###Initial String ["1", '123', "0.5", "10.5", "10.50", " 1 ", "1,000", ".5", "5.", "1.0.0", 1, "Windows 98"] ###Desired String [1, 123, 0.5, 10.5, 10.50, 1, "1,000", ".5", "5.", "1.0.0", 1, "Windows 98"] ###Goal: "Remove quotes surrounding valid integers or floats" ####Find Pattern ["'] *(\d+(.\d+)?) *["']
["']finds beginning of string (either"or').[]wraps character class- You can drop the
[]wrapper and simplify to"or'if all your strings are quoted the same way.
*(space+*) finds optional space(s) before your integer or float.*: "preceding token zero to infinite times"
(\d+(\.\d+)?)(first capture group)(starts the first capture group. this group is represented by$1in the replace pattern.\d+requires that a digit starts this capture group. selects one to infinite digits until it hits a non digit.\escapes following character.\d= digit,d= literal "d"+preceding token one to infinite times- will not match
.5, will match0.5
(\.\d+)?(second capture group)(starts a second capture group inside of the first group.\.\d+says if there is a decimal, it must be followed by at least one digit.\.= literal decimal point,.= "any character except a new line"- will not match
5., will match5.0
)?closes the second capture group and makes it optional.- essentially, if there isn't a decimal this will be ignored.
- this group also mandates that there can only be one decimal if any.
)closes the first capture group
*(*) finds optional space(s) after your integer or float.["']finds end of string. should match the open string pattern.
####Replace With $1
- replace with first capture group
###Demo regex101.com