Displaying 20 results from an estimated 10000 matches similar to: "Multiple merge, better solution?"
2011 Aug 29
2
splitting into multiple dataframes and then create a loop to work
Dear All
Sorry for this simple question, I could not solve it by spending days.
My data looks like this:
# data
set.seed(1234)
clvar <- c( rep(1, 10), rep(2, 10), rep(3, 10), rep(4, 10)) # I have 100
level for this factor var;
yvar <- rnorm(40, 10,6);
var1 <- rnorm(40, 10,4); var2 <- rnorm(40, 10,4); var3 <- rnorm(40, 5, 2);
var4 <- rnorm(40, 10, 3); var5 <- rnorm(40, 15,
2009 Oct 07
1
merging dataframes with an unequal number of variables
Hallo Everyone
I have the kind of problem that one should never have because one must
always plan well and communicate with your team. But now I haven't so here
is my problem.
I have data coming in on a daily basis from surveys in 10 towns. The
questionnaire has 62 variables but some of the regions have used older
versions of the questionnaire that have a few variables less. I want to
combine
2012 Jun 03
2
merging single column from different dataframe
Hi all,
probably really simple to solve, but having no background in programming I
haven't been able to figure this out: I have two dataframes like
df1 <- data.frame(names1=c('aa','ab', 'ac', 'ad'), var1=c(1,5,7,12))
df2 <- data.frame(names2=c('aa', 'ab', 'ac', 'ad', 'ae'),
var2=c(3,6,9,12,15))
Now I want merge
2007 Sep 27
2
create data frame(s) from a list with different numbers of rows
# Hello,
# I have a list with 6 categories and with different numbers of rows.
# I would like to change each of them into a unique data frame in order to
match
# values with other data frames and perform some calculations.
# Or I could make each category or list element have the same number of rows
and create one large data.frame.
# below is a creation of a sample list
# I apologize for the
2011 Jun 06
1
Merge two columns of a data frame
I have the following data:
prefix <- c("cheap", "budget")
roots <- c("car insurance", "auto insurance")
suffix <- c("quote", "quotes")
prefix2 <- c("cheap", "budget")
roots2 <- c("car insurance", "auto insurance")
roots3 <- c("car insurance", "auto
2010 Nov 11
4
How to get a specific named element in a nested list
Hello,
I have a nested named list structure, like the following:
x <- list(
list(
list(df1,df2)
list(df3,
list(df4,df5))
list(df6,df7)))
with df1...d7 as data frames. Every data frame is named.
Is there a way to get a specific named element in x?
so, for example,
x[[c("df5")]] gives me the data frame 5?
Thank you in advance!
Best,
Friedericksen
2010 Dec 16
2
Compare two dataframes
Hello,
I have two dataframes DF1 and DF2 that should be identical but are not
(DF1 has some rows that aren't in DF2, and vice versa). I would like
to produce a new dataframe DF3 containing rows in DF1 that aren't in
DF2 (and similarly DF4 would contain rows in DF2 that aren't in DF1).
I have a solution for this problem (see self contained example below)
but it's awkward and
2009 Mar 08
1
Merge 10 data frames with 3 id columns that are common to all data frames
Hi R users,
Can anyone share some example code using merge_all (from the reshape
package) to merge 10 data frames into 1 file.
Thanks in advance for any help!
--
View this message in context: http://www.nabble.com/Merge-10-data-frames-with-3-id-columns-that-are-common-to-all-data-frames-tp22402493p22402493.html
Sent from the R help mailing list archive at Nabble.com.
2004 Mar 15
1
gzfile & read.table on Win32
Hello ...
Are there any known problems or even gotchas to look out for when using a
gzfile connection in read.csv/read.table in Windows?
In the package PROcess, available at
www.bioconductor.org/repository/devel/package/html/PROcess.html
there are two files in the PROcess/inst/Test directory which are of the
extension *.csv.gz.
With both files, if I open up a gzfile connection, say:
vv <-
2011 Dec 07
2
Dividing rows when time is overlapping
Hi all,
I have dataframe that was created from the fusion of two dataframes. Both
spanned over the same time intervall but contained different information.
When I put them together, the info overlapped since there is no holes in the
time interval of one of the dataframe. Here is an example where the rows
"sp=A and B" are part of a first df and the rows "sp=C" come from a
2010 Jan 20
1
Reshaping data with xtabs giving me 'extra' data
Dear all,
Lets say I have several data frames as follows:
> set.seed(42)
> dates <- as.Date(c("2010-01-19", "2010-01-20"))
> times <- c("09:30:00", "11:30:00", "13:30:00", "15:30:00")
> shows <- c("Red Dwarf", "Being Human", "Doctor Who")
>
> df1 <- data.frame(Date = dates[1],
2013 Jun 11
1
mapply on multiple data frames
Hi all-
I am wondering about using the mapply function to multiple data frames. Specifically, I would like to do a t-test on a subset of multiple data frames. All data frames have the same structure.
Here is my code so far:
f<-function(x,y) {
test<-t.test(x$col1[x$col3=="num",],v$col2[x$col3=="num",],paired=T,alternative="greater")
out<-test$p.value
2018 Nov 10
2
Asignar distancias
Utilizo la función merge desde hace poco, pero no se me ocurre cómo
utilizarla para esto. Yo pienso que se puede hacer con una combinación
de ifelse-s pero no sé cómo. Seguro que hay más de una forma ce hacerlo.
Quoting José María Mateos <chema en rinzewind.org>:
> On Sat, Nov 10, 2018 at 07:54:19PM +0100, Manuel Mendoza wrote:
>> Muy buenas. A ver si alguien puede echarme
2010 Sep 04
4
Please explain "do.call" in this context, or critique to "stack this list faster"
I've been doing some consulting with students who seem to come to R
from SAS. They are usually pre-occupied with do loops and it is tough
to persuade them to trust R lists rather than keeping 100s of named
matrices floating around.
Often it happens that there is a list with lots of matrices or data
frames in it and we need to "stack those together". I thought it
would be a simple
2005 Oct 20
1
Windows 2000 crash while using rbind (PR#8225)
Windows 2000 reports that "Rgui.exe has generated errors and will be =
closed by Windows. You will need to restart the program." when using =
rbind.=20
df1 <- data.frame(cbind(x=3D1, y=3D1:1000), fac=3Dsample(LETTERS[1:3], =
1000, repl=3DTRUE))
df2 <- data.frame(cbind(x=3D1, y=3D1:10), fac=3Dsample(LETTERS[4:6], =
10, repl=3DTRUE))
df3 <- data.frame(cbind(x=3D1,
2004 Mar 09
5
Adding data.frames together
I have a series of data frames that are identical structurally, i.e. -
made with the same code, but I need to add them together so that they
become one, longer, data frame, i.e. - each of the slot vectors are
increased in length by the length of the added data frame vectors.
So if I have df1 with a slot A so that length(df1$A) = 100 and I have
df2 with a slot A so that length(df2$A)=200 then I
2007 Oct 22
3
median value dataframe coming from multiple dataframes
Hi all,
I am not a skillful R programmer and has I am handling with large dataframes (about 30000 x 300) I am in need of an efficient function.
I have 4 dataframes with the same dimension. I need to generate other dataframe with the some dimension than the others where in each position it has the median value of the 4 values in the same position coming from the 4 dataframes.
Grateful by your
2012 Jun 12
4
replacing NA for zero
Dear R users,
I have a very basic query, but was unable to find a proper anwser.
I have the following data.frame
x y
2 0.12
3 0.25
4 0.11
6 0.16
7 0.20
and, due to further calculations, I need the data to be stored as
x y
1 0
2 0.12
3 0.25
4 0.11
5 0
6 0.16
7 0.20
8 0
How do
2018 Mar 22
1
Calculate weighted proportions for several factors at once
Hi,
I have a grouped data set and would like to calculate weighted proportions for a large number of factor variables within each group member. Rather than using dplyr::count() on each of these factors individually, the idea would be to do it for all factors at once. Does anyone know how this would work? Here is a reproducible example:
############################################################
2008 Feb 21
4
How to get names of a list into df:s?
R users,
I have a simple lapply question.
g <- list(a=1:3, b=4:6, c=7:9)
g <- lapply(g, function(x) as.data.frame(x))
lapply(g, function(x) cbind(x, var1 = rep(names(g), each=nrow(x))[1:nrow(x)]))
I get
$a
x var1
1 1 a
2 2 a
3 3 a
$b
x var1
1 4 a
2 5 a
3 6 a
$c
x var1
1 7 a
2 8 a
3 9 a
And I would like to have
$a
x var1
1 1 a
2 2 a
3 3 a