Q:

Given a list of Tetris moves, return the number of completed lines

Description
We consider a slightly simplified version of Tetris where each move consists of:

rotating the piece clockwise, 0 to 3 times
positioning the piece at a given column
fast drop

The goal is to determine the number of completed lines, given a list of such Tetris moves.
Completed rows are removed as pieces are dropped, following standard Tetris rules.
Playfield
The playfield is 10-column wide. There's no Game Over and it is assumed that there's always enough space and time to perform the above actions, no matter the configuration of the playfield. The height of the playfield doesn't really matter here, but you can use the standard 22 rows as an upper limit.
Shapes of Tetrominoes

Input / Output
Input
A comma separated list of Tetris moves encoded with 3 characters. The first two character describe the Tetromino shape to use and the last one describes the position where it's dropped.

Tetromino: I, O, T, L, J, Z or S, in the same order as above.
Number of clockwise rotations: 0 to 3
Column: 0 to 9. This is the column in which the top-left corner of the piece (marked with an x on the above picture) is located after the rotation 1

It is assumed that all moves in the provided list are valid. There's no need to check for invalid entries such as I07 (horizontal I shape put too far on the right).
1 You are free to either implement a real rotation algorithm or to hardcode all the different shapes, as long as the x is located in the column given by the third character of the move.
Output
Number of completed lines.
Example

O00,T24 will generate the first position and O00,T24,S02,T01,L00,Z03,O07,L06,I05 will generate the second position.
Therefore, the following sequence will generate a Tetris and should return 4:
O00,T24,S02,T01,L00,Z03,O07,L06,I05,I19

Test cases
1) "O00,T24,S02,T01,L00,Z03,O07,L06,I05,I19" -> 4
2) "S00,J03,L27,Z16,Z18,I10,T22,I01,I05,O01,L27,O05,S13" -> 5
3) "I01,T30,J18,L15,J37,I01,S15,L07,O03,O03,L00,Z00,T38,T01,S06,L18,L14" -> 4
4) "S14,T00,I13,I06,I05,I19,L20,J26,O07,Z14,Z10,Z12,O01,L27,L04,I03,S07,I01,T25,J23,J27,O01,
    I10,I10" -> 8
5) "O00,T24,L32,T16,L04,Z11,O06,L03,I18,J30,L23,Z07,I19,T05,T18,L30,I01,I01,I05,T02" -> 8

Test page
You can use this JSFiddle to test a move list.

A:

C, 401 392 383 378 374 351 335 324 320 318 316 305 bytes
d[99]={'3Z3Z',983177049,513351321,1016270793,970073931,~0},*L=d+5,V,s;char c;main(x){B:for(c=0;c[++L]|V;V>>=4)c-=(L[c]=*L|(V&15)<<x)>1022;s-=c;if(scanf("%c%1d%d,",&c,&V,&x)>2)for(V=c^79?d[c/4%5]>>24-V*8:27,V=c^73?V/64%4<<8|V/8%8<<4|V%8:V&32?15:4369;;--L)for(c=4;c--;)if(L[c]>>x&V>>c*4&15)goto B;return s;}

Takes comma-separated input on stdin, returns the score in the exit status.
Requires char to be signed (which is the default for GCC) and requires '3Z3Z' to be interpreted as 861549402 (which is the case for GCC on little endian machines, at least).

Usage example:
echo "O00,T24,S02,T01,L00,Z03,O07,L06,I05,I19" | ./tetris; echo "$?";

High-level explanation:
All shapes except the line can fit in a 3x3 grid with one corner missing:
6 7 -
3 4 5
0 1 2

That means it's easy to store them in a byte each. For example:
- - -         0 0 -
- # -   -->   0 1 0   -->   0b00010111
# # #         1 1 1

(we align each piece to the bottom-left of the box to make dropping it easier)
Since we get at least 4 bytes to an int, this means we can store all 4 rotations of each piece in a single integer, with a special case for the line. We can also fit each row of the game grid into an int (only needs 10 bits), and the currently falling piece into a long (4 lines = 40 bits).

Breakdown:
d[99]={             // General memory
  '3Z3Z',           //  Shape of "S" piece (multi-char literal, value = 861549402)
  983177049,        //  Shape of "T" piece
  513351321,        //  Shape of "Z" piece
  1016270793,       //  Shape of "J" piece
  970073931,        //  Shape of "L" piece
  ~0                //  Bottom of game grid (solid row)
},                  //  Rest of array stores lines in the game
*L=d+5,             // Working line pointer (start >= base of grid)
V,                  // Current piece after expansion (also stores rotation)
s;                  // Current score (global to initialise as 0)
char c;             // Input shape identifier (also counts deleted lines)
main(x){            // "x" used to store x location
  B:                                // Loop label; jumps here when piece hits floor
  for(c=0;c[++L]|V;V>>=4)           // Paint latest piece & delete completed lines
    c-=(L[c]=*L|(V&15)<<x)>1022;
  s-=c;                             // Increment score
  if(scanf("%c%1d%d,",&c,&V,&x)>2)  // Load next command
    for(V=c^79?                     // Identify piece & rotation
          d[c/4%5]>>24-V*8          //  Not "O": load from data
        :27,                        //  Else "O": always 27
        V=c^73?                     // If not "I" (line):
          V/64%4<<8|V/8%8<<4|V%8    //  expand shape to 16-bits
        :                           // Else we are a line:
          V&32?                     //  "I" coincides with "J"; use this to check
               15:4369;             //  horizontal/vertical
        ;--L)                       // Loop from top-to-bottom of grid
      for(c=4;c--;)                 //  Loop through rows of piece
        if(L[c]>>x&V>>c*4&15)       //   If collides with existing piece:
          goto B;                   //    Exit loops
  return s;                         // Return score in exit status
}

-4, -1 thanks to @Titus, and -23, -11 with inspiration from their answer

A:

PHP, 405 399 378 372 368 360 354 347 331 330 328 319 309 300 bytes
(with Dave´s block mapping)
<?$f=[~0];L:for($y=0;$f[++$d+$y]|$s;$s>>=4)$y-=1022<$f[$d+$y]=$f[$d]|$s%16<<$x;$c-=$y;if(list($p,$r,$x)=$argv[++$z])for($s=I==$p?$r&1?4369:15:hexdec(decoct(O==$p?27:ord(base64_decode('M1ozWjqaF1kemR6ZPJMPyTnSJ0s')[ord($p)/4%5*4+$r])));;$d--)for($y=4;$y--;)if ($f[$d+$y]>>$x&$s>>$y*4&15)goto L;echo$c;

program, takes moves as separate arguments, prints result
breakdown to function:
takes moves as array, returns result
function t($a)
{
    $f=[~$z=0];             // init field $f (no need to init $z in golfed version)
    L:                      // jump label
                            // A: paint previous piece at line $d+1:
#   if($s)paint($f,$s,$d+1,$x);
    for($y=0;               // $y = working line offset and temporary score
        $f[++$d-$y]|$s;$s>>=4)// next field line; while field or piece have pixels ...
        $s>>=4)                 // next piece line
        $y+=1022<               // 3: decrease temp counter if updated line is full
            $f[$d-$y]=$f[$d]    // 2: use $y to copy over dropped lines
                |$s%16<<$x;     // 1: set pixels in working line
    $c+=$y;                         // add temp score to global score
#   paint($f);var_dump($c);
    if(list($p,$r,$x)=$a[$z++])// B: next piece: $p=name; $r=rotation, $x=column
#   {echo"<b>$z:$p$r-$x</b><br>";
        for(                // C: create base 16 value:
            $s=I==$p
                ? $r&1?4369:15  // I shape (rotated:0x1111, not rotated:0xf)
                : hexdec(decoct(    // 5: convert from base 8 to base 16
                    O==$p ? 27  // O shape (rotation irrelevant: 033)
                    : ord(          // 4: cast char to byte
                                    // 0: 4 bytes for each remaining tetromino
                        base64_decode('M1ozWjqaF1kemR6ZPJMPyTnSJ0s')[
                            ord($p)/4%5 // 1: map STZJL to 01234
                            *4      // 2: tetromino offset
                            +$r]    // 3: rotation offset
            )));;$d--
        )
            for($y=4;$y--;) // D: loop $y through piece lines
                if ($f[$d+$y]>>$x & $s>>$y*4 & 15) // if collision ...
                    goto L;         // goto Label: paint piece, tetris, next piece
#   }
    return$c;               // return score
}

for reference: the old mapping
            hexdec(decoct(          // 3. map from base 8 to base 16
                                    // 0. dword values - from high to low: rotation 3,2,1,0
                [0x991e991e,0xc90f933c,0x4b27d239,0x1b1b1b1b,0,0x5a335a33,0x59179a3a]
                [ord($m[0])/2%9]    // 1. map ZJLO.ST to 0123.56 -> fetch wanted tetromino
                >>8*$m[1]&255       // 2. the $m[1]th byte -> rotated piece
            ))

testing
see my other PHP answer
want to watch?
remove the # from the function source and add this:
function paint($f,$s=0,$d=0,$x=0){echo'<pre>';for($y=count($f)+5;$y--;$g=0)if(($t=(($z=$y-$d)==$z&3)*($s>>4*$z&15)<<$x)|$r=$f[$y]|0){$z=$t?" $r|$t=<b".(1022<($z=$t|$r)?' style=color:green':'').">$z":" <b>$r";for($b=1;$b<513;$b*=2)$z=($t&$b?'<b>'.($r&$b?2:1).'</b>':($r&$b?1:'.')).$z;printf("%02d $z</b>\n",$y);}echo'</pre>';}

some golfing steps
Rev. 5: A big leap (399-21=378) came by simply moving the column shift
from a separate loop to the two existing loops.
Rev. 8: Switching from array to base 16 for the piece ($s) did not give much,
but made way for some more golfing.
Rev. 17: crunched the values with base64_encode(pack('V*',<values>))
and used byte indexing instead of unpack to save 16 bytes
Rev. 25 to 29: inspired by Dave´s code: new hashing (-2), new loop design (-9), goto (-10)
no pre-shift though; that would cost 17 bytes.
more potential
With /2%9, I could save 15 bytes (only 14 bytes with /4%5)
by putting binary data into a file b and then indexing file(b)[0].
Do I want that?
UTF-8 characters would cost a lot for the transformation.
on the hashing
I used ZJLO.ST /2%9 -> 0123.56;
but T.ZJLOS /3%7 -> 0.23456 is as good.
one byte longer: O.STJLZ %13/2 -> 0.23456
and three more: OSTZJ.L %17%12%9 -> 01234.6
I could not find a short hash (max. 5 bytes) that leaves no gap;
but Dave found STZJL /4%5 -> 01234, dropping the O from the list. wtg!
btw: TIJSL.ZO (%12%8) -> 01234.67 leaves room for the I shape
(and a fictional A, M or Y shape). %28%8 and %84%8, do the same (but with E instead of A).

