1. This behavior is dictated by the file system (an operating system feature)
that is in use. Chances of it changing in R are extremely small.
2. While not clearly documented, this behavior is consistent with the
definition of what a "csv" file is. Headers located at other than line 1 are
not v
I
am recently updated to R 3.5.0 and noticed some weird errors in write()
function. Further, I noticed that write.csv, write.table and generally the
functions that derive from write() are all weird.
1. write() function does not accept a path longer than 256 characters
neither on Windows or U
Hello,
Also possible is
is.na (B) <- B >= A
Hope this helps,
Rui Barradas
Enviado a partir do meu smartphone Samsung Galaxy. Mensagem original
De: Bert Gunter Data: 23/06/2018 15:26
(GMT+00:00) Para: javad bayat Cc: R-help
Assunto: Re: [R] Deleting a specific value in a
Bert's solution works if A and B are vectors, but not if they are columns in a
data frame. First, let's make up some data:
set.seed(42)
Dat <- data.frame(A=runif(10), B=runif(1))
Dat
#A B
# 1 0.9148060 0.4577418
# 2 0.9370754 0.7191123
# 3 0.2861395 0.9346722
# 4 0.8304476
Dear List,
I am happy to report that the problem is fixed. as.Date("1998-02-10") as
suggested by David handled the problem with easy. Many thanks to everybody.
as.Date(1998-02-10) really resulted in error. It is my oversight. I really
tried many things the day I was working on that and have forgott
You understand, of course, that all columns in a data frame must be of the
same length; and that "NA" is not the same as NA?
This is pretty basic stuff and suggests you need to spend some time with an
R tutorial or two.
In any case, a construct of the form:
B[B >= A] <- NA
should do.
Cheers,
B
Dear R users;
I have two columns data frame (column names is A and B). I want to write a
function in order to compare column of B with A and if the values of B are
equal or greater than that of A, replace them with "NA" or delete them and
if the values of B are less than values in A, do nothing.
S
On 6/22/2018 4:43 AM, Bill Poling wrote:
Good morning.
I have data in the form:
head(Edit041IA, n=25)
ClaimServiceID ClaimID DiagnosisCode
1 183056004 78044473 C562
2 183056004 78044473 C778
3 183056004 78044473 C784
4 183056004 78044473
Hi Ogbos,
This may help:
# assume your data frame is named "oodf"
oomean<-as.vector(by(oodf$B,oodf$A,mean))
oose<-as.vector(by(oodf$B,oodf$A,std.error))
plot(-5:10,oomean,type="b",ylim=c(5,11),
xlab="days (epoch is the day of Fd)",ylab="strikes/km2/day")
dispersion(-5:10,oomean,oose)
Jim
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