Map IDs to matrix rows in Hadoop/MapReduce - hadoop

I have data about users buying products. I want to create a binary matrix of size |users| x |products| such that the element (i,j) in the matrix is 1 iff user_i has bought product_j, else the value is 0.
Now, my data looks something like
userA, productX
userB, productY
userA, productZ
...
UserIds and productIds are all strings. My problem is, how to map these IDs to row indices (for users) and column indices (for products) in the matrix.
There are over a million unique userIds and roughly 3 million productIds.
To make the problem well defined: given the user1, product1 like input above, how do I convert it to something like
1,1
2,2
1,3
where userA is mapped to row 0 of the matrix, userB is mapped to row 1, productX is mapped to column 0 and so on.
Given the size of data, I would have to use Hadoop Map-Reduce but can't think of a foolproof way of efficiently doing this.
This can be solved if we can do the following:
Dump unique userIds.
Dump unique productIds.
Map each unique userId in (1) to a row index.
Map each unique productId in (2) to a column index.
I can do (1) and (2) easily but having trouble coming up with an efficient approach to solve (3) (4 will be solved if we solve 3).
I have a couple of solutions but they are not foolproof.
Solution 1 (naive) for step 3 above
Map all userIds and emit the same key (say "1") for all map tasks.
Have a long counter initialized to 0 in setup() of the reducer.
In the reduce(), emit the counter value along with the input userId and increment the counter by 1.
This would be very inefficient since all 100 million userIds would be processed by a single reducer.
Solution 2 for step 3 above
While mapping userIds, emit each userId against a key which is an integer uniformly sampled from 1,2,3....N (where N is configurable. N = 100 for example). In a way, we are partitioning the input set.
Within the mapper, use Hadoop counters to count the number of userIds assigned to each of those random partitions.
In the reducer setup, first access the counters in the mapping stage to determine how many IDs were assigned to each partition. Use these counters to determine the start and end values for that partition.
Iterate (while counting) over each userId in reduce and generate matrix rowId as start_of_partition + counter.
context.write(userId, matrix row Id)
This method should work but I am not sure how to handle cases when reducer tasks failed/killed.
I believe there should be ways of doing this which I am not aware of. Can we use hashing/modulo to achieve this? How would we handle collisions at scale?

Related

How do I add specific values of columns to create new columns?

I have a dataset which I want to format in order to perform repeated measures anova. My dataset is of the form:
set.seed(32)
library(tibble)
id<- rep(1:2,each=3)
y_0 <- rep(rnorm(2,mean=50,sd=10),each=3)
time <- rep(c(1,2,3),times=2)
c<-rep(rnorm(2,mean=10,sd=12),each=3)
data <- tibble(id,y,t,c)
I want to bring the dataset in the form of a dataset for repeated measures anova meaning I want to have only one value for id in each column and create 3 more columns. One for y+c in time 1 named y_1,y+c in time 2 named y_2 and y+c in time 3 named y_3. Can anyone provide some assistance?

How to slice the dataset in Python in specific intervals

I have a dataset with n rows, how can I access a specific number of rows every specific number of rows through the whole dataset using Python?
For example, in 100 rows data set and I want to access 10 rows every 10 rows, like 1:10, 20:30, 40:50, 60:70, 80:90
I could think of something like this
df.iloc[np.array([int(x/10) for x in df.index]) % 2 == 0]
It takes the index of the dataframe, divides it by 10 and casts it to an int. This basically just removes the last digit in this example.
With the modulo statement the first 10 rows are True, the next 10 False and so on. This is then used with iloc to get just the lines with the True value.
This requires a continuously increasing index. If for example some rows were already filtered out this is not the case. reset_index can be used to reset the index.

How to understand part and partition of ClickHouse?

I see that clickhouse created multiple directories for each partition key.
Documentation says the directory name format is: partition name, minimum number of data block, maximum number of data block and chunk level. For example, the directory name is 201901_1_11_1.
I think it means that the directory is a part which belongs to partition 201901, has the blocks from 1 to 11 and is on level 1. So we can have another part whose directory is like 201901_12_21_1, which means this part belongs to partition 201901, has the blocks from 12 to 21 and is on level 1.
So I think partition is split into different parts.
Am I right?
Parts -- pieces of a table which stores rows. One part = one folder with columns.
Partitions are virtual entities. They don't have physical representation. But you can say that these parts belong to the same partition.
Select does not care about partitions.
Select is not aware about partitioning keys.
BECAUSE each part has special files minmax_{PARTITIONING_KEY_COLUMN}.idx
These files contain min and max values of these columns in this part.
Also this minmax_ values are stored in memory in a (c++ vector) list of parts.
create table X (A Int64, B Date, K Int64,C String)
Engine=MergeTree partition by (A, toYYYYMM(B)) order by K;
insert into X values (1, today(), 1, '1');
cd /var/lib/clickhouse/data/default/X/1-202002_1_1_0/
ls -1 *.idx
minmax_A.idx <-----
minmax_B.idx <-----
primary.idx
SET send_logs_level = 'debug';
select * from X where A = 555;
(SelectExecutor): MinMax index condition: (column 0 in [555, 555])
(SelectExecutor): Selected 0 parts by date
SelectExecutor checked in-memory part list and found 0 parts because minmax_A.idx = (1,1) and this select needed (555, 555).
CH does not store partitioning key values.
So for example toYYYYMM(today()) = 202002 but this 202002 is not stored in a part or anywhere.
minmax_B.idx stores (18302, 18302) (2020-02-10 == select toInt16(today()))
In my case, I had used groupArray() and arrayEnumerate() for ranking in Populate. I thought that Populate can run query with new data on the partition (in my case: toStartOfDay(Date)), the total sum of new inserted data is correct but the groupArray() function is doesn't work correctly.
I think it's happened because when insert one Part, CH will groupArray() and rank on each Part immediately then merging Parts in one Partition, therefore i wont get exactly the final result of groupArray() and arrayEnumerate() function.
Summary, Merge
[groupArray(part_1) + groupArray(part_2)] is different from
groupArray(Partition)
with
Partition=part_1 + part_2
The solution that i tried is insert new data as one block size, just like using groupArray() to reduce the new data to the number of rows that is lower than max_insert_block_size=1048576. It did correctly but it's hard to insert new data of 1 day as one Part because it will use too much memory for querying when populating the data of 1 day (almost 150Mn-200Mn rows).
But do u have another solution for Populate with groupArray() for new inserting data, such as force CH to use POPULATE on each Partition, not each Part after merging all the part into one Partition?

olap4J - calculations on member grouping

I'm trying to write an olap4j (Mondrian) query that will group the rows by ranges.
Assume we have counts of cards per child and the children ages.
i want to sum the cards amount by age ranges, so i will have counts for ages 0-5,5-10,10-15 and so on.
Is this can be done with olap4j?
You need to define calculated members for that:
With member [Age].[0-4] as [Age].[0]:[Age].[4]
member [Age].[5-9] as [Age].[5]:[Age].[9]
etc.
Alternatively, you may want to re-design your dimension table. I'm guessing you have age as a degenerate dimension in the fact table. I suggest creating a separate dimension dim_age with a structure like this:
age_id, age, age_group
0, null, null
1, 0, 0-4
2, 1, 0-4
(...)
Then it's easy to define a first level on the dimension based on the age_group.

Hive cluster by vs order by vs sort by

As far as I understand;
sort by only sorts with in the reducer
order by orders things globally but shoves everything into one reducers
cluster by intelligently distributes stuff into reducers by the key hash and make a sort by
So my question is does cluster by guarantee a global order? distribute by puts the same keys into same reducers but what about the adjacent keys?
The only document I can find on this is here and from the example it seems like it orders them globally. But from the definition I feel like it doesn't always do that.
A shorter answer: yes, CLUSTER BY guarantees global ordering, provided you're willing to join the multiple output files yourself.
The longer version:
ORDER BY x: guarantees global ordering, but does this by pushing all data through just one reducer. This is basically unacceptable for large datasets. You end up one sorted file as output.
SORT BY x: orders data at each of N reducers, but each reducer can receive overlapping ranges of data. You end up with N or more sorted files with overlapping ranges.
DISTRIBUTE BY x: ensures each of N reducers gets non-overlapping ranges of x, but doesn't sort the output of each reducer. You end up with N or more unsorted files with non-overlapping ranges.
CLUSTER BY x: ensures each of N reducers gets non-overlapping ranges, then sorts by those ranges at the reducers. This gives you global ordering, and is the same as doing (DISTRIBUTE BY x and SORT BY x). You end up with N or more sorted files with non-overlapping ranges.
Make sense? So CLUSTER BY is basically the more scalable version of ORDER BY.
Let me clarify first: clustered by only distributes your keys into different buckets, clustered by ... sorted by get buckets sorted.
With a simple experiment (see below) you can see that you will not get global order by default. The reason is that default partitioner splits keys using hash codes regardless of actual key ordering.
However you can get your data totally ordered.
Motivation is "Hadoop: The Definitive Guide" by Tom White (3rd edition, Chapter 8, p. 274, Total Sort), where he discusses TotalOrderPartitioner.
I will answer your TotalOrdering question first, and then describe several sort-related Hive experiments that I did.
Keep in mind: what I'm describing here is a 'proof of concept', I was able to handle a single example using Claudera's CDH3 distribution.
Originally I hoped that org.apache.hadoop.mapred.lib.TotalOrderPartitioner will do the trick. Unfortunately it did not because it looks like Hive partitions by value, not key. So I patch it (should have subclass, but I do not have time for that):
Replace
public int getPartition(K key, V value, int numPartitions) {
return partitions.findPartition(key);
}
with
public int getPartition(K key, V value, int numPartitions) {
return partitions.findPartition(value);
}
Now you can set (patched) TotalOrderPartitioner as your Hive partitioner:
hive> set hive.mapred.partitioner=org.apache.hadoop.mapred.lib.TotalOrderPartitioner;
hive> set total.order.partitioner.natural.order=false
hive> set total.order.partitioner.path=/user/yevgen/out_data2
I also used
hive> set hive.enforce.bucketing = true;
hive> set mapred.reduce.tasks=4;
in my tests.
File out_data2 tells TotalOrderPartitioner how to bucket values.
You generate out_data2 by sampling your data. In my tests I used 4 buckets and keys from 0 to 10. I generated out_data2 using ad-hoc approach:
import org.apache.hadoop.util.ToolRunner;
import org.apache.hadoop.util.Tool;
import org.apache.hadoop.conf.Configured;
import org.apache.hadoop.fs.Path;
import org.apache.hadoop.io.NullWritable;
import org.apache.hadoop.io.SequenceFile;
import org.apache.hadoop.hive.ql.io.HiveKey;
import org.apache.hadoop.fs.FileSystem;
public class TotalPartitioner extends Configured implements Tool{
public static void main(String[] args) throws Exception{
ToolRunner.run(new TotalPartitioner(), args);
}
#Override
public int run(String[] args) throws Exception {
Path partFile = new Path("/home/yevgen/out_data2");
FileSystem fs = FileSystem.getLocal(getConf());
HiveKey key = new HiveKey();
NullWritable value = NullWritable.get();
SequenceFile.Writer writer = SequenceFile.createWriter(fs, getConf(), partFile, HiveKey.class, NullWritable.class);
key.set( new byte[]{1,3}, 0, 2);//partition at 3; 1 came from Hive -- do not know why
writer.append(key, value);
key.set( new byte[]{1, 6}, 0, 2);//partition at 6
writer.append(key, value);
key.set( new byte[]{1, 9}, 0, 2);//partition at 9
writer.append(key, value);
writer.close();
return 0;
}
}
Then I copied resulting out_data2 to HDFS (into /user/yevgen/out_data2)
With these settings I got my data bucketed/sorted (see last item in my experiment list).
Here is my experiments.
Create sample data
bash> echo -e "1\n3\n2\n4\n5\n7\n6\n8\n9\n0" > data.txt
Create basic test table:
hive> create table test(x int);
hive> load data local inpath 'data.txt' into table test;
Basically this table contains values from 0 to 9 without order.
Demonstrate how table copying works (really mapred.reduce.tasks parameter which sets MAXIMAL number of reduce tasks to use)
hive> create table test2(x int);
hive> set mapred.reduce.tasks=4;
hive> insert overwrite table test2
select a.x from test a
join test b
on a.x=b.x; -- stupied join to force non-trivial map-reduce
bash> hadoop fs -cat /user/hive/warehouse/test2/000001_0
1
5
9
Demonstrate bucketing. You can see that keys are assinged at random without any sort order:
hive> create table test3(x int)
clustered by (x) into 4 buckets;
hive> set hive.enforce.bucketing = true;
hive> insert overwrite table test3
select * from test;
bash> hadoop fs -cat /user/hive/warehouse/test3/000000_0
4
8
0
Bucketing with sorting. Results are partially sorted, not totally sorted
hive> create table test4(x int)
clustered by (x) sorted by (x desc)
into 4 buckets;
hive> insert overwrite table test4
select * from test;
bash> hadoop fs -cat /user/hive/warehouse/test4/000001_0
1
5
9
You can see that values are sorted in ascending order. Looks like Hive bug in CDH3?
Getting partially sorted without cluster by statement:
hive> create table test5 as
select x
from test
distribute by x
sort by x desc;
bash> hadoop fs -cat /user/hive/warehouse/test5/000001_0
9
5
1
Use my patched TotalOrderParitioner:
hive> set hive.mapred.partitioner=org.apache.hadoop.mapred.lib.TotalOrderPartitioner;
hive> set total.order.partitioner.natural.order=false
hive> set total.order.partitioner.path=/user/training/out_data2
hive> create table test6(x int)
clustered by (x) sorted by (x) into 4 buckets;
hive> insert overwrite table test6
select * from test;
bash> hadoop fs -cat /user/hive/warehouse/test6/000000_0
1
2
0
bash> hadoop fs -cat /user/hive/warehouse/test6/000001_0
3
4
5
bash> hadoop fs -cat /user/hive/warehouse/test6/000002_0
7
6
8
bash> hadoop fs -cat /user/hive/warehouse/test6/000003_0
9
CLUSTER BY does not produce global ordering.
The accepted answer (by Lars Yencken) misleads by stating that the reducers will receive non-overlapping ranges. As Anton Zaviriukhin correctly points to the BucketedTables documentation, CLUSTER BY is basically DISTRIBUTE BY (same as bucketing) plus SORT BY within each bucket/reducer. And DISTRIBUTE BY simply hashes and mods into buckets and while the hashing function may preserve order (hash of i > hash of j if i > j), mod of hash value does not.
Here's a better example showing overlapping ranges
http://myitlearnings.com/bucketing-in-hive/
As I understand, short answer is No.
You'll get overlapping ranges.
From SortBy documentation:
"Cluster By is a short-cut for both Distribute By and Sort By."
"All rows with the same Distribute By columns will go to the same reducer."
But there is no information that Distribute by guarantee non-overlapping ranges.
Moreover, from DDL BucketedTables documentation:
"How does Hive distribute the rows across the buckets? In general, the bucket number is determined by the expression hash_function(bucketing_column) mod num_buckets."
I suppose that Cluster by in Select statement use the same principle to distribute rows between reducers because it's main use is for populating bucketed tables with the data.
I created a table with 1 integer column "a", and inserted numbers from 0 to 9 there.
Then I set number of reducers to 2
set mapred.reduce.tasks = 2;.
And select data from this table with Cluster by clause
select * from my_tab cluster by a;
And received result that I expected:
0
2
4
6
8
1
3
5
7
9
So, first reducer (number 0) got even numbers (because their mode 2 gives 0)
and second reducer (number 1) got odd numbers (because their mode 2 gives 1)
So that's how "Distribute By" works.
And then "Sort By" sorts the results inside each reducer.
Use case : When there is a large dataset then one should go for sort by as in sort by , all the set reducers sort the data internally before clubbing together and that enhances the performance. While in Order by, the performance for the larger dataset reduces as all the data is passed through a single reducer which increases the load and hence takes longer time to execute the query.
Please see below example on 11 node cluster.
This one is Order By example output
This one is Sort By example output
This one is Cluster By example
What I observed , the figures of sort by , cluster by and distribute by is SAME But internal mechanism is different. In DISTRIBUTE BY : The same column rows will go to one reducer , eg. DISTRIBUTE BY(City) - Bangalore data in one column , Delhi data in one reducer:
Cluster by is per reducer sorting not global. In many books also it is mentioned incorrectly or confusingly. It has got particular use where say you distribute each department to specific reducer and then sort by employee name in each department and do not care abt order of dept no the cluster by to be used and it more perform-ant as workload is distributed among reducers.
SortBy: N or more sorted files with overlapping ranges.
OrderBy: Single output i.e fully ordered.
Distribute By: Distribute By protecting each of N reducers gets non-overlapping ranges of the column but doesn’t sort the output of each reducer.
For more information http://commandstech.com/hive-sortby-vs-orderby-vs-distributeby-vs-clusterby/
ClusterBy: Refer to the same example as above, if we use Cluster By x, the two reducers will further sort rows on x:
If I understood it correctly
1.sort by - only sorts the data within the reducer
2.order by - orders things globally by pushing the entire data set to a single reducer. If we do have a lot of data(skewed), this process will take a lot of time.
cluster by - intelligently distributes stuff into reducers by the key hash and make a sort by, but does not grantee global ordering. One key(k1) can be placed into two reducers. 1st reducer gets 10K K1 data, the second one might get 1K k1 data.

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