Combine nested repeated fields into array in Apache Beam - apache-beam

I have a collection of records, for simplicity say as follows (comma separated):
A1, B1, C1
A1, B1, C1'
A1, B2, C2
When I pass it through Beam as a PCollection, I map each into an object using a ParDo. Now I want to combine them into
A1: {
B1: [C1, C1'],
B2: C2
}
For key-value pairs I can use GroupByKey, but what if the structure is extremely nested where repeated fields can be at each level? Is there any transforms to combine repeated fields?

You could naively apply the GroupByKey transform consecutive times. First group by B, then group by A. However, this is highly inefficient because all key-value pairs must remain in memory during each GroupByKey step.
The Combine.PerKey transform is similar to GroupByKey, but it additionally allows you specify an accumulator class with custom logic for combining values. These values could be of any type, for example, repeated fields or extremely nested objects.
Reference:
1. Apache Beam Combine.PerKey

Related

If two persons A & B have two large boolean lists, how to find the first mismatch in the lists with least transfer of data between the two persons?

Let's say the list with person A = [T, T, F, T, F ...] and list with person B = [T, T, F, T, T ...], then we need to say that index 4 is the first mismatch position in the lists.
The number of entries in the lists can be very large (~50 million). How do we perform this search efficiently with the least amount of data (bytes) transferred between the two persons?
You can use Merkle tree structure and find mismatch in O(log n) transfers. For 256 bit hash function (for example SHA256), you can split array by parts of 256 elements. This parts will be leafs of Merkle tree.

Combination Operators on Sensors Arrays Data

Is there any standardized operators over data from arrays of sensors?
I am normally dealing with sensors data in the form of time + channels. The time is a timestamp, and the channels are the available data for these timestamps. All these fields are numeric, no strings involved.
Normally I have to mix those data objects in different ways. Let's suppose M1 is size m1xn1 and M2 is size m2xn2:
Combine rows of data from the same channels and different timestamps (i.e. n1 == n2). This leads to a vertical concatenation [M1; M2].
Combine columns of data from the same timestamps and different channels, (i.e. m1 == m2). This leads to a horizontal concatenation [M1 M2].
These operators are trivial and well defined.
When I have slight differences, for example, a few additional samples in M1 or M2, everything turns complicated and I have to think in weird schemes to perform such operations, such as these:
Cleaning the exceeding samples on M1 or M2, for matching the dimensions.
Calculate an aggregated timestamp, obtaining a unique(sort()) on the timestamps, and then apply a union like in a SQL JOIN sentence.
Aggregate the data on M1 or M2, this is, reducing m1 or m2 to a smaller figure, resampling the timescale, and then apply an aggregation like in a SQL GROUP sentence.
I cannot think of a unique and definite function to combine this sort of data. How can I do this?
Let's say you have an m1-element vector of time values t1 and and n1-element vector of channel values c1 for your m1-by-n1 matrix M1 (and likewise for M2). First and foremost, you will likely need to convert your time and channel values into equivalent index values. You can do this by expanding your time and channel values into grids using ndgrid, then converting them to index values using unique:
[t1, c1] = ndgrid(t1, c1);
[t2, c2] = ndgrid(t2, c2);
[tUnion, ~, tIndex] = unique([t1(:); t2(:)]);
[cUnion, ~, cIndex] = unique([c1(:); c2(:)]);
Now there are two approaches you can take for aggregating the data using the above indices. If you know for certain that the matrices M1 and M2 will never contain repeated measurements (i.e. the same combination of time and channel will not appear in both), then you can build the final joined matrix by creating a linear index from tIndex and cIndex and combining the values from M1 and M2 like so:
MUnion = zeros(numel(tUnion), numel(cUnion));
MUnion(tIndex+numel(tUnion).*(cIndex-1)) = [M1(:); M2(:)];
If the matrices M1 and M2 could contain repeated measurements at the same combination of time and channel values, then accumarray will be the way to go. You will have to decide how you want to combine the repeated measurements, such as taking the mean as shown here:
MUnion = accumarray([tIndex cIndex], [M1(:); M2(:)], [], #mean);

How to split a rdd into 2 based on column name

I have an RDD of two columns A and B
How can i create 2RDD's out if it?
I have a use case where i am taking an input RDD, performs some operations and produces two different output (intermediate (column A), final(column B)) which needs to be 2 ifferent locations. How can i split them?
Thanks

How does aggregate generalise fold and fold generalise reduce?

As far as I understand aggregate is a generalisation of fold which in turn is a generalisation of reduce.
Similarily combineByKey is a generalisation of aggregateByKey which in turn is a generalisation of foldByKey which in turn is a generalisation of reduceByKey.
However I have trouble finding simple examples for each of those seven methods which can in turn only be expressed by them and not their less general versions. For example I found http://blog.madhukaraphatak.com/spark-rdd-fold/ giving an example for fold, but I have been able to use reduce in the same situation as well.
What I found out so far:
I read that the more generalised methods can be more efficient, but that would be a non-functional requirement and I would like to get examples which can not be implemented with the more specific method.
I also read that e.g. the function passed to fold only has to be associative, while the one for reduce has to be commutative additionally: https://stackoverflow.com/a/25158790/4533188 (However, I still don't know any good simple example.) whereas in https://stackoverflow.com/a/26635928/4533188 I read that fold needs both properties to hold...
We could think of the zero value as a feature (e.g. for fold over reduce) as in "add all elements and add 3" and using 3 as the zero value, but that would be misleading, because 3 would be added for each partition, not just once. Also this is simply not the purpose of fold as far as I understood - it wasn't meant as a feature, but as a necessity to implement it to be able to take non-commutative functions.
What would simple examples for those seven methods be?
Let's work through what is actually needed logically.
First, note that if your collection is unordered, any set of (binary) operations on it need to be both commutative and associative, or you'll get different answers depending on which (arbitrary) order you choose each time. Since reduce, fold, and aggregate all use binary operations, if you use these things on a collection that is unordered (or is viewed as unordered), everything must be commutative and associative.
reduce is an implementation of the idea that if you can take two things and turn them into one thing, you can collapse an arbitrarily long collection into a single element. Associativity is exactly the property that it doesn't matter how you pair things up as long as you eventually pair them all and keep the left-to-right order unchanged, so that's exactly what you need.
a b c d a b c d a b c d
a # b c d a # b c d a b # c d
(a#b) c # d (a#b) # c d a (b#c) d
(a#b) # (c#d) ((a#b)#c) # d a # ((b#c)#d)
All of the above are the same as long as the operation (here called #) is associative. There is no reason to swap around which things go on the left and which go on the right, so the operation does not need to be commutative (addition is: a+b == b+a; concat is not: ab != ba).
reduce is mathematically simple and requires only an associative operation
Reduce is limited, though, in that it doesn't work on empty collections, and in that you can't change the type. If you're working sequentially, you can a function that takes a new type and the old type, and produces something with the new type. This is a sequential fold (left-fold if the new type goes on the left, right-fold if it goes on the right). There is no choice about the order of operations here, so commutativity and associativity and everything are irrelevant. There's exactly one way to work through your list sequentially. (If you want your left-fold and right-fold to always be the same, then the operation must be associative and commutative, but since left- and right-folds don't generally get accidentally swapped, this isn't very important to ensure.)
The problem comes when you want to work in parallel. You can't sequentially go through your collection; that's not parallel by definition! So you have to insert the new type at multiple places! Let's call our fold operation #, and we'll say that the new type goes on the left. Furthermore, we'll say that we always start with the same element, Z. Now we could do any of the following (and more):
a b c d a b c d a b c d
Z#a b c d Z#a b Z#c d Z#a Z#b Z#c Z#d
(Z#a) # b c d (Z#a) # b (Z#c) # d
((Z#a)#b) # c d
(((Z#a)#b)#c) # d
Now we have a collection of one or more things of the new type. (If the original collection was empty, we just take Z.) We know what to do with that! Reduce! So we make a reduce operation for our new type (let's call it $, and remember it has to be associative), and then we have aggregate:
a b c d a b c d a b c d
Z#a b c d Z#a b Z#c d Z#a Z#b Z#c Z#d
(Z#a) # b c d (Z#a) # b (Z#c) # d Z#a $ Z#b Z#c $ Z#d
((Z#a)#b) # c d ((Z#a)#b) $ ((Z#c)#d) ((Z#a)$(Z#b)) $ ((Z#c)$(Z#d))
(((Z#a)#b)#c) # d
Now, these things all look really different. How can we make sure that they end up to be the same? There is no single concept that describes this, but the Z# operation has to be zero-like and $ and # have to be homomorphic, in that we need (Z#a)#b == (Z#a)$(Z#b). That's the actual relationship that you need (and it is technically very similar to a semigroup homomorphism). There are all sorts of ways to pick badly even if everything is associative and commutative. For example, if Z is the double value 0.0 and # is actually +, then Z is zero-like and # is associative and commutative. But if $ is actually *, which is also associative and commutative, everything goes wrong:
(0.0+2) * (0.0+3) == 2.0 * 3.0 == 6.0
((0.0+2) + 3) == 2.0 + 3 == 5.0
One example of a non-trival aggregate is building a collection, where # is the "append an element" operator and $ is the "concat two collections" operation.
aggregate is tricky and requires an associative reduce operation, plus a zero-like value and a fold-like operation that is homomorphic to the reduce
The bottom line is that aggregate is not simply a generalization of reduce.
But there is a simplification (less general form) if you're not actually changing the type. If Z is actually z and is an actual zero, we can just stick it in wherever we want and use reduce. Again, we don't need commutativity conceptually; we just stick in one or more z's and reduce, and our # and $ operations can be the same thing, namely the original # we used on the reduce
a b c d () <- empty
z#a z#b z
z#a (z#b)#c
z#a ((z#b)#c)#d
(z#a)#((z#b)#c)#d
If we just delete the z's from here, it works perfectly well, and in fact is equivalent to if (empty) z else reduce. But there's another way it could work too. If the operation # is also commutative, and z is not actually a zero but just occupies a fixed point of # (meaning z#z == z but z#a is not necessarily just a), then you can run the same thing, and since commutivity lets you switch the order around, you conceptually can reorder all the z's together at the beginning, and then merge them all together.
And this is a parallel fold, which is really a rather different beast than a sequential fold.
(Note that neither fold nor aggregate are strictly generalizations of reduce even for unordered collections where operations have to be associative and commutative, as some operations do not have a sensible zero! For instance, reducing strings by shortest length has as its "zero" the longest possible string, which conceptually doesn't exist, and practically is an absurd waste of memory.)
fold requires an associative reduce operation plus either a zero value or a reduce operation that's commutative plus a fixed-point value
Now, when would you ever use a parallel fold that wasn't just a reduceOrElse(zero)? Probably never, actually, though they can exist. For example, if you have a ring, you often have fixed points of the type we need. For instance, 10 % 45 == (10*10) % 45, and * is both associative and commutative in integers mod 45. Thus, if our collection is numbers mod 45, we can fold with a "zero" of 10 and an operation of *, and parallelize however we please while still getting the same result. Pretty weird.
However, note that you can just plug the zero and operation of fold into aggregate and get exactly the same result, so aggregate is a proper generalization of fold.
So, bottom line:
Reduce requires only an associative merge operation, but doesn't change the type, and doesn't work on empty collecitons.
Parallel fold tries to extend reduce but requires a true zero, or a fixed point and the merge operation must be commutative.
Aggregate changes the type by (conceptually) running sequential folds followed by a (parallel) reduce, but there are complex relationships between the reduce operation and the fold operation--basically they have to be doing "the same thing".
An unordered collection (e.g. a set) always requires an associative and commutative operation for any of the above.
With regard to the byKey stuff: it's just the same as this, except it only applies it to the collection of values associated with a (potentially repeated) key.
If Spark actually requires commutativity where the above analysis does not suggest it's needed, one could reasonably consider that a bug (or at least an unnecessary limitation of the implementation, given that operations like map and filter preserve order on ordered RDDs).
the function passed to fold only has to be associative, while the one for reduce has to be commutative additionally.
It is not correct. fold on RDDs requires the function to be commutative as well. It is not the same operation as fold on Iterable what is pretty well described in the official documentation:
This behaves somewhat differently from fold operations implemented for non-distributed
collections in functional languages like Scala.
This fold operation may be applied to
partitions individually, and then fold those results into the final result, rather than
apply the fold to each element sequentially in some defined ordering. For functions
that are not commutative, the result may differ from that of a fold applied to a
non-distributed collection.
As you can see order of merging partial values is not part of the contract hence function which is used for fold has to be commutative.
I read that the more generalised methods can be more efficient
Technically speaking there should be no significant difference. For fold vs reduce you can check my answers to reduce() vs. fold() in Apache Spark and Why is the fold action necessary in Spark?
Regarding *byKey methods all are implemented using the same basic construct which is combineByKeyWithClassTag and can be reduced to three simple operations:
createCombiner - create "zero" value for a given partition
mergeValue - merge values into accumulator
mergeCombiners - merge accumulators created for each partition.

How to turn a known structured RDD to Vector

Assuming I have an RDD containing (Int, Int) tuples.
I wish to turn it into a Vector where first Int in tuple is the index and second is the value.
Any Idea how can I do that?
I update my question and add my solution to clarify:
My RDD is already reduced by key, and the number of keys is known.
I want a vector in order to update a single accumulator instead of multiple accumulators.
There for my final solution was:
reducedStream.foreachRDD(rdd => rdd.collect({case (x: Int,y: Int) => {
val v = Array(0,0,0,0)
v(x) = y
accumulator += new Vector(v)
}}))
Using Vector from accumulator example in documentation.
rdd.collectAsMap.foldLeft(Vector[Int]()){case (acc, (k,v)) => acc updated (k, v)}
Turn the RDD into a Map. Then iterate over that, building a Vector as we go.
You could use justt collect(), but if there are many repetitions of the tuples with the same key that might not fit in memory.
One key thing: do you really need Vector? Map could be much more suitable.
If you really need local Vector, you first need to use .collect() and then do local transformations into Vector. Of course you shall have enough memory for this. But here the real problem is where to find Vector which can be built efficiently from pairs of (index, value). As far as I know Spark MLLib has itself class org.apache.spark.mllib.linalg.Vectors which can create Vector from array of indices and values and even from tuples. Under the hood it uses breeze.linalg. So probably it would be best start for you.
If you just need order, you just can use .orderByKey() as you already have RDD[(K,V)]. This way you have ordered stream. Which does not strictly follow your intention but maybe it could suit even better. Now you can drop elements with the same key by .reduceByKey() producing only resulting elements.
Finally if you really need large vector, do .orderByKey and then you can produce real vector by doing .flatmap() which maintain counter and drops more than one element for the same index / inserts needed amount of 'default' elements for missing indices.
Hope this is clear enough.