Displaying 20 results from an estimated 70 matches similar to: "help need on working in subset within a dataframe"
2018 May 26
3
Grouping by 3 variable and renaming groups
ALCON
I'm trying to figure out how to rename groups in a data frame after groups
by selected variabels. I am using the dplyr library to group my data by 3
variables as follows
# group by lat (StoreX)/long (StoreY)
priceStore <- LapTopSales[,c(4,5,15,16)]
priceStore <- priceStore[complete.cases(priceStore), ] # keep only non NA
records
priceStore_Grps <- priceStore %>%
2018 May 26
0
Grouping by 3 variable and renaming groups
Hello,
See if this is it:
priceStore_Grps$StoreID <- paste("Store",
seq_len(nrow(priceStore_Grps)), sep = "_")
Hope this helps,
Rui Barradas
On 5/26/2018 2:03 PM, Jeff Reichman wrote:
> ALCON
>
>
>
> I'm trying to figure out how to rename groups in a data frame after groups
> by selected variabels. I am using the dplyr library to group my
2011 Dec 14
1
Hi
First PRoblem solved
> read.csv("KT80.csv")
X x
1 1 0.01331361
>
I ommitt "" in calling the file name...
2011/12/14 Trying To learn again <tryingtolearnagain@gmail.com>
> Hi all,
>
> I have 100 csv files always with this information (I Attach two example
> excels)
>
> KT80.csv contains:
> ,"x"
>
2018 May 26
1
Grouping by 3 variable and renaming groups
Hello,
Sorry, but I think my first answer is wrong.
You probably want something along the lines of
sp <- split(priceStore_Grps, priceStore_Grps$StorePC)
res <- lapply(seq_along(sp), function(i){
sp[[i]]$StoreID <- paste("Store", i, sep = "_")
sp[[i]]
})
res <- do.call(rbind, res)
row.names(res) <- NULL
Hope this helps,
Rui Barradas
On 5/26/2018
2003 Nov 19
1
heap error while trying to run TrueSync Dekstop
Hello,
I'm trying to run the "TrueSync Desktop" app (the equivalent of Hotsync
for non-Palm handhelds, but without any Linux port, unfortunately), and
I get the following error:
err:heap:HEAP_ValidateInUseArena Heap 40360000: in-use arena 403e7eb8
next block has PREV_FREE flag
wine: Unhandled exception (thread 0009), starting debugger...
The loading screen "TrueSync
2005 Jun 09
1
krig.image help
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Mike J Smith
> Sent: 09 June 2005 09:58
> To: r-help at stat.math.ethz.ch
> Subject: [R] krig.image help
>
>
> Hi
>
> I have recently been experimenting with the use of kriging, primarily
> through Goldensoftware's Surfer.
2012 Oct 09
1
other way of making a table?
I'm making tables for prediction results of classifiers (2 classes) that show the usual numbers, true positives, false positives, etc
I used the command
table(predictedLabels,realLabels)
to make those.
I just had a case though ,where one of the label vectors had only one class in it. This will result in only half a table.
Compare:
x<-c(1,1,1,0,0)
y<-c(1,1,1,0,1)
table(x,y)
to
2009 Jun 20
1
how to apply the dummy coding rule in a dataframe with complete factor levels to another dataframe with incomplete factor levels?
Dear R helpers:
Sorry to bother for a basic question about model.matrix.
Basically, I want to apply the dummy coding rule in a dataframe with
complete factor levels to another dataframe with incomplete factor levels.
I used model.matrix, but could not get what I want.
The following is an example.
#Suppose I have two dataframe A and B
2010 Jul 13
2
Checking for duplicate rows in data frame efficiently
I wrote something to check for duplicate rows in a data frame, but it is too inefficient. Is there a way to do this without the nested loops?
This code correctly indicates rows 1-7, 1-8, 2-9 and 7-8 are duplicates.
> m <- matrix(c(1,1,1,1,1, 2,2,2,2,2, 6,6,6,6,6, 3,3,3,3,3, 4,4,4,4,4, 5,5,5,5,5, 1,1,1,1,1, 1,1,1,1,1, 2,2,2,2,2, 7,7,7,7,7), ncol=5, byrow=TRUE)
> df <- data.frame(m)
2012 Jan 20
4
extract fixed width fields from a string
Hi,
I have a data frame with one column containing string of the form "ABC...|XYZ..."
where ABC etc are fields of 6 alphanumeric characters each
and XYZ etc are fields of 8 alphanumeric characters each;
"|" is a mandatory separator;
I do not know in advance how many fields of each kind will each row contain.
I need to extract these fields from the string.
=== How do I do that?
2008 Jul 24
2
What is wrong with this contrast matrix?
Dear all,
I am fitting a multivariate linear model with 7 response variables and 1 explanatory variable.
The following matrix P:
P <- cbind(
c(1,-1,0,0,0,0,0),
c(2,2,2,2,2,-5,-5),
c(1,0,0,-1,0,0,0),
c(-2,-2,0,-2,2,2,2),
c(-2,1,0,1,0,0,0),
c(0,-1,0,1,0,0,0))
should consist of orthogonal elements (as can be shown using %*% on the individual columns).
However, when I use
2005 May 31
1
GLM question
I am unfamiliar with R and I’m trying to do few statistical things like GLM and GAM with it. I hope my following questions will be clear enough:
My datas ( y(i,j ))are run off triangles for example :
J=1
J=2
J=3
I=1
1
2
3
I=2
4
5
I=3
6
My model is :
E[y(i,j)] =m(i,j)
Var[y(i,j)] =constant *m(i,j)
Log(m(i,j)) = eta (i,j)
eta (i,j) = c + alpha(i)
2005 Apr 13
1
logistic regression weights problem
Hi All,
I have a problem with weighted logistic regression. I have a number of
SNPs and a case/control scenario, but not all genotypes are as
"guaranteed" as others, so I am using weights to downsample the
importance of individuals whose genotype has been heavily "inferred".
My data is quite big, but with a dummy example:
> status <- c(1,1,1,0,0)
> SNPs <-
2008 Jun 21
2
Generating groupings of ordered observations
Dear List,
I have a problem I'm finding it difficult to make headway with.
Say I have 6 ordered observations, and I want to find all combinations
of splitting these 6 ordered observations in g groups, where g = 1, ...,
6. Groups can only be formed by adjacent observations, so observations 1
and 4 can't be in a group on their own, only if 1,2,3&4 are all in the
group.
For example,
2004 Apr 02
1
which on array
Good morning !
Today I found a strange, for my poor knowledge of R, behaviour of
'which' on a matrix:
HAL9000> str(cluster.matrix)
num [1:227, 1:6300] 2 2 2 2 2 2 2 2 2 2 ...
HAL9000> class(cluster.matrix)
[1] "matrix"
HAL9000> ase <- cluster.matrix[1:5,1:5]
HAL9000> ase
[,1] [,2] [,3] [,4] [,5]
[1,] 2 2 2 0 -2
[2,] 2 2 2 0 -2
[3,]
2007 Oct 05
1
creating objects of class "xtabs" "table" in R
I have an application that would generate a cross-tabulation in array
format in R. In particular, my application would give me a result
similar to that of :
array(5,c(2,2,2,2,2))
The above could be seen as a cross-tabulation of 5 variables with 2
levels each (could be 0 and 1). In this case, the data were such that
each cell has exactly 5 observations. I
Now, I want the output to look like the
2006 Apr 02
3
speeding up a recursive function
Hi All,
is there any general advice about speeding up recursive functions
(not mentioning 'don't use them')?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax +44 (0)20 75943193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t]
2004 Jan 04
1
array problem
Dear all,
I define , for n=5 or any integer greater than 0.
A<-array((1/2)^n , c(rep(2,n)))
then for any i not equal to j, and 1<=i,j<=n,
B<-apply(a,c(i,j),sum)
now B is a 2 by 2 matrix, I also define another costant 2 by 2 matrix G,
How can I change the values of each elements of array A, according the rule
that,
for example, i=3,j=5,
2011 May 04
2
first occurrence of a value?
Hello,
A simple question perhaps, but how do I, within each row, find the first
occurence of the number 1 in the df below? I want to use this position to
programmatically create the variable 'year'. I'v come up with a solution, but I
find it downright ugly. Is there a simpler way? I was hoping for a useful
built-in function that I don;t yet know about.
df <-
2009 Jan 13
2
NA-values and logical operation
Dear list,
as a result of a logical operation I want to assign
a new variable to a DF with NA-values.
z <- data.frame( x = c(5,6,5,NA,7,5,4,NA),
y = c(1,2,2,2,2,2,2,2) )
p <- (z$x <= 5) & (z$y == 1)
p
z[p, "p1"] <-5
z
# ok, this works fine
z <- z[,-3]
p <- (z$x <= 5) & (z$y == 2)
p
z[p, "p2"] <-5
z
# this failed... - how