h.
>>
>> I´m trying to read a file, but the name of this file changes every day, for
>> example: today is May 24, 2012
>>
>> bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t")
>>
>> So, tomorrow I want to read the fi
Yes I exactly followed what you all suggested:
X<-(82:92) ; Y<-(364:369) # for sellected region
> extract <- double(365)
> setwd("C:\\Users\\aalyaari\\Desktop\\New folder (10)\\")
> listfile<-dir()
> for (i in 1:365) {
+ conne <- file(listfile[i], "rb")
+ file1<- readBin(co
On Thu, May 24, 2012 at 08:24:38PM -0700, igorre25 wrote:
> Hello,
>
> I need to build certain interpolation logic using R. Unfortunately, I just
> started using R, and I'm not familiar with lots of advanced or just
> convenient features of the language to make this simpler. So I struggled
> for
Dear Wolfgang,
Thank you very much for the quick reply. I already assumed that it might
get too complicated,
so I will just stick to using a moderator (as you also suggested).
Best,
Anke
On 24.05.2012 17:05, Viechtbauer Wolfgang (STAT) wrote:
At the moment, there is no possibility of speci
Hello,
I need to build certain interpolation logic using R. Unfortunately, I just
started using R, and I'm not familiar with lots of advanced or just
convenient features of the language to make this simpler. So I struggled
for few days and pretty much reduced the whole exercise to the following
?paste
On 25/05/12 11:45, Minerva Mora wrote:
Hi, I apologize for my english.
I´m trying to read a file, but the name of this file changes every day, for
example: today is May 24, 2012
bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t")
So, tomorrow
I have a data set that has some comma separated strings in each row.
I'd like to create a vector consisting of all distinct strings that
occur. The number of strings in each row may vary.
Thanks for any help.
__
R-help@r-project.org mailing list
https:
On Thu, May 24, 2012 at 9:31 AM, wrote:
> Dear R-Help,
>
> I have a clustering problem with hclust that I hope someone can help
> me with. Consider the classic hclust example:
>
> hc <- hclust(dist(USArrests), "ave")
> plot(hc)
>
> I would like to cut the tree up in such a way so as to av
Can't put my finger on it but something about your idea rubs me the
wrong way. Maybe it's that the tree depends on the hierarchical
clustering algorithm and the choice on how to trim it should be based
on something more defensible than "avoid singletons". In this example
Hawaii is really different
On May 24, 2012, at 7:45 PM, Minerva Mora wrote:
Hi, I apologize for my english.
I´m trying to read a file, but the name of this file changes every
day, for
example: today is May 24, 2012
bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t")
So, tomo
On May 24, 2012, at 7:45 PM, Minerva Mora wrote:
Hi, I apologize for my english.
I´m trying to read a file, but the name of this file changes every
day, for
example: today is May 24, 2012
bonos<- read.table(*20120524*.csv", header=TRUE, sep="\t")
yr = "2012&quo
Thank you both,
I will try using a zero inflated negative binomial as suggested. I had success
with negative binomial on previous runs but only when I had fewer covariates
and only ran a portion (10%) of the data.
I may also try to reduce the number of covariates in the model (i.e., combine
Sent: Friday, 25 May 2012 11:45a
To: r-help@r-project.org
Subject: [R] How to open a file with a name changed?
Hi, I apologize for my english.
I´m trying to read a file, but the name of this file changes every day, for
example: today is May 24, 2012
bonos<- read.table("C:/Bonos/*20120524
On May 24, 2012, at 6:37 PM, Hans Thompson wrote:
The function I am giving for context is cbind. Are you asking how I
would
like to apply the answer to my question?
I am trying to take the results of a Fluidigm SNP microarray,
organized by
assay into a list (each component is the results
On May 24, 2012, at 3:41 PM, Nathan Svoboda wrote:
Re: [R] R Error: System is computationally singular
Hi David,
My apologies, I am not sure if this makes a big difference in your
assessment of the problem, but the results I just sent were only
from a portion (1/15) of the data. The datase
Hi, I apologize for my english.
I´m trying to read a file, but the name of this file changes every day, for
example: today is May 24, 2012
bonos<- read.table("C:/Bonos/*20120524*.csv", header=TRUE, sep="\t")
So, tomorrow I want to read the file again, but i don´t want to
The function I am giving for context is cbind. Are you asking how I would
like to apply the answer to my question?
I am trying to take the results of a Fluidigm SNP microarray, organized by
assay into a list (each component is the results of one assay), find
coordinate midpoints ([1,] and [2,] of
Yes. This gives me:
[,1] [,2] [,3] [,4]
[1,]1234
[2,]2345
[,1] [,2] [,3] [,4]
[1,]6789
[2,]789 10
BUT, how can I have it still within components like
[[1]]
[,1] [,2] [,3] [,4]
[1,]1234
[2,]23
I am running 64-bit R 2.15.0 on a Windows Server 2008 R2 Amazon EC2 instance.
grid does not produce output. For example, the following code should print the
R logo to the window() device:
library(grid)
library(png)
img.path <- system.file("img", "Rlogo.png", package="png")
bg <- readPNG(img.pa
On Fri, May 25, 2012 at 4:10 AM, lasciel wrote:
>
> So, I could use ‘rbind’ to stack the two years of data together with their
> appropriate weights into a single data frame.
> svychisq(~MyVar+MyVar, BothYears, statistic=”Chisq”)
>
> But now I have a problem that the variable name is the same acr
Nathan,
This does help, as in the first cut you provided, there was no variability in
LOCS for LCOVER >= 5 and you have very few values of LOCS > 0 (you still do,
relative to the scale of the total).
Have you tried using a zero inflated negative binomial model (dist = "negbin")
rather than poi
Hi David,
My apologies, I am not sure if this makes a big difference in your assessment
of the problem, but the results I just sent were only from a portion (1/15) of
the data. The dataset is rather large and the computer I am currently using to
set up the models is limited in its capabilities
Thank you for your quick reply,
When I run the code you provide I get this output:
LCOVER
LOCS 1 2 3 4 5 6 7 9
0 214507 79939 69803 778359 22932 32391 99630 8082
1 15 7 1 32 0 0 0 0
2 2 1
Thank you for your quick reply!
When I run the code you provide I get this output:
LCOVER
LOCS 1 2 3 4 5 6 7 9
0 214507 79939 69803 778359 22932 32391 99630 8082
1 15 7 1 32 0 0 0 0
2 2 1
On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:
Greetings,
I am trying to fit a zero-inflated Poisson model using zeroinfl()
from the
pscl library. I have 5 covariates (4 continuous, 1 categorical); the
categorical variable has 7 levels. I have had success fitting
models that
contain
Greetings,
I am trying to fit a zero-inflated Poisson model using zeroinfl() from the
pscl library. I have 5 covariates (4 continuous, 1 categorical); the
categorical variable has 7 levels. I have had success fitting models that
contain only the continuous covariates; however, when I add the cate
> > X<-c(364:369) ; Y<-c(82:92) # for sellected region
> > ...
> > file2<-matrix(data=file,ncol=720,nrow=360)
> > extract[i]<-mean(file2[X,Y],na.rm=TRUE)
Note that 2-dimensional subscripts are given as
mat[ROWS, COLUMNS]
and you are using the reverse orde
On May 24, 2012, at 12:49 PM, Hans Thompson wrote:
I'm confused why I haven't made clear what I am asking for help with.
I have two different lists with two (or many) components, [[1]] and
[[2]].
One of the list has components with dim=c(2,3) and the other has
dim=c(2,2).
I want to create
On 2012-05-24 03:26, stefan23 wrote:
Hi Folks,
I am currently trying to present some results I obtained by using the
quantreg package developed by Roger Koenker. After calculating
fit<-summary(rq(Y~X1+X2, tau=2:98/100) ) the function plot(fit) presents a
really nice the results by showing the val
You don't make it clear if you want the two lists processed in parallel, but
if so, this may be a step toward the function you want:
Combine <- function(l1, l2) {
pattern <- cbind(c(1, 2, 2, 3), c(1, 1, 2, 2))
for (i in 1:length(l1)) {
print((l1[[i]][,pattern[,1]]+l2[[i]][,pattern[,2]])/2)
Yes this helped a lot .
I exactly followed what you suggested:
X<-(82:92) ; Y<-(364:369) # for sellected region
> extract <- double(365)
> setwd("C:\\Users\\aalyaari\\Desktop\\New folder (10)\\")
> listfile<-dir()
> for (i in 1:365) {
+ conne <- file(listfile[i], "rb")
+ file1<-
Thank you William,
Does this mean I should use "" when inner_perc_table is called in the code
then or am I supposed to define a full path?
Regards,
Charles
On Thu, May 24, 2012 at 12:31 PM, William Dunlap wrote:
> You need to supply a character string, e.g., "inner_perc_table", that names
> th
You need to supply a character string, e.g., "inner_perc_table", that names
the C entry point. If you supply a name, e.g., entry_point, or a more general
expression,
e.g, entry_points[j], then it will be evaluated with the usual R evaluation
rules and the result must be a character string.
>
Sorry, forgot the last line in the op code.
write.table(extract, ...) is called every time through the loop, not just
one time after it. Put it after closing '}'.
Rui Barradas
Rui Barradas wrote
>
> Hello,
>
> There are several things with your code, most of which are not errors.
>
> 1. X<-
Hello,
There are several things with your code, most of which are not errors.
1. X<-c(364:369) ; Y<-c(82:92 and later c(1:365)
Expressions of the form m:n are integer sequences, the c() is not needed.
2. extract<-vector()
First you create the vector, then keep extending it throughout the loop.
T
I'm confused why I haven't made clear what I am asking for help with.
I have two different lists with two (or many) components, [[1]] and [[2]].
One of the list has components with dim=c(2,3) and the other has dim=c(2,2).
I want to create a new list with components dim=c(2,4) by binding togeth
--- segue il messaggio inoltrato (the forwarded
message follows) ---
--- Begin Message ---
Dear R users,
i am very new to R. i am working on the following data.
01.01.1967 0.87
02.01.1967 0.87
03.01.1968 0.87
04.01.1968 0.87
05.01.1969 0.87
06.01.1969 0.87
My apologies, feel a little foolish. I forgot the "" for the object. This
worked, however, once I have this information, I am not sure what I should
be using to fix the issue. I can determine the location in the nlme.dll
file but how can I use this information to have the other function work?
Greg Snow <538280 gmail.com> writes:
>
> I believe that what is happening is that when you try to edit the
> entry widget any intermediate values get sent to the slider widget
> which then checks to see if they are in the allowable range and if it
> is not then it sets the value to either the mi
Assuming you mean that the column is constant so that the sd is 0
test <- matrix(rnorm(50), nrow=10, ncol=5)
test[,2] <- 1
test[,4] <- 2
# test[,which(apply(test, 2, sd) > 0)] could fail on rounding errors
Test2 <- test[, which(apply(test, 2, sd) > 1e-10)]
--
Thank you Duncan,
I used the R function getNativeSymbolInfo(inner_perc_table) and the object
is not found:
Error in is.vector(X) : object 'inner_perc_table' not found
Done.
I also tried the defining the directory to the nlme.dll file and it had the
same error.
Is there something I missed while d
On 24/05/2012 11:35 AM, Charles Determan Jr wrote:
Hello,
Does anyone on this list know what inner_perc_table is or where it is
typically found? I am trying to modify some source code and it is used
with the .C() function. When I try and run it, it states that
'inner_perc_table is not found'.
Hi,
Just an addendum.
Suppose if you want to add more columns and do the split, you can try,
inscrutable.df2<-data.frame(inscrutable.df1,V2=(c(rnorm(1),"GTDF","SQ:1234","DFFD","DFDSE")[rep(c(1,2,3,4,5),times=4)]))
inscrutable3.df3<-data.frame(matrix(inscrutable.df1$V1,ncol=2,byrow=TRUE),matrix(i
On Thu, May 24, 2012 at 07:16:20AM -0700, Haio wrote:
> Hello everyone!
> I´m trying to solve a linear optimization problem with r using the LPsolve
> function and i want to set upper and lower bounds for the obejctive
> function. Any idea´s on how to do this??
Hello:
The formulation of an LP pr
Hello,
I’m hoping you have a few minutes to help out someone very new to R. I’ve
done some searching, but cannot find this particular issue.
I have survey data from two different time periods (years). Both years are
stratified samples and have the same variables (and variable names), but are
di
I'll answer line by line so:
"You ask for help from people who have used R.NET, give them time to
answer."
I have.. I'm not sure why you seem so certain I didn't. But it is beside the
point anyway
"Also, what about folks who have not used it? "Does not work"
is not enough information. What
Dear R-Help,
I have a clustering problem with hclust that I hope someone can help
me with. Consider the classic hclust example:
hc <- hclust(dist(USArrests), "ave")
plot(hc)
I would like to cut the tree up in such a way so as to avoid small
clusters, so that we get a minimum number of
?do.call
On Thu, May 24, 2012 at 9:32 AM, Alexander Shenkin wrote:
> Hello Folks,
>
> Is there any way to pass a list into a function such that the function
> will use the list as its arguments? I haven't been able to figure that out.
>
> The background: I'm trying to build a function that will
On May 24, 2012, at 11:51 AM, Jonsson wrote:
Dear All,
The code given bellow is to extract values of one region and write
that to
a text file(there are 365 binary files in the directory).
The problem which I am facing is that all my files are binary with
size of
360 rows and 720 columns.
Dear All,
The code given bellow is to extract values of one region and write that to
a text file(there are 365 binary files in the directory).
The problem which I am facing is that all my files are binary with size of
360 rows and 720 columns.
I specified that in this line:file2<-matrix(data=f
Hello,
Does anyone on this list know what inner_perc_table is or where it is
typically found? I am trying to modify some source code and it is used
with the .C() function. When I try and run it, it states that
'inner_perc_table is not found'. It is only called in such a way and isn't
defined at
Hello Folks,
Is there any way to pass a list into a function such that the function
will use the list as its arguments? I haven't been able to figure that out.
The background: I'm trying to build a function that will apply another
function multiple times, each time with a different set of specif
Dear All,
I am running R in a system with the following configuration
*Processor: Intel(R) Xeon(R) CPU X5650 @ 2.67GHz
OS: Ubuntu X86_64 10.10
RAM: 24 GB*
The R session info is
*
R version 2.14.1 (2011-12-22)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8
This method worked perfectly! The rle() function was key and I was
completely unfamiliar with it. Thanks so much,
Max
On Thu, May 24, 2012 at 8:52 AM, Rui Barradas wrote:
> Hello,
>
> Assuming that 'd' is your original data.frame and that you've set entire
> rows to NA, try this
>
>
> d$leak_num
The "do.call" function may be what you want, but it is not completely
clear. If that does not solve your problem then try giving us an
example of what your list looks like (the dput function can help) and
what you want your end result to look like.
On Thu, May 24, 2012 at 5:38 AM, Hans Thompson
At the moment, there is no possibility of specifying the weights with the rma()
function. While the main model fitting part could be easily adapted to
incorporate user-specified weights, the problem comes in with all the
additional statistics that can be computed based on a fitted model. How sho
I believe that what is happening is that when you try to edit the
entry widget any intermediate values get sent to the slider widget
which then checks to see if they are in the allowable range and if it
is not then it sets the value to either the minimum and maximum and
sends that back to the entry
> Does anyone have any thoughts on how to code this, perhaps using the NA
> values as a "break point"?
You can count the cumulative number of NA breakpoints in a vector
with cumsum(is.na(vector)), as in
> cbind(d, LeakNo=with(d, cumsum(is.na(lon)|is.na(lat)|is.na(CH4
lon lat
See insertion below
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 5/24/12 1:43 AM, "Jim Lemon" wrote:
>On 05/24/2012 05:10 PM, Soheila Khodakarim wrote:
>> Dear All
>>
>> How can I transform 1 column to 2 columns in R?
>>
Hello everyone!
I´m trying to solve a linear optimization problem with r using the LPsolve
function and i want to set upper and lower bounds for the obejctive
function. Any idea´s on how to do this??
--
View this message in context:
http://r.789695.n4.nabble.com/Upper-an-lower-bound-in-linear-p
Dear R-experts,
Dear Wolfgang,
Weighted model fitting in metafor uses the inverse of the study specific
variances as weights.
I am wondering if it is possible to specify different weights.
In my meta-analysis, there are two types of studies with (intrinsic)
differences in their range of sample
I see. It is decided then. As it might be possible that the license might be
violated, I certainly will not include R in my programme.
Thank you all for your help, you've been very helpful.
---
Giannis Mamalikidis
e-mai
On May 24, 2012, at 9:06 AM, David Winsemius wrote:
On May 24, 2012, at 4:43 AM, davv.vikas wrote:
Hi All,
I am writing a macro which perform following functionality
1) take a file say excel or csv
2)read all the labels
3) dynamically genrate square,square-root,log of each column with
appr
Will this do it for you:
> x <- list(1,2,3)
> x[[1]]
[1] 1
> cat(x[[1]], '\n')
1
>
On Thu, May 24, 2012 at 5:50 AM, Manish Gupta wrote:
> HI,
>
> I am printing list element.
>
> print(desc[[2]])
> [1] XXXxx
>
> I want to remove [1]
> Output should be
> XXXxx
>
>
Hello,
Try
?subset
subset(dtfr, Col1 == "xxx")
?"["
dtfr[Col1 == "xxy", ]
Hope this helps,
Rui Barradas
Em 24-05-2012 11:00, Manish Gupta escreveu:
Date: Thu, 24 May 2012 01:09:58 -0700 (PDT)
From: Manish Gupta
To:r-help@r-project.org
Subject: [R] How to retrieve value form dataframe by Key
Your answers were very helpful, thanks :)
But, you see, the program I'm writing will be a team effort - and while I
have not problem on making this program an open source - we don't really see
eye to eye with the team on this.
So no source code will be available - just freeware.
R will not be
On 05/23/2012 07:24 AM, rmje wrote:
I have a data frame like this:
T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h
C0h C0.25h C0.5h C1h C2h C3h C6h C12h C24h C48h
NM_001001130 68 9556 43 66 62 68 90 63 89 65
8558 49 81 49 76 7
On May 24, 2012, at 4:43 AM, davv.vikas wrote:
Hi All,
I am writing a macro which perform following functionality
1) take a file say excel or csv
2)read all the labels
3) dynamically genrate square,square-root,log of each column with
appropiate column name and write it to a new file,
i am st
Giannis,
While your economic situation is appreciated, that you plan to give away your
application and not generate any income from it, is irrelevant to the question.
The nuances as to how you are interfacing with R and that you plan to include R
(importantly, possibly only a part of it) as a c
Hello,
Assuming that 'd' is your original data.frame and that you've set entire
rows to NA, try this
d$leak_num <- NA
ix <- !is.na(d[, 1]) # any column will do, entire row is NA
## alternative, if other rows may have NAs, due to something else
#ix <- apply(d, 1, function(x) all(!is.na(x)))
r
On 12-05-24 8:23 AM, Giannis Mamalikidis wrote:
Your answers were very helpful, thanks :)
But, you see, the program I'm writing will be a team effort - and while I
have not problem on making this program an open source - we don't really see
eye to eye with the team on this.
So no source code wil
The combination of column means in cbind is what I am trying to do. I just
don't know how to do it for every component (2 in this case) of the list.
I've been able to work with R to apply a function across all components when
there is just one list but I am having trouble working with multiple li
I figured out how to use quotes and parentheses when using Rscript -e (on a
bash shell):
Rscript -e write\(1,\"a.txt\"\)
--> Question 1: why do the parentheses need to be escaped in the shell?
(More a shell than an R question)
Then I figured out how to use quotes and parentheses when calling Rsc
HI,
I am printing list element.
print(desc[[2]])
[1] XXXxx
I want to remove [1]
Output should be
XXXxx
How can i implement it?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-avoid-numbering-while-printing-list-element-tp46311
Hi, I am working under Windows and I am using R2.11
I want to use tkscale in my GUI. As the interval is quite big, I can't set
the scale to a certain specific value. Therefore I want to add tkentry to
allow the user to set tkscale to a certain value.
Here is the code
library(tcltk)
tt<-tktopleve
Hi All,
I am writing a macro which perform following functionality
1) take a file say excel or csv
2)read all the labels
3) dynamically genrate square,square-root,log of each column with
appropiate column name and write it to a new file,
i am stuck at dynamically genrateing the column names like
Hi Folks,
I am currently trying to present some results I obtained by using the
quantreg package developed by Roger Koenker. After calculating
fit<-summary(rq(Y~X1+X2, tau=2:98/100) ) the function plot(fit) presents a
really nice the results by showing the values for all "regressors" in the
given i
Dear List:
I write codes to get"Mean and standard deviations (Std.) for EM
estimatesâ500 samples from the FM-ST model"(R package mixsmsn).It has a
mistake:"number of cluster centres must lie between 1 and nrow(x) ". my
codes below:
mu1 <- c(2,2)
Sigma1 <- matrix(c(1.5,0,0,1.5), 2,2)
shape1 <-c(-
Hi again Joshua.
I tried your function. I think it's what I need. It works well in the small
example of my first post. But I have difficulties to adapt it to my data.
I'll try to give you another fake example with my real script and kind of
data (you can just copy and paste it to try):
ST1 <-
dat
On Wed, May 23, 2012 at 10:56:43PM -0700, kylmala wrote:
> Hi,
>
> and thanks for replying this. Yes, you are right that the term
> min(24/bb,26/cc) is actually min(bb/24,cc/26) - my mistake. But still I
> don't get it. If the objective function is
>
> #min(m1,m2,m3)
> f.obj <- c(1, 1, 1)
>
> #
On 05/24/2012 05:10 PM, Soheila Khodakarim wrote:
Dear All
How can I transform 1 column to 2 columns in R?
211217_s_at
GO:0005249
211217_s_at
GO:0005251
211217_s_at
GO:0005515
211217_s_at
GO:0015271
211217_s_at
GO:0030955
211217_s_at
GO:0005249
211217_s_at
GO:0005251
Hi,
I have one datagrame
Col1 Col2Col3
Col3
xxx erThis is third record of 1st line.
This is 4th record of first line.
xxy erThis is third record of 1st l
Hi,
I am getting following error message while retrieving records from mysql.
*Error in if (count > 0L) { : missing value where TRUE/FALSE needed
Calls: -> do.call -> ->
Execution halted*
con2 <- dbConnect(MySQL(), user="XXX", password="",dbname="",
host="XXX.X.X.X")
#if(nchar(disname)
take a look at the distr package
library(distr)
U <- Unif()
U2 <- U + U
# or
U2 <- convpow(U, 2)
plot(U2)
# or
curve(d(U2)(x), from = 0, to = 2)
Best
Matthias
On 24.05.2012 08:29, c...@mail.tu-berlin.de wrote:
Hi,
I want to convolve 2 uniform distributed [0,1] variables, so that I get
this gr
Dear All
How can I transform 1 column to 2 columns in R?
211217_s_at
GO:0005249
211217_s_at
GO:0005251
211217_s_at
GO:0005515
211217_s_at
GO:0015271
211217_s_at
GO:0030955
211217_s_at
GO:0005249
211217_s_at
GO:0005251
211217_s_at
GO:0005515
211217_s_at
GO:0015271
Hi,
I want to convolve 2 uniform distributed [0,1] variables, so that I
get this graphic:
http://de.wikipedia.org/wiki/Benutzer:Alfred_Heiligenbrunner/Liste_von_Verteilungsdichten_der_Summe_gleichverteilter_Zufallsvariabler (second
graphic)
if I do
u1<-seq(0,1,length=100)
fu1=dunif(u1,min
Hi,
and thanks for replying this. Yes, you are right that the term
min(24/bb,26/cc) is actually min(bb/24,cc/26) - my mistake. But still I
don't get it. If the objective function is
#min(m1,m2,m3)
f.obj <- c(1, 1, 1)
#and we now that
# a+aa<=15
# b+bb+bbb<=35
# cc+ccc<=40
# so
# 1a + 0b +
Dear List,
Couple of issues while using functions from “BCA” library:
1. I am trying to use “lift.chart” function from “BCA” library, but facing
issues while using model where model formula is passed as formula object in
glm.
When model formula is written as text, then it works fine. In my case
88 matches
Mail list logo
