Remove 1 character from both sides of a recurring substring where one digit changes in each occurrence

I need to remove apostrophes from both sides of a sub-string. The substring occurs numerous times within a starting string, and one digit changes within the substring for each occurrence.

starting_string = "{'color':'Highcharts.getOptions().colors[0]','color':'Highcharts.getOptions().colors[1]','color':'Highcharts.getOptions().colors[2]'}"

substring = Highcharts.getOptions().colors[i]

desired_string = "{'color':Highcharts.getOptions().colors[0],'color':Highcharts.getOptions().colors[1],'color':Highcharts.getOptions().colors[2]}"

Above, in ‘substring’, ‘i’ represents the digit that changes in each occurrence of the substring.

The number of times ‘substring’ occurs in ‘starting_string’ will vary. This example is simplified.

>Solution :

gsub("'(Highcharts\\.getOptions\\(\\)\\.colors\\[[0-9]+\\])'",
     "\\1", starting_string)
# [1] "{'color':Highcharts.getOptions().colors[0],'color':Highcharts.getOptions().colors[1],'color':Highcharts.getOptions().colors[2]}"

Explanation of the regex:

• the parens (Hig...) define a group that we’ll reference later using \\1;
• the enveloping ' are the literal single quotes; note that these are outside the paren-group, as we will want to drop them once we find them;
• I took the liberty of inferring that i means "any number", so I replaced it with [0-9]+ which means "one or more digit".
• many characters have special meaning in regex, so they are backslash-escaped; here, they are (, ), [, ], and .. For the record, I might have been able to omit all of the backslashes and used instead fixed=TRUE, except that we want to be able to match on arbitrary numbers in [i].