[R] summing up a column.
Jeff Newmiller
jdnewmil at dcn.davis.ca.us
Tue Jun 14 05:48:04 CEST 2016
You did half of what Petr asked... but your email still looks unreadable
because you did not send it as text only. Look at [1] to see what we see,
and why we want you to send plain text. What you see is not what we see.
This is an outer join... an inherently inefficient operation according to
relational database theory. Most solutions to this type of problem are
likely to be slow, but minimizing unnecessary use of memory can help and
to do that you can overwrite existing values instead of successively
concatenating longer vectors of results as you go.
To understand the solution below, you should execute individual
expressions interactively as you step through the code to see what the
intermediate values look like. In particular, expressions on the right
side of assignments can be interactively entered at the console without
changing the variables in the environment so do it as much as you need to
in order to see what is happening.
# alignment not required, but HTML run-together lines not wanted
A <- structure( list( posA = c( 1L, 2L, 5L, 4L, 9L)
, posB = c( 9L, 7L, 12L, 7L, 13L)
)
, .Names = c("posA", "posB")
, class = "data.frame"
, row.names = c(NA, -5L)
)
B <- structure( list( pos = c( 4L, 2L, 7L, 1L, 13L, 12L, 9L)
, a = c(0.4, 0.1, 0.5, 0.4, 0.1, 0.2, 0.3)
, b = c(7L, 5L, 8L, 1L, 6L, 11L, 12L)
, c = c(0.8, 0.4, 0.32, 0.1, 0.13, 0.01, 0.23))
, .Names = c("pos", "a", "b", "c")
, class = "data.frame"
, row.names = c(NA, -7L)
)
# sort B
sB <- B[ order( B$pos ), ]
# performance: set aside memory to remember results
A$count07 <- NA
A$mina <- NA
for ( i in seq.int( nrow( A ) ) ) {
# logical indexing vector
idx <- A[ i, "posA" ] <= sB$pos & sB$pos <= A[ i, "posB" ]
# only extract desired vector once
a <- sB[ idx, "a" ]
# sum adds logical values as if TRUE=1
A[ i, "count07" ] <- sum( cumsum( a ) < 0.7 )
A[ i, "mina" ] <-min( a )
}
print( A )
#=== sample interactive session for study after A and B are defined
#=== execute lines one at a time and study them!
order( B$pos )
B[ order( B$pos ), ]
sB <- B[ order( B$pos ), ]
A$count07 <- NA
A$mina <- NA
i <- 1
A[ i, "posA" ]
A[ i, "posB" ]
sB$pos
A[ i, "posA" ] <= sB$pos
sB$pos <= A[ i, "posB" ]
A[ i, "posA" ] <= sB$pos & sB$pos <= A[ i, "posB" ]
idx <- A[ i, "posA" ] <= sB$pos & sB$pos <= A[ i, "posB" ]
sB[ idx, "a" ]
a <- sB[ idx, "a" ]
cumsum( a )
cumsum( a ) < 0.7
sum( cumsum( a ) < 0.7 )
A[ i, "count07" ] <- sum( cumsum( a ) < 0.7 )
min( a )
------
[1] https://stat.ethz.ch/pipermail/r-help/2016-June/439404.html
On Mon, 13 Jun 2016, oslo via R-help wrote:
> Hi Petr;
> Thanks so much. Here are the questions and the dput(A) and dput(B). Basicaly I have two questions;
>> dput(A)structure(list(posA = c(1L, 2L, 5L, 4L, 9L), posB = c(9L, 7L, 12L, 7L, 13L)), .Names = c("posA", "posB"), class = "data.frame", row.names = c(NA, -5L))> dput(B)structure(list(pos = c(4L, 2L, 7L, 1L, 13L, 12L, 9L), a = c(0.4, 0.1, 0.5, 0.4, 0.1, 0.2, 0.3), b = c(7L, 5L, 8L, 1L, 6L, 11L, 12L), c = c(0.8, 0.4, 0.32, 0.1, 0.13, 0.01, 0.23)), .Names = c("pos", "a", "b", "c"), class = "data.frame", row.names = c(NA, -7L))
> Q1) Values in A represent the region of chromosome. I need choose these regions in B (all region in A are exist in B in a single column) and then summing up the column "a in B and count the numbers that gives >0.7. For example, consider the first row in A. They are 1 and 9. After sorting the first column in B then summing column "a" only between 1 to 9 in sorted B and cut off at >0.7. Then count how many rows in sorted B gives >0.7. For example there are only 5 rows between 1 to 9 in sorted B and only summing first 2 of them gives>0.7 . Then my answer is going to be 2
> Q2) What is the min value of B$a for given each intervals in A
> Regards,
> Oslo
>
> On Monday, June 13, 2016 8:05 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote:
>
>
> Ok.
>
> Instead of explaining what you have, please send a result of
>
> dput(B) and dput(A)
>
> And set you mail client to send plain text mail otherwise your code is barely readable.
>
> What do you want to do with printed values?
>
> What is B? From this it seems that it is data frame but then you try to put sorted data frame into a data frame column.
>
> B$possort=B[order(B$pos),]
>
> With such code and data frame I get an error.
>
> So please try to keep above mentioned when posting a query.
>
> Regards
> Petr
>
>
>> -----Original Message-----
>> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of oslo via R-
>> help
>> Sent: Friday, June 10, 2016 11:09 PM
>> To: oslo <oslo at yahoo.com>; Jeff Newmiller <jdnewmil at dcn.davis.ca.us>;
>> oslo via R-help <r-help at r-project.org>; oslo <hokut1 at yahoo.com>
>> Subject: Re: [R] summing up a column.
>>
>> Jeff;
>> thanks for this. My question was job related. No from my course. I need
>> finish a job for the place I work. I am so sorry for causing misunderstanding.
>> thanks,
>> Oslo
>>
>> On Friday, June 10, 2016 5:08 PM, oslo via R-help <r-help at r-project.org>
>> wrote:
>>
>>
>> Jeff thanks for this. My question was job related. No from my course. I need
>> finish a job for the place I work. I am so sorry for causing misunderstanding.
>> thanks,
>> Oslo
>>
>> On Friday, June 10, 2016 5:02 PM, Jeff Newmiller
>> <jdnewmil at dcn.davis.ca.us> wrote:
>>
>>
>> Multiple posting happens when you are learning a new system, but reading
>> the posting guide can keep the bleeding down.
>>
>> 1) There is a no-homework policy on this list... different educational
>> organizations have different standards for what is acceptable outside help,
>> so you should be using the support offered by your instructor or educational
>> institution.
>>
>> 2) Once you have completed your course, you CAN learn to post data with
>> your code so that it is self-contained... that is, reproducible on our vanilla R
>> session. Using the dput function is one excellent strategy.
>>
>> 3) This is not a problem that needs a loop... as Bert (not Bret) said, you can do
>> this in one or two statements if you simply use basic logical indexing. If your
>> instructor wants you to do it with a loop for sine reason then you really really
>> should not be here... you should be talking to him/her.
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On June 10, 2016 1:34:03 PM PDT, oslo via R-help <r-help at r-project.org>
>> wrote:
>> Dear All;
>> I had difficulty to post a mail along with appropriate of data structure. I do
>> sincerely apologize for multiple posting
>>
>>
>> I would like to sum up the B$a column and cut off at 0.7 for the each row of
>> intervals giving in file=A.For example the interval at the first row in A$posA
>> and A$posB is 1 and 9. So, I need adding up the B$a and cut off B$a>.7 from
>> the 1 to 9 in B$pos. And then I need to the same using the intervals in the
>> second, third..... rows in A. Obviously my loop is wrong and does not work
>> properly. Please help for my this first experience. Regards Here are my
>> codes #sorting B$possort=B[order(B$pos),] #Running loop for(i in 1:nrow(A))
>> {if(sum(B[a$B, i:A[1:2])>0.7) {print(A[1:i,]) } } Reply, R R-help at r-project.org
>> mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-
>> guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-
>> guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
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>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-
>> guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
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