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[SPARK-22614] Dataset API: repartitionByRange(...) #19828
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08527b0
repartitionByRange() + repartition() with SortOrders
adrian-ionescu 950b3dc
avoid changing semantics for .repartition()
adrian-ionescu 671e9e4
review feedback: val partition = ... + unit test
adrian-ionescu fe690fc
move test to DataFrameSuite + minor ws
adrian-ionescu 66b192d
handle empty partition-by expr list as RoundRobin
adrian-ionescu 012d617
update docs
adrian-ionescu 60ec0e3
fix typo: hash -> range
adrian-ionescu e0bda1d
Revert "fix typo: hash -> range"
adrian-ionescu 2f0f8f4
Revert "update docs"
adrian-ionescu f6cd388
Revert "handle empty partition-by expr list as RoundRobin"
adrian-ionescu 67ca139
avoid changing repartition() and make repartitionByRange() throw on e…
adrian-ionescu 012baa0
fix triple quotes
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Revert "handle empty partition-by expr list as RoundRobin"
This reverts commit 66b192d
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commit f6cd38873e06380faf4dfb771b3839c68278996c
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Still the same question. What happened when the SortOrder is not at the root node.
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It's going to follow the
HashPartitioningpath and eventually lead to a "Cannot evaluate expression" exception, just like it would presently do if you tried runningdf.repartition($"col".asc + 1)ordf.sort($"col".asc + 1)