[R] Why is merge sorting even when sort = F?

[Jeff Newmiller]

If you are still wondering, try re-reading my answer. FALSE is more efficient, TRUE is sorted. Lack of sorting has nothing to do with preserving order.
-- 
Sent from my phone. Please excuse my brevity.

On March 8, 2017 8:55:06 AM PST, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote:
>Thank you. I was just curious what sort=FALSE had no impact.
>Wondering what it is there for then...
>
>On Wed, Mar 8, 2017 at 11:43 AM, Jeff Newmiller
><jdnewmil at dcn.davis.ca.us> wrote:
>> Merging is not necessarily an order-preserving operation, but sorting
>can make the operation more efficient. The sort=TRUE argument forces
>the result to be sorted, but sort=FALSE is in not a promise that order
>will be preserved. (I think the imperfect sorting occurs when there are
>multiple keys but am not sure.) You can add columns to the input data
>that let you restore some semblance of the original ordering afterward,
>or you can roll your own possibly-less-efficient merge using match and
>indexing:
>>
>> info[ match( grades2$grade, info$grade ), ]
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On March 8, 2017 8:07:27 AM PST, Dimitri Liakhovitski
><dimitri.liakhovitski at gmail.com> wrote:
>>>Hello!
>>>I have a vector 'grades' and a data frame 'info':
>>>
>>>grades2 <- data.frame(grade = c(1,2,2,3,1))
>>>info <- data.frame(
>>>  grade = 3:1,
>>>  desc = c("Excellent", "Good", "Poor"),
>>>  fail = c(F, F, T)
>>>)
>>>
>>>I want to get the info for all grades I have in info:
>>>
>>>This solution resorts everything in the order of column 'grade':
>>>merge(grades2, info, by = "grade", all.x = T, all.y = F)
>>>
>>>Could you please explain why this solution also resorts - despite
>sort
>>>= FALSE?
>>>merge(grades2, info, by = "grade", all.x = T, all.y = F, sort =
>FALSE)
>>>
>>>Thanks a lot!