Another question,
How do I extract ID based on the third and fourth letter:
I have for example, AA14004, AB15035, CB14024, PA14009, PA14009 etc
I would like to extract ID no. of AB14..., CB14..., PA14...
On Mon, Mar 13, 2017 at 12:37 PM, roslinazairimah zakaria <
roslina...@gmail.com> wrote:
>
Hi Bert,
Thank you so much for your help. However I don't really sure what is the
use of y values. Can we do without it?
x <- as.character(FKASA$STUDENT_ID)
y <- c(1,786)
My.Data <- data.frame (x,y)
My.Data[grep("^AA14", My.Data$x), ]
I got the following data:
x y
1 AA14068 1
1. Your code is incorrect. All entries are character strings and must be quoted.
2. See ?grep and note in particular (in the "Value" section):
"grep(value = TRUE) returns a character vector containing the selected
elements of x (after coercion, preserving names but no other
attributes)."
3. Wh
Dear r-users,
I have this list of student ID,
dt <- c(AA14068, AA13194, AE11054, AA12251, AA13228, AA13286, AA14090,
AA13256, AA13260, AA13291, AA14099, AA15071, AA13143, AA14012, AA14039,
AA15018, AA13234, AA13149, AA13282, AA13218)
and I would like to extract all student of ID AA14... only.
I
Thanks so much, Richard. That works.
Hanna
2017-03-12 14:31 GMT-04:00 Richard M. Heiberger :
> I think you need either
>
> mod4 <- lm( y ~ -1 + area / (month*group), data=two)
>
> mod5 <- lm( y ~ area / (month*group), data=two)
>
> With either of those, area:month and area:group and Residuals
Dear all,
Thanks to Ista and Olivier. The solution of Ista works for some characters
but not all. The solution of Olivier works, at least for the characters
that I've tried.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team
Hi
This does not help with a solution, but ...
The standard PDF device only does single-byte character sets (sbcs in
the warning message). The conversion from a multi-byte character set
(mbcs), which is what UTF8 is, will only work if the characters map to a
single-byte character set (things
I think you need either
mod4 <- lm( y ~ -1 + area / (month*group), data=two)
mod5 <- lm( y ~ area / (month*group), data=two)
With either of those, area:month and area:group and Residuals add up.
On Sun, Mar 12, 2017 at 10:39 li li wrote:
> Hi All,
> I have a dataset which contains 4 varia
Dear Hannah
You say they are different (mod3 from mod1 and mod2) but in what way are
the results different?
On 12/03/2017 15:38, li li wrote:
Hi All,
I have a dataset which contains 4 variables: area, group, time, y,
Area is a factor that has two levels A and B, group is a factor that is
ne
Hi All,
I have a dataset which contains 4 variables: area, group, time, y,
Area is a factor that has two levels A and B, group is a factor that is
nested within area. There are four groups within each area.
y is the response variable, and time refers to different days.
Below is the how data lo
Hello,
Just to add to what others have said, I believe you haven't understood
the dots argument. Where you to change the function in the package and
the correct form would be
myplot <- function(x,y, ...) { plot(x,y,...) }
Hope this helps,
Rui Barradas
Em 11-03-2017 22:11, Mohammad Tanvir A
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