Re: Sum elements in a list with conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg128430] Re: Sum elements in a list with conditions
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Fri, 19 Oct 2012 02:41:03 -0400 (EDT)
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• References: <20121018063647.0023568A4@smc.vnet.net>

```list1 = {{1, a}, {2, b}, {1, c}, {2, d}, {2, e}, {3, f}};

list2 = Total[#]/{Length[#], 1} & /@
GatherBy[list1, First]

{{1, a + c}, {2, b + d + e}, {3, f}}

list2 = list1 //. {s___, {n_, x_}, m___, {n_, y_},
e___} :>
{s, {n, x + y}, m, e}

{{1, a + c}, {2, b + d + e}, {3, f}}

Bob Hanlon

On Thu, Oct 18, 2012 at 2:36 AM, Guillermo Sanchez
<guillermo.sanchez at hotmail.com> wrote:
> I have a list of sublist of pairs. I wish sum the second elements of
> the sublists when the first elements are equals
> Example:
> Int[]:={{1, a}, {2, b}, {1, c}, {2, d}, {2, e}, {3, f}}
> I hope
> Out[]:= {{1, a + c}, {2, b + c + e}, {3, f}}
>
> I can do it as  is it is shown below but could any body find and
> easier solution.
>
>
> list1 = {{1, a}, {2, b}, {1, c}, {2, d}, {2, e}, {3, f}};
> list2 = GatherBy[list1, First];
> {l1, l2} = Transpose[Plus @@@ GatherBy[list, First]];
> Transpose[{l1/Length /@ list2 , l2}]
>

```

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