Difficulties with adding spaces around equal signs using regular expressions with Notepad++

Question

I am currently trying to improve the readability of some of my python scripts by adding spaces around the equal signs. For example, currently an assignment looks like this:

Answer 1

  (?<=[\w\]\)])(=)(?=\w)

You can use this . Replacement by

 " \1 "

See Demo.

http://regex101.com/r/vY0rD6/2

Your earlier regex had no captured pattern.So just added that.

Answer 2

With your initial regex,

Regex:

(?<=[\w\]\)])=(?=\w)

REplacement string:

 =<space>

I added <space> instead of . Lookarounds would do a zero width match. It won't match any characters.

DEMO

Answer 3

The accepted answer to that question only contains the search expression. The question asker left a comment pointing out that the replacement string should be left empty. What that does is delete whatever is matched.

What you're looking to do here is not to delete the = operators outright (as that would break your scripts quite catastrophically!), but simply to space-pad them.

What your original regex does:

(?<=[\w\]\)])=(?=\w)

Is find any = character that

• is preceded by one of \w, ] or ), as in (?<=[\w\]\)]), and
• is followed by a \w, as in (?=\w).

The (?<=) and (?=) portions are lookbehind and lookahead assertions respectively, not capture groups. Since your original regex doesn't capture anything, replacing with \1 = \2 won't work as expected.

In your working regex:

([\w\]\)])=(\w)

The ?<= and ?= tokens are removed which makes the two parentheticals capture groups. This allows the \1 = \2 backreferences to work, inserting spaces between the first capture, the = sign, and the second capture appropriately.

On a side note (since you've already found a working solution), you can still make the original regex work, but the replacement string should simply be = surrounded by spaces. The stuff that is tested by the lookbehind and lookahead assertions is never actually picked up — the only character that is matched is the = sign itself — so it will not disappear when you replace.

Answer 4

In your first Regular Expression(?<=[\w\]\)])=(?=\w) there's no capturing groups so you can't use back references (replacing with \1 = \2)

But you can use space=space as replacement with the above regex where space is a real white space character.