notepad++ - Remove quotation marks around a number (while leaving other quotes)

I want to replace all text like this:

"latitude": "32.336533",

with this:

"latitude": 32.336533,

I'm using Notepad++.

Using Regex use the following pattern:

"([0-9]+\.{0,1}[0-9]*)"

and Replace with:

\1

This worked for me with replace all function of notepad++. It would find "12." too and remove the double quotes. For a more comprehensive search use this Regex pattern:

"(\-{0,1}[0-9]+(\.[0-9]+){0,1})"

which actually will find negative numbers as well and matches floats only with digits after the decimal point.

Explanation:

it will match

    "             ; a leading double quote
    (             ; followed by the outer subpattern (in backreference \1:
      \-{0,1}       ; an optional minus sign
      [0-9]+        ; followed by 1 or more decimal digits (could be replaced by \d)
      (             ; followed by the next subpattern
         \.           ; a decimal point
         [0-9]+       ; followed by 1 or more digits
      ){0,1}        ; maximal 1 occurrence of this subpattern, and it's optional
    )             ; end of the outer subpattern
    "             ; followed by the trailing double quote

Backreference \1 includes everything in the outer subpattern including the inner one, if it exists. You could use \d for the [0-9] classes and use the question mark ? instead of the last {0,1} group. Remember that the use of the ? may change the greediness of patterns.

Example:

the text in notepad++ with following lines

"latitude": "-32.336533",
"latitude": "32.336533",
"foo": "14"
"bla": "12."
"to7": "12.a"

will be changed after applying "Replace all" to

"latitude": -32.336533,
"latitude": 32.336533,
"foo": 14
"bla": "12."
"to7": "12.a"

Regex match

"([\d\.]+)"

Replace with

\1