Sum selected rows/columns and replace them with the sum

Question

I have the code:

pos = {2, 3, 5};
tbl = Table[i*j, {i, 6}, {j, 3}];
totRow = Total /@ tbl[[pos]];

which defines some row indices (pos) and a table and some of all rows given by indices pos. However, I would like to do in mathematica way (so probably no fors) sum those rows and replace them by that single total row, which in my example has the name totRow. What is the most effective way. I was thinking succession of drop, join but is there a better way?

To supply an example:

pos = {2, 3, 5};
tbl = Table[i*j, {i, 6}, {j, 3}];
totRow = Total /@ tbl[[pos]];
finTbl = Join[Delete[tbl, Partition[pos, 1]], {totRow}];
totRow
TableForm[tbl]
TableForm[finTbl]

gives an idea, the three rows from tbl at indices given by pos are deleted and the totRow added. So this works but looks clumsy, I am just curious whether there is proper or better way.

UPDATE: I changed the examples so that the original table is made of i*j for better legibility.

Answer 1

Answer:

(*Generate table*)
table = Table[RandomReal[], {6}, {3}];

(*Declare positions*)
pos = {2, 3, 5};

(*Carry out addition, removal and modification*)
Append[Delete[table, Partition[pos, 1]], #] &[Total@# &[table[[#]] & /@ pos]]

Reference:
Append
Table
RandomReal
Delete
Partition
Total
@ | /@ | # etc.

Answer 2

The way you are doing it seems just fine. If you want to make it less clear what you are doing, and add some shorthand notation you could rewrite it like this:

finTbl = tbl~Delete~Transpose@{pos}~Join~{Total /@ tbl[[pos]]}

(it hurts my brain to try and read code written using infix notation, but I was trying to do it with the smallest number of characters)

Answer 3

As an oneliner. You don't need Table to create a random matrix

Append[Delete@##, Plus @@@ Extract@##] & @@ {RandomReal[1, {6, 3}], List /@ {2, 3, 5}};

Answer 4

f[table_, pos_] := Append[Delete[table, List /@ pos], Tr /@ table[[pos]]]

finTbl = f[tbl, pos]