summing up a column.
Jeff; Thanks a lot. I do appreciate for this. I will try your codes. Regards, Oslo
On Monday, June 13, 2016 11:47 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
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. ? ? ? [[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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