Displaying 20 results from an estimated 3000 matches similar to: "Returning the coef from two coordinates"
2012 Nov 11
2
Cropping a matrix by rows
Hello r-help,
I've been banging my head against the computer in an attempt to learn how
to divide my matrix into segments by rows. I want to be able to return each
segment as a newly named object. I've tried looking at the apply functions
and creating a for loop but brain no work. Here's the basic starting
objects that I believe would be needed to separate the matrix.
mat <-
2012 Apr 06
4
Order sapply
Good Afternoon,
I have the following code, but it seems that something must be doing
wrong, because it is giving the results I want.
The idea is to create segments while the value of Commutation is less than
1000.
for example, from the small set of data below
text="
val_user pos v v_star v_end commutation v_source
v_destine
1 1 96-96 1173438391 1173438391 0
2018 May 19
0
Split a data.frame
Hello,
Maybe something like the following.
splitDF <- function(data, col, s){
n <- nrow(data)
inx <- which(data[[col]] %in% s)
lapply(seq_along(inx), function(i){
k <- if(inx[i] < n) (inx[i] + 1):(inx[i + 1])
data[k, ]
})
}
splitDF(DF, "name", split_str)
Hope this helps,
Rui Barradas
On 5/19/2018 12:07 PM, Christofer Bogaso
2018 May 19
5
Split a data.frame
Hi,
I am struggling to split a data.frame as will below scheme :
DF = data.frame(name = c('a', 'v', 'c'), val = 0); DF
split_str = c('a', 'c')
Now, for each element in split_str, R should find which row of DF contains
that element, and return DF with all rows starting from next row of the
corresponding element and ending with the preceding value of the
2012 Oct 15
2
Chopping a two column data frame by rows into a three dimensional array.
If I have a two column data frame like:
> dat <- cbind("x"=c(1:100),"y"=c(100:1))
How can I create an array that splits every ten rows of that data frame
into a third dimension of an array so that:
> newarray[,,1]
,,1
x y
1 100
2 99
3 98
... ...
10 91
,,2
x y
11 90
12 89
... ...
...
Thanks.
[[alternative HTML version deleted]]
2012 Mar 27
2
R extract parts
Good Afternoon,
I believe that my to the problem, the R has a more effective solution.
in place the use the loop
I have the following set of data, and needs to extract some sections.
user pos communications source v_destine
7 1 109 22 22
7 2 100 22 22
7 3 214 22 22
7 4 322 22 22
7 5 69920 22 161
7 6
2012 Jul 24
3
Collapsing a vector/data-frame based on the previous values
Hello
I have a data frame like this:
dput(states)
structure(list(Date = c("24/07/2012", "25/07/2012", "26/07/2012",
"27/07/2012", "28/07/2012", "24/07/2012", "25/07/2012", "26/07/2012",
"27/07/2012", "28/07/2012"), State = c(1L, 1L, 1L, 1L, 1L, -1L,
-1L, -1L, 1L, -1L)), .Names = c("Date",
2010 Jul 07
2
What does `_data` mean in transform()?
Hi All,
I meant to take the min row by row. But the result is apparently not
what I want. Changing min to pmin solve the problem.
> df=data.frame(X=1:10, Y=1:10)
> transform(df, Z=min(X,10-Y))
X Y Z
1 1 1 0
2 2 2 0
3 3 3 0
4 4 4 0
5 5 5 0
6 6 6 0
7 7 7 0
8 8 8 0
9 9 9 0
10 10 10 0
I try to look at the source code to understand what transform() does.
I know
2023 Mar 02
1
transform.data.frame() ignores unnamed arguments when no named argument is provided
Dear r-devel,
See below:
transform(data.frame(a = 1), 2, 3)
#> a
#> 1 1
transform(data.frame(a = 1), b=2, 3)
#> a b X3
#> 1 1 2 3
We need a small modification to make it work consistently, see below:
transform.data.frame <- function (`_data`, ...) {
e <- eval(substitute(list(...)), `_data`, parent.frame())
tags <- names(e)
## NEW LINE
2012 Aug 21
2
Sequence detection longer than a certain value
Hello,
I have 2 variable: one is an "id" sequence from 1:1000 and the other is
variable with real values "VI" from -15.0 to 20.0 and I want to detect id
values that have indicator values less than a certain threshold, for
example (x=1) BUT that are in sequence equal or longer than 5.
For instance, in the following column I want to recognize the sequence from
"id" 4
2010 Aug 10
3
grep problem decimal points looping
Hi R Users,
I have been trying to work out how to rename column names using grep,
basically I have generated these column names using tapply:
[1] "NAME" "X1.1" "X2.1" "X3.1" "X4.1" "X5.1" "X6.1" "X7.1" "X8.1"
[10] "X1.2" "X2.2" "X3.2" "X4.2"
2012 Aug 10
2
Regular Expressions + Matrices
Hi all,
My code looks like the following:
inname = read.csv("ID_error_checker.csv", as.is=TRUE)
outname = read.csv("output.csv", as.is=TRUE)
#My algorithm is the following:
#for line in inname
#if first string up to whitespace in row in inname$name = first string up
to whitespace in row + 1 in inname$name
#AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the
2018 May 22
4
remove rows of a matrix by part of its row name
Dear R-experts,
How can I remove a certain feature or observation by a part of its name. To be clear, I have a matrix with 766 observations as a rows. The row names are like this
70/556
71.1/280
72.1/556
72.1/343
73.1/390
73.1/556
Now I would like to remove all the rows that contain the text 73.1
Any ideas or suggestion please ?
Regards
**********************
Ahmed Serag
Analytical
2008 Feb 05
2
wav to flac corruption
Hello,
I'm attempting to convert fairly large WAV files (90 - 800 MB each) using
flac but the files do not work after the encoding. (The play fine in wav
format)
Command I'm using:
flac --verify -8 file.wav
Attempting to run the file with either flac123 or the default player for
Ubuntu (Movie Player?) results in the extremely terse messages:
Default Player: "An Error Occurred:
2012 Sep 23
3
Confused by code?
x<-matrix(c(1,0,0,0,1,0,0,0,1),nrow=3)
> y<-matrix(c(0,0,0,1,0,0,1,1,0),nrow=3)
> z<-matrix(c(0,1,0,0,1,0,1,0,0),nrow=3)
> x[z]<-y[z]
The resultant matrix x is all zeros except for the last two diagonal cells
which are 1's.
While y is lower triangualr 0's with the remaining cells all ones.
I really don't understand how this deceptively simple looking piece of
2012 Aug 01
1
Error message: $ operator is invalid for atomic vectors
HI,
The code was working perfectly fine yesterday and today, until half an hour ago.? Couldn't find any problems in the code. Still, I am getting error message.
myMatrix <- data.matrix(read.table(text="
Name??????????? Age
ANTONY??????? 27
IMRAN????????? 30
RAJ????????????????? 22
NAHAS????????? 32
GEO??????????????? 42
", header=TRUE))
MinMaxArray? <- data.frame(MIN =
2012 Apr 22
4
How to take ID of number > 7.
I figured out something new that I would like to see if I can do this more
easy with R then Excel.
I have these huge files with data.
For example:
DataFile.csv
ID Name log2
1 Fantasy 5.651
2 New 7.60518
3 Finding 8.9532
4 Looeka -0.248652
5 Vani 0.3548
With like header1: ID, header 2: Name, header 3: log2
Now I need to get the $ID out who have a &log2 value higher then 7.
I
2012 Aug 11
3
replace funcion
Hello everybody,
I would like to replace or recode a list of numbers between 1 and 12 (total
100). I have tried to make it with recode, but i have two types of
replacements. For 1,2,3,4,11,12 => invierno and for 5,6,7,8,9 and 10 =>
verano.
recode(datos.mx1[,7], "1='invierno'; 2='invierno';
2018 May 22
0
remove rows of a matrix by part of its row name
Hello,
Use grep to get the row indices and then subset with a *negative* index
to remove those rows.
rn <- scan(what = character(), text = "
70/556
71.1/280
72.1/556
72.1/343
73.1/390
73.1/556
")
mat <- matrix(rnorm(6*6), nrow = 6)
row.names(mat) <- rn
inx <- grep("73\\.", row.names(mat))
new_mat <- mat[-inx, ]
new_mat
Hope this helps,
Rui Barradas
On
2012 Jan 06
6
cbind alternate
I have two one dimensional list of elements and want to perform cbind and
then write into a file. The number of entries are more than a million in
both lists. R is taking a lot of time performing this operation.
Is there any alternate way to perform cbind?
x = table1[1:1000000,1]
y = table2[1:1000000,5]
z = cbind(x,y) //hanging the machine
write.table(z,'out.txt)
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