在重新分区一个pyspark密集矩阵 [英] repartition a dense matrix in pyspark

问题描述

我在pyspark一个密集矩阵（100 * 100），我想再分配它分成十个组，每组包含10行。

I have a Dense matrix(100*100) in pyspark, and I want to repartition it into ten groups with each containing 10 rows.

from pyspark import SparkContext, SparkConf
from pyspark.mllib import *
sc = SparkContext("local", "Simple App")
dm2 = Matrices.dense(100, 100, RandomRDDs.uniformRDD(sc, 10000).collect())
newRdd = sc.parallelize(dm2.toArray())
rerdd = newRdd.repartition(10)

以上code结果 rerdd 包含100个元素。我想present这个矩阵 DM2 为逐行分区块（例如，10行的分区）。

the above code results in rerdd containing 100 elements. I want to present this matrix dm2 as row-wise partitioned blocks (e.g., 10 rows in a partition).

推荐答案

我没有多大意义，但你可以例如做这样的事情。

I doesn't make much sense but you can for example do something like this

mat =  Matrices.dense(100, 100, np.arange(10000))

n_par = 10
n_row = 100

rdd = (sc
    .parallelize(
        # Add indices
        enumerate(
            # Extract and reshape values
            mat.values.reshape(n_row, -1)))
    # Partition and sort by row index
    .repartitionAndSortWithinPartitions(n_par, lambda i: i // n_par))

每个分区的分区和行的支票号码：

Check number of partitions and rows per partition:

rdd.glom().map(len).collect()
## [10, 10, 10, 10, 10, 10, 10, 10, 10, 10

检查第一行包含所需数据：

Check if the first row contains desired data:

assert np.all(rdd.first()[1] == np.arange(100))

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