Hi Fridtjof, I might miss something, but can’t you assign the ids once before starting the iteration and then reuse them throughout the iterations? Of course you would have to add another field to your input data but then you don’t have to run the zipWithIndex for every iteration.
Cheers, Till On Mon, Feb 1, 2016 at 11:37 AM, Fridtjof Sander < fsan...@mailbox.tu-berlin.de> wrote: > (tried to reformat) > > > Hi, > > I have a problem which seems to be unsolvable in Flink at the moment > (1.0-Snapshot, current master branch) > and I would kindly ask for some input, ideas on alternative approaches or > just a confirmatory "yup, that doesn't work". > > ### Here's the situation: > > I have a dataset and its elements are totally ascending sorted by some key > (Int). Each element has a "next-pointer" to its successor, which is just > another field with the key of the following element: > > x0 -> x1 -> x2 -> x3 -> ... -> xn > > The keys are not necessarily increasing by 1, so it may be that: x0 has > key 2 and x1 has key 10, x2 has 11, x3 has 25 and so on. I need to process > that set in the following way: > > iterate: > > find all pairs of elements where "next == key" BUT make sure no element > appears in multiple pairs > > example: do pair (x0, x1), (x2, x3), (x4, x5), ... but don't pair (x1, > x2), (x3, x4), ... > > then, if some condition is met, combine a pair > > run above procedure again with switched pairing-condition: > > example: do pair (x1, x2), (x3, x4), (x5, x6), ... do not pair (x0, x1), > (x2, x3), .. > > I hope the problem is clear... > > > ### Now my approach: pseudo-scala-code: > > > val indexed = input.zipWithIndex > > val flagged = indexed.map((i, el) => el.setFlag(i % 2 == 0)) > > val left = flagged.filter(el => el.flag) > > val right = flagged.filter(el => !el.flag) > > left.fullOuterJoin(right) > > .where(el.next) > > .equalTo(el.key) > > ... > > > I attach my elements with a temporary key, that is increasing by 1, with > zipWithIndex. Then, I map that tempKey to a boolean joinFlag: true if key > is even, false if key is odd. Then I filter all elements with true, and put > them in a dataset that is the left side of the next == key join. The right > side are all elements with flag == false In the second run, I switch the > flag construction to el.setFlag(i % 2 != 0). > > That actually works, there is only one problem: > > > ### The problem: > > > In my approach, I must not loose the total ordering of the data, because > only if that ordering is preserved, the assignment of alternating > join-flags works. Initially it is done by range-partitioning and > partition-sorting. However, that ordering is destroyed, when data is > shuffled for the join. And I can not restore it, because I have to run the > whole thing in an iteration, and range-partitioning is not supported within > iterations. > > > ### Help? > > It sounds all very complicated, but the only thing I really have to solve > is that join without any element appearing in multiple pairs (as described > in "the situation"). If anyone has any idea how to solve this, that person > would make my day so hard... > > Anyways, thanks for your time! > > Best, Fridtjof > > > > Am 01.02.16 um 11:32 schrieb Fridtjof Sander: > > Hi, > > I have a problem which seems to be unsolvable in Flink at the moment > (1.0-Snapshot, current master branch) > and I would kindly ask for some input, ideas on alternative approaches or > just a confirmatory "yup, that doesn't work". > > ### Here's the situation: > > I have a dataset and its elements are totally ascending sorted by some key > (Int). Each element has a "next-pointer" to its successor, which is just > another field with the key of the following element: x0 -> x1 -> x2 -> x3 > -> ... -> xn The keys are not necessarily increasing by 1, so it may be > that: x0 has key 2 and x1 has key 10, x2 has 11, x3 has 25 and so on. I > need to process that set in the following way: iterate: find all pairs of > elements where "next == key" BUT make sure no element appears in multiple > pairs example: do pair (x0, x1), (x2, x3), (x4, x5), ... but don't pair > (x1, x2), (x3, x4), ... then, if some condition is met, combine a pair run > above procedure again with switched pairing-condition: example: do pair > (x1, x2), (x3, x4), (x5, x6), ... do not pair (x0, x1), (x2, x3), .. I hope > the problem is clear... ### Now my approach: pseudo-scala-code: > > val indexed = input.zipWithIndex val flagged = indexed.map((i, el) => > el.setFlag(i % 2 == 0)) val left = flagged.filter(el => el.flag) > val right = flagged.filter(el => !el.flag) left.fullOuterJoin(right) > .where(el.next) .equalTo(el.key) ... I attach my elements with a temporary > key, that is increasing by 1, with zipWithIndex. Then, I map that tempKey > to a boolean joinFlag: true if key is even, false if key is odd. Then I > filter all elements with true, and put them in a dataset that is the left > side of the next == key join. The right side are all elements with flag == > false In the second run, I switch the flag construction to el.setFlag(i % 2 > != 0). That actually works, there is only one problem: ### The problem: In > my approach, I must not loose the total ordering of the data, because only > if that ordering is preserved, the assignment of alternating join-flags > works. Initially it is done by range-partitioning and partition-sorting. > However, that ordering is destroyed, when data is shuffled for the join. > And I can not restore it, because I have to run the whole thing in an > iteration, and range-partitioning is not supported within iterations. ### > Help? It sounds all very complicated, but the only thing I really have to > solve is that join without any element appearing in multiple pairs (as > described in "the situation"). If anyone has any idea how to solve this, > that person would make my day so hard... Anyways, thanks for your time! > Best, Fridtjof > > >
