Erin,
I do not think there is an R package that will enable you to get the
data you would like from spotcrime.com.
You could write code, in R, or some other language, to extract the data
you want but that is going to be a changeling task and if
the website changes its format then your code
then my solution should work.
Bob
On 1/30/2017 9:44 AM, greg holly wrote:
> Hi Robert;
>
> I do appreciate your advice. Only the first column of the data is
> text. The rest columns are numeric.
>
> Regards,
>
> Greg
>
> On Mon, Jan 30, 2017 at 9:36 AM, Rober
Greg,
I am assuming that your data is in a text file. R is a good tool but not
the tool I would use for this job. The tool I would
use is grep. The following command should get you want you want:
grep -v "^rs"
Bob
On 1/30/2017 9:23 AM, greg holly wrote:
Hi all;
I have a file
Here is one thought. Assign each month a value of 0, 1 or 2. Then do a
simple linear regression analysis where the value of the month
is the independent variable. You can also do multiple linear regression
with the value you assigned to the month plus the other factors that
you believe are
I only see one for loop in your code. I am wondering if you want a
second for loop based upon the length of newdata.
I would also think that you do not need the second call to set.seed.
Bob
On 1/12/2017 4:44 PM, Jennifer Sheng wrote:
Dear friends, I am working on a double loop using for.
Hi Ragia,
First, when you wrote mad, I assume you mean made. Also, when you say
it is a multi core prog, does that mean it is using threads? running two
or more items in parallel? By any chance are you using this package?
I am not up on the internals of R but there does seem some run for
parallelism. Are we talking about special hardware? or running this on
an Intel Box? If it is the second, then I am thinking threads would be
the way to go. Please consider the following
R statements:
for( i in 1:30 )
This should work:
if ( age > 4 && age < 8 && infection > 0 ) replacement = 2
Bob
On 2/24/2016 7:08 AM, Polychronis KOSTOULAS wrote:
Hi there,
apologies if this is easy. I want to write this condition:
If age is more than 4 years and less or equal to 8 years and infection
is positive
I would like to write a function in R that would take a variable number
of integers as parameters. I do not have a pressing reason to do this, I
am just trying to learn R. I thought a good first step would be to print
out the arguments. So I wrote the following function:
f1 = function (...)
;
}
It does not use nested functions and it works also. Is there a reason
why your solution is better?
Bob
On 2/7/2016 7:14 PM, Ben Tupper wrote:
Hi,
On Feb 7, 2016, at 6:24 PM, Duncan Murdoch <murdoch.dun...@gmail.com> wrote:
On 07/02/2016 6:12 PM, Robert Sherry wrote:
I would lik
AM, S Ellison wrote:
On 23.01.2016 01:21, Robert Sherry wrote:
In R, I run the following commands:
df = data.frame( x=runif(10), y=runif(10) )
df2 = df[order(x),]
You use another x from your workspace, you actually want to
df2 = df[order(df[,"x"]),]
or
df[order(df$x),]
I think this mailing list is wonderful and it has helped me a lot. In
fact, I am not sure I would be using R today if it was not for this
list.
Bob
On 1/24/2016 4:42 PM, Michael Friendly wrote:
On 1/23/2016 7:28 AM, Jean-Luc Dupouey wrote:
Dear members,
Not a technical question:
But one
In R, I run the following commands:
df = data.frame( x=runif(10), y=runif(10) )
df2 = df[order(x),]
The first, as I would expect, creates a data frame with two columns and
10 rows. I expect the second to sort the data based upon
the columns x and produce a new data frame, df2, with the
When I use a table, from a Schaum book, I see that for the 95
percentile, with v_1 = 1 and v_2 = 1 the value is 161. In the modern era,
looking values up in a table is less than ideal. Therefore, I would
expect R to have a function to do this and based upon my
reading of the documentation, I
I am thinking that R is running out of memory. Therefore, I would look
to increase the size of my virtual memory. Here are two links
that might help you with that:
http://windows.microsoft.com/en-us/windows/change-virtual-memory-size#1TC=windows-7
I am trying to use the package quantmod to get option quotes in R.
Therefore, I executed the following two commands:
library ("quantmod" )
getOptionChain("AAPL")
The first one worked but the second one produced the following error
message:
Error in
ckage ‘RSXML’ is not available (for R version 3.1.2)
I am wondering why it is not work. Please help.
Thanks
Bob
On 11/7/2015 6:41 PM, Hasan Diwan wrote:
> Bob,
>
> On 7 November 2015 at 15:27, Robert Sherry <rsher...@comcast.net
> <mailto:rsher...@comcast.net>> wrote:
&g
I have created what I believe to be a data frame. It is called
env1$SPY. The r statement head( env1$SPY ) produces the following output:
SPY.Open SPY.High SPY.Low SPY.Close SPY.Volume SPY.Adjusted
1995-01-03 45.7031 45.8437 45.6875 45.7812 324300 31.55312
1995-01-04 45.9843
Here is a question that I might ask. What are the alternatives to R and
how does R compare? That is, for what class of problems is R the best
tool around?
Bob
On 1/11/2015 1:16 PM, Barry Rowlingson wrote:
Ask if they have a favourite R programmer. This will tell you how much into
the R
I am trying to get the current price of gold for my application. I am
using the library quantmod. The
R commands I use are:
getMetals(c('XAU'), from=Sys.Date(), autoassign = FALSE )
XAUUSD$XAU.USD[1,1]
I would expect the value in XAUUSD$XAU.USD[1,1] to be a scalar but it
comes back
I am using R and quantmod to get stock and option quotes. However, it
has stopped working. I expect the following
function call to produce a list of options:
getOptionChain( XOM, Exp = 2015-01-20 )
However, I get the following error messages:
Error in lapply(strsplit(opt, tr),
I have the following data set:
xy p
11 1/2
22 1/4
39 1/4
In this case, p represents the probability of the values occurring. I
compute the covariance of x and y by hand and come up with a value of 41/16.
When computing the covariance, I am
David,
Thanks for the response. I believe you have solved my problem.
Bob
On 7/26/2014 3:50 PM, David Winsemius wrote:
On Jul 26, 2014, at 11:07 AM, Robert Sherry wrote:
I have the following data set:
xy p
11 1/2
22 1/4
39 1/4
Please consider the following R Script:
x = c(1,2,3)
y = c(1,2,9)
cor(x,y)
These three lines will produce, as I expected, the correlation between
the variables x and y. However, R is going to assume that the
probability that x = 1 is the same as the probability that x = 2.
I created the following file:
symbol,shares
XOM,1000
APA,400
CVX,200
I then read the file in R using the command:
stockList=read.table(/NotesOnR/stockList, header = T, sep=,)
I would then expect the following expression to evaluate to the simple
string APA:
stockList$symbol[2]
Suppose that a data frame has been created by the user. Perhaps it has been
created using the library quantmod. Is there any command to find out what
the members of the data frame is?
Thanks
Bob
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