Displaying 20 results from an estimated 6000 matches similar to: "DPLYR Multiple Mutate Statements On Same DataFrame"
2024 Oct 18
1
DPLYR Multiple Mutate Statements On Same DataFrame
?s 22:50 de 17/10/2024, Sparks, John escreveu:
> Hi R Helpers,
>
> I have been looking for an example of how to execute different dplyr mutate statements on the same dataframe in a single step. I show how to do what I want to do by going from df0 to df1 to df2 to df3 by applying a mutate statement to each dataframe in sequence, but I would like to know if there is a way to execute this
2017 Dec 14
2
help with recursive function
Hi, I accidently left out few lines of code from the calclp function. Updated function is pasted below.
I am still getting the same error ?Error: !(any(data1$norm_sd >= 1)) is not TRUE?
I would appreciate any help.
Nilesh
dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>%
2017 Dec 14
3
help with recursive function
If you are trying to understand why the "stopifnot" condition is met you
can replace it by something like:
if ( any(dat2$norm_sd >= 1) )
browser()
This will put you in a debugging session where you can examine your
variables, e.g.
> dat$norm_sd
HTH,
Eric
On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger <ericjberger at gmail.com> wrote:
> The message is coming from
2017 Dec 14
2
help with recursive function
My own typo ... whoops ...
!( any(dat2$norm_sd >= 1 ))
On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger <ericjberger at gmail.com> wrote:
> You seem to have a typo at this expression (and some others like it)
>
> Namely, you write
>
> any(!dat2$norm_sd) >= 1
>
> when you possibly meant to write
>
> !( any(dat2$norm_sd) >= 1 )
>
> i.e. I think your !
2017 Dec 14
0
help with recursive function
Eric: I will try and see if I can figure out the issue by debugging as you suggested. I don?t know why my code after stopifnot is not getting executed where I like the code to run the funlp2 function when the if statement is TRUE but when it is false, I like it to keep running until the stopifnot condition is met.
When the stopifnot condition is met, I like to get the output from if statement
2017 Dec 14
0
help with recursive function
The message is coming from your stopifnot() condition being met.
On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] <
nilesh.dighe at monsanto.com> wrote:
> Hi, I accidently left out few lines of code from the calclp function.
> Updated function is pasted below.
>
> I am still getting the same error ?Error: !(any(data1$norm_sd >= 1)) is
> not TRUE?
>
>
>
2017 Dec 14
0
help with recursive function
Eric: Thanks for taking time to look into my problem. Despite of making the change you suggested, I am still getting the same error. I am wondering if the logic I am using in the stopifnot and if functions is a problem.
I like the recursive function to stop whenever the norm_sd column has zero values that are above or equal to 1. Below is the calclp function after the changes you suggested.
2017 Dec 14
2
help with recursive function
Hi, I need some help with running a recursive function. I like to run funlp2 recursively.
When I try to run recursive function in another function named "calclp" I get this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".
I have never built a recursive function before so having trouble executing it in this case. I would appreciate any help or guidance to resolve this issue.
2017 Dec 14
1
help with recursive function
Your code contains the lines
stopifnot(!(any(data1$norm_sd >= 1)))
if (!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
stop() "throws an error", causing the current function and all functions in
the call
stack to abort and return nothing. It does not mean to stop now and return
a result.
Does the function give
2009 May 15
1
data summary and some automated t.tests.
I would like to preform a t.test to each of the measured variables
(sand.silt etc.) with a mean and sd for each of the treatments (up or
down), and out put this as a table.... I am having a hard time
starting- maybe it is to close to lunch. Any suggestions would be
greatly appreciated.
Stephen Sefick
x <- (structure(list(sample. = structure(c(1L, 7L, 8L, 9L, 10L, 11L,
12L, 13L, 14L, 2L, 3L,
2017 Dec 14
0
help with recursive function
You seem to have a typo at this expression (and some others like it)
Namely, you write
any(!dat2$norm_sd) >= 1
when you possibly meant to write
!( any(dat2$norm_sd) >= 1 )
i.e. I think your ! seems to be in the wrong place.
HTH,
Eric
On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362] <
nilesh.dighe at monsanto.com> wrote:
> Hi, I need some help with running a
2010 Jul 02
2
Problem with aggregating data across time points
Hello-
I have a dataset which basically looks like this:
Location Sex Date Time Verbal Self harm
Violence_objects Violence
A 1 1-4-2007 1800 3 0
1 3
A 1 1-4-2007 1230 2 1
2 4
D 2 2-4-2007 1100 0
2016 Apr 22
1
Unique Ordering
Hi R-Help,
data at bottom
I've been struggling with a problem where I need to order based on 1) the
Frequency "Freq" and 2) keeping each group of 3 of the same type together
"Var2" but I want across all groups it to go "high to low" based on the
earn factor.
Thank you!
structure(list(Var1 = structure(c(1L, 3L, 2L, 1L, 2L, 3L, 1L,
3L, 2L, 3L, 1L, 2L, 3L, 1L,
2009 May 28
2
ggplot2 legend
Hi:
I need some help with the legend. I got 14 samples(Muestreo) and I
am trying to plot a smooth line for each sample. I am able to accomplish that but the problem is that the legend only displays every other sample. How can I force the legend to show all of my Muestreos? Thanks in advance.
fish_ByMuestreo <- structure(list(data = structure(list(SampleDate = structure(c(3L,
3L, 3L, 3L,
2018 Feb 26
0
alternative for multiple if_else statements
That many ifelse statements is obviously rather a pain.
Would you not have got what you want with
... paste("survey", year, sep="_")
?
If that is not what you're looking for (eg because 'year' is the observation year and not the study start year), perhaps something that picks the minimum year for a subject or other relevant group might work? For example
2009 Nov 26
1
lattice --- different properties of lines corresponding to type=c("l", "a") respectively
I think the subject says it all. I want to make a simple lattice plot,
using xyplot with the
argument type=c("l","a").
The problem then is that in the resulting plot it is
difficult/impossible to see which plot corresponds to the average
and which to the individual profiles. I triedthings like extra
arguments lwd=c(1,3) or col=c("blue","red")
hoping
2008 Aug 26
1
processing subset lists and then plot(density())
d <- structure(list(Site = structure(c(8L, 12L, 7L, 6L, 11L, 5L, 10L,
4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L,
1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L,
12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L,
11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L,
4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L,
1L, 9L,
2013 Feb 08
2
Can not melt data.frame
I realize it's -12C and we're having the next best thing to a blizzard but why can I not melt this data frame.
I am missing something terribly obvious but I just don't understand what the error message is saying.
John Kane
Kingston ON Canada
Code and aata below
#================================================#
library(reshape2)
melt(mydata, id.vars = c("date"))
2011 Jun 13
1
Heatmap in R and/or ggplot2
I have a dataframe df with columns x, y, and height. I want to create a
heatmap-like plot that creates a grid of x by y, and then color codes the
grid depending on the value of height.
Is there a ggplot2 object to do this? I'm able to easily do this in Excel
with pivot tables and conditional formatting so I'm including an image that
is close to the output I want. I want to be able to
2011 Feb 19
1
problem in plotting numeric x by POSIXt class with lattice
# hi all,
# I'm trying to plot temperatures by date in a trellis plot by their
stations
# I'm plotting the following data.frame
library(lattice)
h <- structure(list(station_name = structure(c(3L, 4L, 2L, 10L, 11L,
12L, 6L, 7L, 5L, 8L, 9L, 3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L, 5L,
8L, 9L, 3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L), .Label = c("Ashqelon",
"Beer Sheva",