Thank you. It worked. Case closed
On Thu, 11 Jun 2020, 16:48 Rasmus Liland, wrote:
> On 2020-06-10 10:20 +0200, Luigi Marongiu wrote:
> > I have been trying to convert European
> > short dates formatted as dd/mm/yy into the
> > ISO 8601 but the function as.Dates
> > interprets them as American o
> solving Linear Programming Problems with O(L^1.5)
> computational complexity
I'm not an expert on this topic.
However, a quick glance at the topic suggests that these sorts of
algorithms are usually exponential in "n", here the number of
variables/dimensions.
Apparently, "L" is the number of inp
Thank you so much it is getting better (see attach) when I do:
ggplot( data = tmp.tidy) +geom_point( aes(y = -log10(value),x =
CHR,color=key) ,position = "jitter", size=0.5 )
is there is a way to have separation between every chromosome shown
better and also every number of chromosome shown on the
I tried it,
ggplot( data = tmp.tidy) +geom_point( aes(y = BP,x = CHR,color=key)
,position = "jitter" )
I got the attached
On Thu, Jun 11, 2020 at 4:18 PM wrote:
>
> Try adding
> position = "jitter" to the geom_point(...
>
>
>
> On 11 Jun 2020 21:41, Ana Marija wrote:
>
> Hello,
>
> I tried your
Hello,
I tried your code and this is what I got
I really need two groups side by side shown per chromosome as it is here:
https://imgur.com/a/pj40c
on the image there are 4 groups I do have only two
On Thu, Jun 11, 2020 at 11:52 AM wrote:
>
> On 2020-06-11 15:59, Ana Marija wrote:
> > yes all i
On 2020-06-11 15:59, Ana Marija wrote:
yes all in one plot.
So I want key (and therefore color)to be "Pold" and "Pnew" as those I
am comparing per CHR
so I used facet_wrap(~CHR) to create a graph per chromosome (on x-axis)
On the end x-axis would have two strikes of Pold and Pnew (different
color
You're correct. It does work. I was looking at some other result printed. My
bad.
Looks like we can also get the same result using 'SELECT changes()'.
Approach 1:
con <- data.frame(V1 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))
sqldf()
suppressWarnings(sqldf(c(" update con set V1 = 0 where V1 >
yes all in one plot.
So I want key (and therefore color)to be "Pold" and "Pnew" as those I
am comparing per CHR
so I used facet_wrap(~CHR) to create a graph per chromosome (on x-axis)
On the end x-axis would have two strikes of Pold and Pnew (different
colors) per one chromosome, and CHR would go f
On 2020-06-10 10:20 +0200, Luigi Marongiu wrote:
> I have been trying to convert European
> short dates formatted as dd/mm/yy into the
> ISO 8601 but the function as.Dates
> interprets them as American ones
> (mm/dd/yy)
Dear Luigi,
?strptime says:
‘%D’ Date format such as ‘%m/%d/%y’:
There is no real difference between your example and the example I provided.
Both use a data.frame in R and both work for me under R 3.5 with RSQLite 2.2.0
See log below.
Note that there is a bug in R 4.0 related to tcltk that could possibly affect
sqldf as it uses tcltk. A fix has been announced
> Rich Shepard
> on Thu, 11 Jun 2020 06:29:13 -0700 writes:
> On Thu, 11 Jun 2020, Martin Maechler wrote:
>> > Look at Hadley Wickham's 'tidyverse' collection as >
>> described in R for Data Science. There are date,
>> datetime, > and time functions that will do just w
On 2020-06-11 14:54, Ana Marija wrote:
Hello,
I expected it to look like this:
https://imgur.com/a/pj40c
Ah - so all on the one plot? - so you don't want a facet. It puts two
plots side by side (or 22)
where x-axis would be CHR, there is 22 of them
unique(tmp.tidy$CHR)
[1] 1 2 3 4
Hello,
I expected it to look like this:
https://imgur.com/a/pj40c
where x-axis would be CHR, there is 22 of them
> unique(tmp.tidy$CHR)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
I also have two phenotypes (keys) which I would like to compare
> unique(tmp.tidy$key)
[1
Thanks for the response Gabor. Looks like the below example will work when
using SQLite, but in my case I'm just creating a dataframe in R and trying to
update it using sqldf as below and it doesn't seem to work ...
con <- data.frame(V1 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))
sqldf()
sqldf(c("pragm
On Thu, 11 Jun 2020, Martin Maechler wrote:
> Look at Hadley Wickham's 'tidyverse' collection as
> described in R for Data Science. There are date, datetime,
> and time functions that will do just what you want.
I strongly disagree that automatic guessing of date format is a
good ide
On Thu, 11 Jun 2020, Richard O'Keefe wrote:
I would add to this that in an important data set I was working with, most
of the dates were dd/mm/yy but some of them were mm/dd/yy and that led to
the realisation that I couldn't *tell* for about 40% of the dates which
they were. If they were all one
Here is an example. Ignore the warning or use the workaround discussed here
https://github.com/ggrothendieck/sqldf/issues/40
to avoid the warning.
library(sqldf)
sqldf() # use same connection until next sqldf()
sqldf(c("pragma count_changes = 1", "update BOD set demand = 99
where Time > 4"
Hello all, When I execute a SQL using SQLDF, how do I get the number of
records affected? I mean, if I run an UPDATE on a data frame, it doesn't
tell me if and how many records got updated. I've read through the
documentation and there don't seem to be a way to get this info unless it's
done on a
I would add to this that in an important data set I was working with,
most of the dates were dd/mm/yy but some of them were mm/dd/yy and
that led to the realisation that I couldn't *tell* for about 40% of
the dates which they were. If they were all one or the other, no
worries, but when you have p
" not having any access to the resources "
You have internet (as is proven by sending this email)
you even have R cli these days on mobile (I do anyway).
Learn R , code, let me know how it turns out
On Wed, Jun 10, 2020 at 8:46 PM Keshava PRASADa Halemane <
k.prasa...@gmail.com> wrote:
> Frien
> Rich Shepard
> on Wed, 10 Jun 2020 07:44:49 -0700 writes:
> On Wed, 10 Jun 2020, Jeff Newmiller wrote:
>> Fix your format specification? ?strptime
>>> I have been trying to convert European short dates
>>> formatted as dd/mm/yy into the ISO 8601 but the function
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