Hi Vin,
If I read your Output (Wanted) section correctly, you want a new
list in which each element is the concatenation of the corresponding
elements of the original two. If so, maybe this shorter example will
help:
list1-list(id=as.data.frame(matrix(rep(c(563,623,581),3),nrow=3,byrow=TRUE)),
From: newrnew...@hotmail.com
To: r-help@r-project.org
Date: Mon, 11 May 2015 20:16:40 +
Subject: [R] binding two lists of lists of dataframes together
Hi,
I'm new to R and am stumped. I'm trying to bind List 1 to List 2 and have
the corresponding Output.
I've found the
My name is Erick Robinson and I am a SAS administrator at PNC
Bank. I have been tasked with installing R on two of our SAS servers,
but am having trouble downloading the source. I have little background
in UNIX (our servers run On Solaris under Unix) and cannot get to the
site to
Hi list,
Could anybody help me to explain the following error message and fix it? Thank
you very much!
Tao
sessionInfo()
R version 3.1.2 (2014-10-31)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8
Downloading should be the easy bit. Just go to cran.r-project.org (or a mirror)
and take the first link under source code for all platforms.
However, you do need to study the R Installation and Administration manual,
especially the platform notes. Solaris is a bit outside the mainstream these
Antonio Silva aolinto.lst at gmail.com writes:
Thanks for your attention Jim
Following your idea of adding one more step to construct the diagram, I used
text(spe2.rdaspe, row.names(spe2.rdaspe), pos=3, col=red,cex=0.8)
and I could add species names.
It seems that there's nothing
Dear Glenn,
I think that you are confusing the signature of a method and the default
values of of function. It looks like you want the signature(rates.data =
character, method = character). But you are setting character as
default value of both function arguments. Since you define a default value
Yes, it is possible with the mmc function in the HH package.
install.packages(HH) ## if you don't have it yet.
library(HH)
?MMC
Look at the maiz example, the long last example in ?MMC.
When you are ready for followup questions, reply to this message to
R-help (not directly to me and
not to
I have been asked by a couple researchers to install Rcmdr on our HPC cluster.
The cluster is a CentOS 6.5 64-bit linux OS and I am having issues running
Rcmdr on 3.2.0 (and other versions, but I fear the problem I am experiencing is
affecting all the versions I have installed). I have done a
Hi,
Apologies for the newbie error. Thanks John, Jim, Jeff!
Please find the dput output below. I'm trying to take list1 and list 2 and
bind them together at an individual list level by column - all the structures
and col names will be the same, but the number of rows could be different
Ip, g ftehgytreehjjjijiputv
Sent from my iPadgypyrrytutytytfedewaqyŷiijj
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hello everybody.
I have an esthetic question. I have managed to create a stacked and
grouped bar plot but I don't manage with putting the text in the middle of
the bar plots. Do you know how to write the numbers in that position?
Thank you so much.
Example code:
test - data.frame(variables =
Dear R-help,
can you please post the following message to the r-nabble forum.
Thanks!
-
Dear R users,
I have been attempting to carry out post-hoc tests on a
Hi,
Apologies for the newbie error. Thanks John, Jim, Jeff!
Please find the dput output below. I'm trying to take list1 and list 2 and
bind them together at an individual list level by column - all the structures
and col names will be the same, but the number of rows could be different for
Well, the stacking makes it tricky. AFAIK you have to calculate the
positions yourself, e.g.,
melted - melted[order(melted$variables, melted$cat, melted$cO), ]
melted$rec2 - melted$recuento/2
melted[melted$cO == B, rec2] - melted[melted$cO == B, rec2]
+ melted[melted$cO == A, recuento]
Jean: Thanks a lot!! The changes you made to the code gave me what I needed.
I truly appreciate your time in correcting the code.
Nilesh
From: Adams, Jean [mailto:jvad...@usgs.gov]
Sent: Tuesday, May 12, 2015 2:14 PM
To: DIGHE, NILESH [AG/2362]
Cc: r-help@r-project.org
Subject: Re: [R] help
On May 12, 2015, at 9:24 AM, Vin Cheng wrote:
list1-list(structure(list(id = c(493L, 564L, 147L, 83L, 33L, 276L, 402L,
285L, 30L, 555L),
WgtBand = c(1, 1, 2, 3, 4, 5, 6, 7, 8, 9),
Wgt = c(NaN, NaN, NA, NA, NA, NA, NA, NA, NA, NA)),
Hi,
I have an anonymous function called function(x) that will run anova, run
HSD.test on the model, and then sort the results. I am passing this anonymous
function to the by function to get results by Isopair factor which is my
index variable. Since I want to run the anova on multiple
Nilesh,
I found a couple errors in your code. First, in your by() statement you
have a function to operate on the selected subset of data, which you refer
to as
x
but then, in your aov statement you refer to
data_set
not
x
Second, your funC() statement is a function of
Hi
I have been reading the documentation, and it seems there is no difference
in the caliper option in the packages genmatch and match. In both the
simplistic way to use them is to specific a scalar which reflects the
maximum number of standard deviations the control can be away from the
treated.
Hi,
Thanks David! Your solution 3 worked well with the sample data, but when I run
solution 2 3 on the actual data - it creates a list of 1 for each data point
and it doesn't end up looking like the desired output for some reason.
I've run a dput on the actual data and pasted below.
On May 12, 2015, at 12:56 PM, Vin Cheng wrote:
Hi,
Thanks David! Your solution 3 worked well with the sample data, but when I
run solution 2 3 on the actual data - it creates a list of 1 for each data
point and it doesn't end up looking like the desired output for some reason.
I've
Hi everyone:
I'm trying to do a phylogenetic ANOVA analysis, but i'm stuck with both
possibilities: phylANOVA from phytools package, and aov.phylo from geiger
package.
Here is what i have
Input data:
CSV file with columns:
Spp (categorical, for taxonomic species,, 29 species)
Fam (categorical,
This may or may not be helpful, but you can create a dataframe with the
geom_text exactly where you want it to be (although it will probably be a
lot of extra work and will take time fiddling with it to see if you like
how it looks - and I think it looks fine the way you have it already).
Example
Dear R-help,
I am interested in plotting some pedigrees and came across the kinship2
package. What follows is an example of the pedigrees I am working with.
Now, when running
## check package availability
if(!require(kinship2)) install.packages('kinship2')
require(kinship2)
## data to plot
d
Hi David,
Would it be possible to modify it a little so that I can keep V1, V2, and V3
intact?
I also don't quite know where to put list2 in the solution you provided below?
list2 can be an exact copy of list1 for this example.
list1 has (48 observations of 3 variables(V1, V2,V3)) each
On May 12, 2015, at 3:12 PM, Vin Cheng wrote:
Hi David,
Would it be possible to modify it a little so that I can keep V1, V2, and V3
intact?
I also don't quite know where to put list2 in the solution you provided
below? list2 can be an exact copy of list1 for this example.
In
Dear all
I'm trying to examine the effect that the additive genetic covariance (B)
between the sexes has on the predicted response to selection on life-history
traits in the sexes.
Using the method of Argrawal and Stinchcombe (2009) this is done by calculating
the response to selection
Hello R help,
I am trying to run a permutation (or randomization) test on a mixed
effects model. The randomization is somewhat complex because I want to
maintain the nested and unbalanced structure of the data. I am trying
two different methods. The first is the PermTest() function in the
I'm helping colleagues with analysis of frog calls at night, and they
want to plot call statistics against time. This means we hit a
problem: we want the x-axis to start at (say) 18:00 and end at (say)
06:00. I'm reluctant to use the date as well, because we have data
from several dates, but only
Urgent - Need help in segmentation using KCluster means.
Please send me the analysis if someone can - talktoneil2...@ymail.com
Data link (Droopbox)
https://www.dropbox.com/s/kict2ezfokeyq7h/MAD_ZPR_User-Base_10K_Data_CONSOLIDATED%20_COPY2.xlsx?dl=0
--
View this message in context:
Urgent - Need help in segmentation using KCluster means.
Please send me the analysis if someone can - talktoneil2...@ymail.com
Data link (Droopbox)
https://www.dropbox.com/s/kict2ezfokeyq7h/MAD_ZPR_User-Base_10K_Data_CONSOLIDATED%20_COPY2.xlsx?dl=0
--
View this message in context:
Thanks Peter.
Erick Robinson
Software Engineer Lead
Retail MIS SAS Support
PNC Technology and Operations
23000 Mill Creek Blv.
Hignland Hills, Oh 44122
440-342-3100
e.robin...@pnc.com
From: peter dalgaard pda...@gmail.com
To: e.robin...@pnc.com,
Cc: R-help@r-project.org
On 12 May 2015, at 16:34 , Bob O'Hara rni@gmail.com wrote:
I'm helping colleagues with analysis of frog calls at night,
What do you do during the daytime then?
and they
want to plot call statistics against time. This means we hit a
problem: we want the x-axis to start at (say) 18:00
Try this.
From the full data-time value subtract 18:00:00.
This places the times you are interested in into the range 00:00:00 - 12:00:00
Remove the date from these adjusted date-time values and plot y
against the new times.
Take control of the tick-labels and display 18:00 - 0600 instead of
the
Dear Bob,
Is this useful?
library(lubridate)
library(ggplot2)
n - 100
h - sample(c(18:23, 0:9), size = n, replace = TRUE)
m - sample(0:59, size = n, replace = TRUE)
d - sample(1:3, size = n, replace = TRUE)
dataset - data.frame(
Time = as.POSIXct(paste0(2015-01-, d, , h, :, m, :0)),
Thing =
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