The short answer

Since the element we want to replace has a unique name, what we’re looking for is this:

nav {
    ul {
        li {
            a {
                color: red;

                @at-root #{selector-replace(&, 'ul', 'ul.opened')} {
                    color: green;
                }
            }
        }
    }
}

The long answer

Manipulating selectors is extremely dirty, and I would advise against it unless you absolutely had to. If you’re overqualifying your selectors by specifying things like table tr td or ul li, then start by simplifying: tr and ul are both redundant in these selectors (unless you’re trying to avoid styling elements under an ordered list). Adjust your nesting to be simpler, etc.

Starting with Sass version 3.4, there are 2 important features that allow you to modify selectors.

• Selector functions
• The parent selector can be stored in a variable

Example:

.foo ul > li a, .bar {
    $sel: &;
    @debug $sel;
}

You’ll always get a list of list of strings because selectors can be chained together with a comma, even when you have only one selector.

.foo ul > li a, .bar { ... }
 (1  2  3  4 5), (1)

You’ll note that the descendant selector is being counted here (lists in Sass can be either space or comma delimited). This is extremely important to remember.

---

When selector-replace() doesn’t work

The selector-replace() function does not work in the following cases:

• The selector you want to replace is not unique (eg. ul ul li)
• You want to insert one or more selectors (eg. ul ul li -> ul ul ul li)
• You want to remove a selector (eg. ul > li -> ul li)

In this case, you’ll need to loop over the selectors and you’ll need to know which position you want to modify. The following function will take a function and apply it to a specific position in your selector using the magic of the call() function.

@function selector-nth($sel, $n, $f, $args...) {
    $collector: ();
    @each $s in $sel {
        $modified: call($f, nth($s, $n), $args...);
        $collector: append($collector, set-nth($s, $n, $modified), comma);
    }

    @return $collector;
}

Append a class (when the selector isn’t unique or you don’t know its name)

The function we need here takes 2 arguments: the original selector and the selector you’d like to append to it. Uses simple interpolation to do the job.

@function append-class($a, $b) {
    @return #{$a}#{$b};
}

.foo, .bar {
    ul > li a {
        color: red;

        @at-root #{selector-nth(&, -2, append-class, '.baz')} {
            color: blue;
        }
    }
}

Output:

.foo ul > li a, .bar ul > li a {
  color: red;
}
.foo ul > li.baz a, .bar ul > li.baz a {
  color: blue;
}

Insert a selector

This function also takes 2 arguments: the original selector and the selector you’d like to insert before it.

@function insert-selector($a, $b) {
    @return $b $a;
}

.foo, .bar {
    ul > li a {
        color: red;

        @at-root #{selector-nth(&, -2, insert-selector, '.baz')} {
            color: blue;
        }
    }
}

Output:

.foo ul > li a, .bar ul > li a {
  color: red;
}
.foo ul > .baz li a, .bar ul > .baz li a {
  color: blue;
}

Remove a selector

Removing a selector is as simple as replacing your selector with an empty string.

@function remove-selector($sel) {
    @return '';
}

.foo, .bar {
    ul > li a {
        color: red;

        @at-root #{selector-nth(&, -2, remove-selector)} {
            color: blue;
        }
    }
}

Output:

.foo ul > li a, .bar ul > li a {
  color: red;
}
.foo ul > a, .bar ul > a {
  color: blue;
}

---

TL;DR

Selectors are just a lists. Any list manipulation functions will work on it and you can loop over it to modify it as necessary.

So yeah, don’t do it unless you really really really need to. If you’ve decided you still need it, I’ve packaged these functions up into the selector-nth library.