Try the code below:
df <- read_delim("C:/Users/lruan1/Desktop/1112.csv", "|", escape_double =
FALSE, trim_ws = TRUE)
df_new <- subset(df,df$IPC == 'H04M001/02'| df$IPC == 'C07K016/26' )
You can add more condition with "|" in the subset function. Good luck!
On Wed, Jan 3, 2018 at 2:53 PM, Sapto
Hello,
If you want to select rows with just one IPC, use `==`.
If you want to select rows with several IPC's, use `%in%`.
See the code below for the two ways of doing this.
oecd <- read.table(text = "
Appln_id|Prio_Year|App_year|IPC
1|1999|2000|H04Q007/32
1|1999|2000|G06K019/077
1|1999|2000|H01
> On Jan 3, 2018, at 11:57 AM, Andras Farkas via R-help
> wrote:
>
> Dear All,
> using the example from the help of summary.rms
>
> library(rms)
> n <- 1000# define sample size
> set.seed(17) # so can reproduce the results
> age<- rnorm(n, 50, 10)
> blood.pressure <- rnorm(n
Dear All,
using the example from the help of summary.rms
library(rms)
n <- 1000# define sample size
set.seed(17) # so can reproduce the results
age<- rnorm(n, 50, 10)
blood.pressure <- rnorm(n, 120, 15)
cholesterol<- rnorm(n, 200, 25)
sex<- factor(sample(c('fem
Hello,
I have a data of Patents from OECD in delimited text format with IPC being
one column, I want to filter the data by selecting only certain IPC in that
column and delete other rows which do not have my required IPCs. Please,
can anybody guide me doing it, also the IPC codes are string variab
Hello,
I believe the following regex will do it.
x <- "\":\"03-JAN-2018 16:00:00\""
sub('^.*(\\d{2}-\\w{3}-\\d{4} \\d{2}:\\d{2}:\\d{2})[:"]', '\\1', x)
Hope this helps,
Rui Barradas
On 1/3/2018 2:26 PM, Christofer Bogaso wrote:
Hi,
I was working on following expression :
"\":\"03-JAN-201
Hi Lindsey,
Yeah, I do realize that this answer gets to you 12 years too late, however, it
may be helpful to others in future.
I have been struggling with the same issue. When analysing the source code
(https://svn.r-project.org/R-packages/trunk/nlme/R/nlme.R)
I found out that the E matrix needs
On Wed, Jan 03, 2018 at 07:56:27PM +0530, Christofer Bogaso wrote:
> Hi,
>
> I was working on following expression :
>
> "\":\"03-JAN-2018 16:00:00\""
>
>
> This is basically a combination of Date and Time mixed with some Noise.
>
> I want to extract only Date and Time part i.e. "03-JAN-2018 1
Hi,
I was working on following expression :
"\":\"03-JAN-2018 16:00:00\""
This is basically a combination of Date and Time mixed with some Noise.
I want to extract only Date and Time part i.e. "03-JAN-2018 16:00:00
I tried following :
gsub("![0-9][0-9]-[a-zA-Z][a-zA-Z][a-zA-Z]-[0-9][0-9][0-9
On 3 January 2018 at 08:36, Jeff Newmiller wrote:
> Function arguments are not pass-by-reference... they are pass-by-value. You
> need to return the altered object to the caller when you are done messing
> with it.
Thanks.
Of course. Now it is working.
> Note that R is a data processing langu
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