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[SPARK-23291][SQL][R] R's substr should not reduce starting position by 1 when calling Scala API #20464

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[SPARK-23291][SQL][R] R's substr should not reduce starting position by 1 when calling Scala API #20464

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a2ffdc1
R's substr should not reduce starting position by 1 when calling Scal…
viirya Feb 1, 2018
95c8a4e
Add a note to migration guide of R doc.
viirya Feb 1, 2018
d994d76
Fix doc.
viirya Feb 2, 2018
0ebdf74
Improve doc.
viirya Mar 6, 2018
8c1a8ec
Improve doc clarity.
viirya Mar 7, 2018
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10 changes: 8 additions & 2 deletions R/pkg/R/column.R
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Expand Up @@ -164,12 +164,18 @@ setMethod("alias",
#' @aliases substr,Column-method
#'
#' @param x a Column.
#' @param start starting position.
#' @param start starting position. It should be 1-base.
#' @param stop ending position.
#' @examples
#' \dontrun{
#' df <- createDataFrame(list(list(a="abcdef")))
#' collect(select(df, substr(df$a, 1, 4))) # the result is `abcd`.
#' collect(select(df, substr(df$a, 2, 4))) # the result is `bcd`.
#' }
#' @note substr since 1.4.0
setMethod("substr", signature(x = "Column"),
function(x, start, stop) {
jc <- callJMethod(x@jc, "substr", as.integer(start - 1), as.integer(stop - start + 1))
jc <- callJMethod(x@jc, "substr", as.integer(start), as.integer(stop - start + 1))
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@felixcheung felixcheung Feb 1, 2018

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I'm a bit concern with changing this. As you can see it's been like this from the very beginning...

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@viirya viirya Feb 1, 2018

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This API behavior should be considered as wrong and performs inconsistently. Because for starting position 1, we get substring from 1st element, but for position 2, we still get the substring from 1. So we will get the following inconsistent results:

> collect(select(df, substr(df$a, 1, 5)))
  substring(a, 0, 5)
1              abcde
> collect(select(df, substr(df$a, 2, 5)))
  substring(a, 1, 4)
1               abcd

For such change, we might need to add a note in the doc as @HyukjinKwon suggested.

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@felixcheung felixcheung Feb 2, 2018

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question:

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@viirya viirya Feb 2, 2018

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Just because I think it is another issue regarding 0/negative indices. I can deal it here if you strongly feel it is better.

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@viirya viirya Feb 2, 2018

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is there a way to make the behavior the same before this change for any caller calling substr with common index like 0

Should we keep the behavior when calling substr with 0 as start index?

> df <- createDataFrame(list(list(a="abcdef")))
> collect(select(df, substr(df$a, 0, 5)))
  substring(a, -1, 6)
1                   f
> substr("abcdef", 0, 5)
[1] "abcde"

I think the previous behavior is pretty unreasonable..

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@felixcheung felixcheung Mar 4, 2018

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I think we should do two things:

  1. add to the func doc that the start param should be 0-base and to add to the example with the result
    collect(select(df, substr(df$a, 0, 5))) # this should give you...

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@viirya viirya Mar 6, 2018 •
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I think you mean 1-base.

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@viirya viirya Mar 6, 2018

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Added to the func doc.

column(jc)
})

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1 change: 1 addition & 0 deletions R/pkg/tests/fulltests/test_sparkSQL.R
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Expand Up @@ -1649,6 +1649,7 @@ test_that("string operators", {
expect_false(first(select(df, startsWith(df$name, "m")))[[1]])
expect_true(first(select(df, endsWith(df$name, "el")))[[1]])
expect_equal(first(select(df, substr(df$name, 1, 2)))[[1]], "Mi")
expect_equal(first(select(df, substr(df$name, 4, 6)))[[1]], "hae")
if (as.numeric(R.version$major) >= 3 && as.numeric(R.version$minor) >= 3) {
expect_true(startsWith("Hello World", "Hello"))
expect_false(endsWith("Hello World", "a"))
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4 changes: 4 additions & 0 deletions docs/sparkr.md
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Expand Up @@ -663,3 +663,7 @@ You can inspect the search path in R with [`search()`](https://stat.ethz.ch/R-ma
- The `stringsAsFactors` parameter was previously ignored with `collect`, for example, in `collect(createDataFrame(iris), stringsAsFactors = TRUE))`. It has been corrected.
- For `summary`, option for statistics to compute has been added. Its output is changed from that from `describe`.
- A warning can be raised if versions of SparkR package and the Spark JVM do not match.

## Upgrading to Spark 2.4.0

- The `start` parameter of `substr` method was wrongly subtracted by one, previously. In other words, the index specified by `start` parameter was considered as 0-base. This can lead to inconsistent substring results and also does not match with the behaviour with `substr` in R. It has been fixed so the `start` parameter of `substr` method is now 1-base, e.g., therefore to get the same result as `substr(df$a, 2, 5)`, it should be changed to `substr(df$a, 1, 4)`.