{"action":"closed","issue":{"url":"https://api.github.com/repos/ramda/ramda/issues/1284","repository_url":"https://api.github.com/repos/ramda/ramda","labels_url":"https://api.github.com/repos/ramda/ramda/issues/1284/labels{/name}","comments_url":"https://api.github.com/repos/ramda/ramda/issues/1284/comments","events_url":"https://api.github.com/repos/ramda/ramda/issues/1284/events","html_url":"https://github.com/ramda/ramda/issues/1284","id":95525620,"number":1284,"title":"reduce with transform","user":{"login":"tolmasky","id":23753,"avatar_url":"https://avatars.githubusercontent.com/u/23753?v=3","gravatar_id":"","url":"https://api.github.com/users/tolmasky","html_url":"https://github.com/tolmasky","followers_url":"https://api.github.com/users/tolmasky/followers","following_url":"https://api.github.com/users/tolmasky/following{/other_user}","gists_url":"https://api.github.com/users/tolmasky/gists{/gist_id}","starred_url":"https://api.github.com/users/tolmasky/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/tolmasky/subscriptions","organizations_url":"https://api.github.com/users/tolmasky/orgs","repos_url":"https://api.github.com/users/tolmasky/repos","events_url":"https://api.github.com/users/tolmasky/events{/privacy}","received_events_url":"https://api.github.com/users/tolmasky/received_events","type":"User","site_admin":false},"labels":[],"state":"closed","locked":false,"assignee":null,"assignees":[],"milestone":null,"comments":4,"created_at":"2015-07-16T20:50:59Z","updated_at":"2016-09-12T21:34:32Z","closed_at":"2016-09-12T21:34:32Z","body":"Maybe as much of a question as a suggestion (I feel like I've seen a way to do this before). I often find myself wanting to perform an action on the second argument of the reducing function. For example:\r\n\r\nvar sumAbs = R.reduce(function(acc, val) { return acc + abs(val) }, 0, list)\r\n\r\nNormally I'd do: R.reduce(R.add, 0, R.map(abs, list))\r\n\r\nBut it would be nice to not have to iterate twice. Again, I *feel* like I've seen a good workaround using nthArg before, but can't quite remember. "}}
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