Thanks Peter for a very promising tip.
On Tue, Aug 25, 2020 at 11:40 AM peter dalgaard wrote:
> If you don't worry too much about an additive constant, then half the
> negative squared deviance residuals should do. (Not quite sure how weights
> factor in. Looks like they are accounted for.)
>
>
Hi Luigi,
Maybe just:
plot(as.numeric(factor(x,levels=x)),y,xaxt="n",
main="Concentration by effect",
xlab="Concentration",ylab="Effect")
axis(1,at=1:4,labels=x)
Jim
On Thu, Aug 27, 2020 at 10:16 PM Luigi Marongiu
wrote:
>
> Hello,
> I have a dataframe as follows:
> ```
> x = c("0 pmol", "10
Hello,
The plots that you say give bars (or my equivalent version below) don't
give bars, what they give are boxplots with just one value and the
median Q1 and Q3 are all equal.
plot(y ~ factor(x), df, pch = 16) # boxplot
Is the following what you are looking for?
plot(y ~ as.integer(fac
I clearly didn't read well enough. As Petr pointed out, there is also
the col_names argument.
```
# Solution 4a
map_dfr(files, function(cur_file, ranges){
map_dfc(ranges, function(cur_range, df){
read_excel(cur_file, sheet = 1, col_names = cur_range, range =
cur_range)
}, df = df)
},
Hi Thomas,
I am not familiar with the use of the range argument, but it seems to me
that the cell value becomes the column name. This might be fine, but you
might get into trouble if you have repeated cell values since
as.data.frame() will fix these.
I am also not sure about what you want, b
exactly what I needed!
Thank you so much
Luigi
On Thu, Aug 27, 2020 at 2:43 PM Jim Lemon wrote:
>
> Hi Luigi,
> Try this:
>
> brkdn.plot(y~x+z,data=Q)
>
> Jim
>
> On Thu, Aug 27, 2020 at 9:57 PM Luigi Marongiu
> wrote:
> >
> > Hello,
> > I have a dataframe as follows
> > ```
> > x = c(rep("1000
Thank you, better than before...
On Thu, Aug 27, 2020 at 2:37 PM PIKAL Petr wrote:
>
> Hi.
> It is probably somewhere in docs, but factors are actually numerics vith
> labels.
>
> So with your original data frame
>
> df$x <- factor(df$x)
> plot(as.numeric(df$x), df$y)
>
> gives you points. You ne
Hi Luigi,
Try this:
brkdn.plot(y~x+z,data=Q)
Jim
On Thu, Aug 27, 2020 at 9:57 PM Luigi Marongiu wrote:
>
> Hello,
> I have a dataframe as follows
> ```
> x = c(rep("1000 pmol", 2), rep("100 pmol", 2), rep("10 pmol", 2),
> rep("0 pmol", 2))
> y = c(2.7642, 2.8192, 2.1976, 2.2816, 1.8929, 1.8883,
Hi.
It is probably somewhere in docs, but factors are actually numerics vith
labels.
So with your original data frame
df$x <- factor(df$x)
plot(as.numeric(df$x), df$y)
gives you points. You need to set labels to x axis though.
Cheers
Petr
> -Original Message-
> From: R-help On Behalf
Hello,
I have a dataframe as follows:
```
x = c("0 pmol", "10 pmol", "100 pmol", "1000 pmol")
y = c(0.9306, 1.8906, 2.2396, 2.7917)
df = data.frame(x, y)
> str(df)
'data.frame': 4 obs. of 2 variables:
$ x: chr "0 pmol" "10 pmol" "100 pmol" "1000 pmol"
$ y: num 0.931 1.891 2.24 2.792
```
I wou
Hello,
I have a dataframe as follows
```
x = c(rep("1000 pmol", 2), rep("100 pmol", 2), rep("10 pmol", 2),
rep("0 pmol", 2))
y = c(2.7642, 2.8192, 2.1976, 2.2816, 1.8929, 1.8883, 1.0051, 0.8561)
z = c(rep("Sample",6), rep("Control", 2))
Q = data.frame(x, y, z, stringsAsFactors = FALSE)
```
I am try
Daniel,
I have used this package with success:
https://cran.r-project.org/web/packages/SASxport/SASxport.pdf
On Mon, Aug 24, 2020 at 3:57 PM Daniel Nordlund
wrote:
> It is still not clear to me (1) if you just want the printed output in
> your SAS list file, or (2) if you want the actual numer
Thanks a lot. I’ve got it just now.
On Wed, Aug 26, 2020 at 6:03 PM peter dalgaard wrote:
> It is because you don't know whether you want it or not.
>
> It is a bit more obvious with integer indexing, as in color[race]: if race
> is NA you don't know what color to put in, but the result should b
Thanks for your reply.
You're right, here is what I did:
> library(foreign)
> sz201401=read.spss("/Users/e.daadmehr/Desktop/Term/LastLast/untitled
folder/2014/1.sav", to.data.frame=TRUE)
Warning message:
In read.spss("/Users/e.daadmehr/Desktop/Term/LastLast/untitled
folder/2014/1.sav", :
/
Thanks guys. but I'm a bit confused. the input is the first column (z[,1]
and z1[,1]).
How is it possible that a subset of a non-NA vector, contains NA?
On Wed, Aug 26, 2020 at 4:58 PM Eric Berger wrote:
> Good point! :-)
>
>
> On Wed, Aug 26, 2020 at 5:55 PM peter dalgaard wrote:
>
>> Offhand,
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