First, a histogram would not be appropriate (your data appear to be
categorical - a histogram is for continuous numeric vales) - you would need
a bar plot. You should make two vectors (one for the category names and the
other for the frequencies) and use the barplot function.
On Fri, Nov 9, 2018 a
Hello,
I've just tested Jeff's solution, it works but the second code line
should be
dsh <- sh[ , -length( sh ) ]
(dsh doesn't exist yet.)
Hope this helps,
Rui Barradas
Às 02:46 de 10/11/2018, Jeff Newmiller escreveu:
Your file has 5 commas in the first data row, but only 4 in the header.
Your file has 5 commas in the first data row, but only 4 in the header. R
interprets this to mean your first column is intended to be row names (has
no corresponding column label) rather than data. (Row names are "outside"
the data frame... use str(dsh) to get a better picture.)
Basically, you
HI all,
I am trying to read a csv file, but have a problem in the row names.
After reading, the name of the first column is now "row.names" and
all other column names are shifted to the right. The value of the last
column become all NAs( as an extra column).
My sample data looks like as follow,
f
Yes, exactly (heh, heh).
?fisher.test
is probably what is wanted.
For arbitrary rxc tables with fixed marginals, this is a difficult problem.
Mehta's efficient network algorithm to solve it can be found by a web
search on "algorithm for Fisher exact test."
-- Bert
On Fri, Nov 9, 2018 at 12:3
Seems like you are trying to recreate the calculations needed to perform
an exact test. Why not look at the code for that or even easier, just
use the function.
--
David.
On 11/8/18 8:05 PM, li li wrote:
Hi all,
I am trying to list all the 4 by 2 tables with some fixed margins.
For exa
Don't give up on for loops entirely... some of the largest time savings in
optimizing loops are achieved by managing memory effectively. [1]
[1] https://www.r-bloggers.com/r-tip-use-vectormode-list-to-pre-allocate-lists
On November 8, 2018 8:05:39 PM PST, li li wrote:
>Hi all,
> I am trying t
Hello,
You probably want a bar plot, not a histogram.
old.sci <- options(scipen=999)
with(mydata, barplot(number, space = 0, names.arg = name, beside = TRUE))
options(scipen = old.sci)
#-
mydata <- read.table(text = "
name number
ds6277
lk 24375
ax46049
dd70656
What would be the correct code (simplest version) (without gplot())
for histogram (with 7 bars), which would include 7 names of bars under
the X-axis. The data are:
name number
ds6277
lk 24375
ax46049
dd70656
az216544
df 220620
gh641827
(I'm attaching mydata.r, making
Hi,
The missRanger package performs predictive mean matching which should
generate positive values if the non-missing values are positive.
Regards,
Denes
On 11/09/2018 01:11 PM, Rebecca Bingert wrote:
Hi!
How can I generate only positive data with randomForrest-imputation? I'm
working with
Hi!
How can I generate only positive data with randomForrest-imputation? I'm
working with laboratory values which are always positive.
Can anybody help out?
Thanks!
(P.S.: I needed to send this request again because there were problems by
delivering it)
__
Dear all,
I am pleased to announce a new CRAN package, "pgsc" version 1.0.0 <
https://cran.r-project.org/package=pgsc>.
This package implements the extension of the synthetic control method to
allow for continuous and time-varying treatments described in Powell (2017)
. Functions for both estima
12 matches
Mail list logo
