful

(?<=\b[A-Za-z]+) won't work in any flavor but .NET and JGSoft. You had it right the first time. – Scudo

@Alan - good point. I've added a warning on that. Either way, I did say it was rather pointless :) – Bashful

And (depending on the regex flavor and whether you're planning on matching other characters than unaccented letters between a and z) you might want to use \p{L} or [^\W\d_] instead of [a-z]. – Percentage

How would that work when you had something like "Jack's toothbrush"? – Dibucaine

@NathanArthur - That is a good question... The underlying question is even more difficult: What is a word? The pattern above assumes a word is made of alphanumeric characters, which is wrong. In fact, I do not believe I can solve that problem reliably with a simple pattern - there are just too many edge cases. Still, as for your question: In .Net, you can add an apostrophe to the pattern above: (?<=\b[A-Za-z']+)[a-z]. On other flavors I think (?<=\B|\b')[a-z] can work. Either way, it requires some thinking. – Bashful

Answer 2

You can try this

\B\w

this will replace all characters except for the first letter of every word

from this ==Hello==World== into ==H****==W****==

Curule

Answer 3

Try this possibly:

(\w{1})\w*

Uird

Answer 4

This is an old question. Adding an answer since the others don't seem to solve this problem completely or clearly. The simplest regular expression that handles this is /(\B[a-z])/g. This adds 'g' as a global flag, so the single character search will be repeated throughout the string.

string = "There is an enormous apple tree in my backyard."
answer = string.replace(/\B[a-z]/g, "*");

string = "There is an enormous apple tree in my backyard."
$("#stringDiv").text(string);

answer = string.replace(/\B[a-z]/g, "*");
$("#answerDiv").text(answer);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="stringDiv"></div>
<div id="answerDiv"></div>

Mien