<DT><CODE><VAR>s</VAR>[<VAR>i</VAR>] = <VAR>x</VAR></CODE><DD>item <VAR>i</VAR> of <VAR>s</VAR> is replaced by <VAR>x</VAR>
<DT><CODE><VAR>s</VAR>[<VAR>i</VAR>:<VAR>j</VAR>] = <VAR>t</VAR></CODE><DD>slice of <VAR>s</VAR> from <VAR>i</VAR> to <VAR>j</VAR> is replaced by <VAR>t</VAR>
<DT><CODE>del <VAR>s</VAR>[<VAR>i</VAR>:<VAR>j</VAR>]</CODE><DD>same as <CODE><VAR>s</VAR>[<VAR>i</VAR>:<VAR>j</VAR>] = []</CODE>
<DT><CODE><VAR>s</VAR>.append(<VAR>x</VAR>)</CODE><DD>same as <CODE><VAR>s</VAR>[len(<VAR>s</VAR>):len(<VAR>s</VAR>)] = [<VAR>x</VAR>]</CODE>
<DT><CODE><VAR>s</VAR>.count(<VAR>x</VAR>)</CODE><DD>return number of <VAR>i</VAR>'s for which <CODE><VAR>s</VAR>[<VAR>i</VAR>] == <VAR>x</VAR></CODE>
<DT><CODE><VAR>s</VAR>.index(<VAR>x</VAR>)</CODE><DD>return smallest <VAR>i</VAR> such that <CODE><VAR>s</VAR>[<VAR>i</VAR>] == <VAR>x</VAR></CODE> --- (1)
<DT><CODE><VAR>s</VAR>.insert(<VAR>i</VAR>, <VAR>x</VAR>)</CODE><DD>same as <CODE><VAR>s</VAR>[<VAR>i</VAR>:<VAR>i</VAR>] = [<VAR>x</VAR>]</CODE>
if <CODE><VAR>i</VAR> >= 0</CODE>
<DT><CODE><VAR>s</VAR>.remove(<VAR>x</VAR>)</CODE><DD>same as <CODE>del <VAR>s</VAR>[<VAR>s</VAR>.index(<VAR>x</VAR>)]</CODE> --- (1)
<DT><CODE><VAR>s</VAR>.reverse()</CODE><DD>reverses the items of <VAR>s</VAR> in place
<DT><CODE><VAR>s</VAR>.sort()</CODE><DD>permutes the items of <VAR>s</VAR> to satisfy