Generate full width character string in R - r

In R, how can I make the following:
convert this string: "my test string"
to something like this ( a full width character string): "my  test  string"
is there a way to do this through hexidecimal character encodings?
Thanks for your help, I'm really not sure how to even start. Perhaps something with {stringr}
I'm trying to get an output similar to what I would expect from this online conversion tool:
http://www.linkstrasse.de/en/%EF%BD%86%EF%BD%95%EF%BD%8C%EF%BD%8C%EF%BD%97%EF%BD%89%EF%BD%84%EF%BD%94%EF%BD%88%EF%BC%8D%EF%BD%83%EF%BD%8F%EF%BD%8E%EF%BD%96%EF%BD%85%EF%BD%92%EF%BD%94%EF%BD%85%EF%BD%92

Here is a possible solution using a function from the archived Nippon package. This is the han2zen function, which can be found here.
x <- "my test string"
han2zen <- function(s){
stopifnot(is.character(s))
zenEisu <- paste0(intToUtf8(65295 + 1:10), intToUtf8(65312 + 1:26),
intToUtf8(65344 + 1:26))
zenKigo <- c(65281, 65283, 65284, 65285, 65286, 65290, 65291,
65292, 12540, 65294, 65295, 65306, 65307, 65308,
65309, 65310, 65311, 65312, 65342, 65343, 65372,
65374)
s <- chartr("0-9A-Za-z", zenEisu, s)
s <- chartr('!#$%&*+,-./:;<=>?#^_|~', intToUtf8(zenKigo), s)
s <- gsub(" ", intToUtf8(12288), s)
return(s)
}
han2zen(x)
# [1] "my test string"

Related

paste specific text to strings that do not have it

I would like to paste "miR" to strings that do not have "miR" already, and skipping those that have it.
paste("miR", ....)
in
c("miR-26b", "miR-26a", "1297", "4465", "miR-26b", "miR-26a")
out
c("miR-26b", "miR-26a", "miR-1297", "miR-4465", "miR-26b", "miR-26a")
One way could be by removing "miR" if it is present in the beginning of the string using sub and pasting it to every string irrespectively.
paste0("miR-", sub("^miR-","", x))
#[1] "miR-26b" "miR-26a" "miR-1297" "miR-4465" "miR-26b" "miR-26a"
data
x <- c("miR-26b", "miR-26a", "1297", "4465", "miR-26b", "miR-26a")
vec <- c("miR-26b", "miR-26a", "1297", "4465", "miR-26b", "miR-26a")
sub("^(?!miR)(.*)$", "miR-\\1", vec, perl = T)
#[1] "miR-26b" "miR-26a" "miR-1297" "miR-4465" "miR-26b" "miR-26a"
If you want to learn more:
type ?sub into R console
learn regex, have a closer look at negative look ahead, capturing groups LEARN REGEX
I've used perl = T because I get an error if I don't. READ MORE

Read the VW raw scores from (CS)OAA

VowpalWabbit writes raw predictions from (CS)OAA model as a sequence of lines like this:
1:-2.31425 2:-3.98557 3:-3.97967 4:-2.63708 5:-3.18749 6:-2.43984 7:-4.99018 8:-3.49138 9:-3.07816 10:-6.15126 11:-6.01152 12:-5.76039 13:-5.13096 14:-5.18472 15:-5.37358 16:-5.24147 17:-5.21512 18:-5.67961 19:-4.62929 20:-4.61404 000db8cd6aef4e5fa459126d36e0fa1f-none
1:-2.65864 2:-3.33924 3:-2.8116 4:-1.83108 5:-2.05677 6:-1.29879 7:-6.7446 8:-3.05036 9:-2.82138 10:-5.19605 11:-4.5119 12:-5.28309 13:-4.35789 14:-4.76992 15:-4.16866 16:-4.6897 17:-3.76224 18:-4.13129 19:-4.4489 20:-4.32605 000e0e58a4cb4a218bbc6cae0b1af201-none
How do I read it into R?
Here is my code:
## load raw vw (CS)OAA scores
read.vw.oaa.scores <- function (myfile) {
v <- sapply(strsplit(readLines(myfile),' ',fixed=TRUE), function (r) {
m <- matrix(unlist(strsplit(head(r,-1),':',fixed=TRUE)),ncol=2,byrow=TRUE)
stopifnot(identical(1:nrow(m),as.integer(m[,1])))
c(tail(r,1),m[,2])
})
f <- as.data.frame(t(v),stringsAsFactors=FALSE)
names(f) <- c("id",head(names(f),-1))
for (n in tail(names(f),-1))
f[[n]] <- as.numeric(f[[n]])
f
}
Are there any obvious bugs/inefficiencies?
Is there a better way?
PS. This data format looks like CRS but it is not it.
See if the following works for you (probably really slow). Assumes all desired values are in numeric:value format. And uses raw which requires each line to be stored as a character array.
raw = c("1:-2.31425 2:-3.98557 3:-3.97967 4:-2.63708 5:-3.18749 6:-2.43984 7:-4.99018 8:-3.49138 9:-3.07816 10:-6.15126 11:-6.01152 12:-5.76039 13:-5.13096 14:-5.18472 15:-5.37358 16:-5.24147 17:-5.21512 18:-5.67961 19:-4.62929 20:-4.61404 000db8cd6aef4e5fa459126d36e0fa1f-none",
"1:-2.65864 2:-3.33924 3:-2.8116 4:-1.83108 5:-2.05677 6:-1.29879 7:-6.7446 8:-3.05036 9:-2.82138 10:-5.19605 11:-4.5119 12:-5.28309 13:-4.35789 14:-4.76992 15:-4.16866 16:-4.6897 17:-3.76224 18:-4.13129 19:-4.4489 20:-4.32605 000e0e58a4cb4a218bbc6cae0b1af201-none")
Function to clean
clean = function(t, n) {as.numeric(gsub("^[0-9]+:", "", unlist(strsplit(t, split=" "))[1:n]))}
lapply(raw, clean, n = 20)
[[1]]
[1] -2.31425 -3.98557 -3.97967 -2.63708 -3.18749 -2.43984 -4.99018 -3.49138 -3.07816 -6.15126 -6.01152 -5.76039
[13] -5.13096 -5.18472 -5.37358 -5.24147 -5.21512 -5.67961 -4.62929 -4.61404
[[2]]
[1] -2.65864 -3.33924 -2.81160 -1.83108 -2.05677 -1.29879 -6.74460 -3.05036 -2.82138 -5.19605 -4.51190 -5.28309
[13] -4.35789 -4.76992 -4.16866 -4.68970 -3.76224 -4.13129 -4.44890 -4.32605

Converting geo coordinates from degree to decimal

I want to convert my geographic coordinates from degrees to decimals, my data are as follows:
lat long
105252 30°25.264 9°01.331
105253 30°39.237 8°10.811
105255 31°37.760 8°06.040
105258 31°41.190 8°06.557
105259 31°41.229 8°06.622
105260 31°38.891 8°06.281
I have this code but I can not see why it is does not work:
convert<-function(coord){
tmp1=strsplit(coord,"°")
tmp2=strsplit(tmp1[[1]][2],"\\.")
dec=c(as.numeric(tmp1[[1]][1]),as.numeric(tmp2[[1]]))
return(dec[1]+dec[2]/60+dec[3]/3600)
}
don_convert=don1
for(i in 1:nrow(don1)){don_convert[i,2]=convert(as.character(don1[i,2])); don_convert[i,3]=convert(as.character(don1[i,3]))}
The convert function works but the code where I am asking the loop to do the job for me does not work.
Any suggestion is apperciated.
Use the measurements package from CRAN which has a unit conversion function already so you don't need to make your own:
x = read.table(text = "
lat long
105252 30°25.264 9°01.331
105253 30°39.237 8°10.811
105255 31°37.760 8°06.040
105258 31°41.190 8°06.557
105259 31°41.229 8°06.622
105260 31°38.891 8°06.281",
header = TRUE, stringsAsFactors = FALSE)
Once your data.frame is set up then:
# change the degree symbol to a space
x$lat = gsub('°', ' ', x$lat)
x$long = gsub('°', ' ', x$long)
# convert from decimal minutes to decimal degrees
x$lat = measurements::conv_unit(x$lat, from = 'deg_dec_min', to = 'dec_deg')
x$long = measurements::conv_unit(x$long, from = 'deg_dec_min', to = 'dec_deg')
Resulting in the end product:
lat long
105252 30.4210666666667 9.02218333333333
105253 30.65395 8.18018333333333
105255 31.6293333333333 8.10066666666667
105258 31.6865 8.10928333333333
105259 31.68715 8.11036666666667
105260 31.6481833333333 8.10468333333333
Try using the char2dms function in the sp library. It has other functions that will additionally do decimal conversion.
library("sp")
?char2dms
A bit of vectorization and matrix manipulation will make your function much simpler:
x <- read.table(text="
lat long
105252 30°25.264 9°01.331
105253 30°39.237 8°10.811
105255 31°37.760 8°06.040
105258 31°41.190 8°06.557
105259 31°41.229 8°06.622
105260 31°38.891 8°06.281",
header=TRUE, stringsAsFactors=FALSE)
x
The function itself makes use of:
strsplit() with the regex pattern "[°\\.]" - this does the string split in one step
sapply to loop over the vector
Try this:
convert<-function(x){
z <- sapply((strsplit(x, "[°\\.]")), as.numeric)
z[1, ] + z[2, ]/60 + z[3, ]/3600
}
Try it:
convert(x$long)
[1] 9.108611 8.391944 8.111111 8.254722 8.272778 8.178056
Disclaimer: I didn't check your math. Use at your own discretion.
Thanks for answers by #Gord Stephen and #CephBirk. Sure helped me out.
I thought I'd just mention that I also found that measurements::conv_unit doesn't deal with "E/W" "N/S" entries, it requires positive/negative degrees.
My coordinates comes as character strings "1 1 1W" and needs to first be converted to "-1 1 1".
I thought I'd share my solution for that.
df <- c("1 1 1E", "1 1 1W", "2 2 2N","2 2 2S")
measurements::conv_unit(df, from = 'deg_min_sec', to = 'dec_deg')
[1] "1.01694444444444" NA NA NA
Warning message:
In split(as.numeric(unlist(strsplit(x, " "))) * c(3600, 60, 1), :
NAs introduced by coercion
ewns <- ifelse( str_extract(df,"\\(?[EWNS,.]+\\)?") %in% c("E","N"),"+","-")
dms <- str_sub(df,1,str_length(df)-1)
df2 <- paste0(ewns,dms)
df_dec <- measurements::conv_unit(df2,
from = 'deg_min_sec',
to = 'dec_deg'))
df_dec
[1] "1.01694444444444" "-1.01694444444444" "2.03388888888889" "-2.03388888888889"
as.numeric(df_dec)
[1] 1.016944 -1.016944 2.033889 -2.033889
Have a look at the command degree in the package OSMscale.
As Jim Lewis commented before it seems your are using floating point minutes. Then you only concatenate two elements on
dec=c(as.numeric(tmp1[[1]][1]),as.numeric(tmp2[[1]]))
Having degrees, minutes and seconds in the form 43°21'8.02 which as.character() returns "43°21'8.02\"", I updated your function to
convert<-function(coord){
tmp1=strsplit(coord,"°")
tmp2=strsplit(tmp1[[1]][2],"'")
tmp3=strsplit(tmp2[[1]][2],"\"")
dec=c(as.numeric(tmp1[[1]][1]),as.numeric(tmp2[[1]][1]),as.numeric(tmp3[[1]]))
c<-abs(dec[1])+dec[2]/60+dec[3]/3600
c<-ifelse(dec[1]<0,-c,c)
return(c)
}
adding the alternative for negative coordinates, and works great for me . I still don't get why char2dms function in the sp library didn't work for me.
Thanks
Another less elegant option using substring instead of strsplit. This will only work if all your positions have the same number of digits. For negative co-ordinates just multiply by -1 for the correct decimal degree.
x$LatDD<-(as.numeric(substring(x$lat, 1,2))
+ (as.numeric(substring(x$lat, 4,9))/60))
x$LongDD<-(as.numeric(substring(x$long, 1,1))
+ (as.numeric(substring(x$long, 3,8))/60))

List and description of all packages in CRAN from within R

I can get a list of all the available packages with the function:
ap <- available.packages()
But how can I also get a description of these packages from within R, so I can have a data.frame with two columns: package and description?
Edit of an almost ten-year old accepted answer. What you likely want is not to scrape (unless you want to practice scraping) but use an existing interface: tools::CRAN_package_db(). Example:
> db <- tools::CRAN_package_db()[, c("Package", "Description")]
> dim(db)
[1] 18978 2
>
The function brings (currently) 66 columns back of which the of interest here are a part.
I actually think you want "Package" and "Title" as the "Description" can run to several lines. So here is the former, just put "Description" in the final subset if you really want "Description":
R> ## from http://developer.r-project.org/CRAN/Scripts/depends.R and adapted
R>
R> require("tools")
R>
R> getPackagesWithTitle <- function() {
+ contrib.url(getOption("repos")["CRAN"], "source")
+ description <- sprintf("%s/web/packages/packages.rds",
+ getOption("repos")["CRAN"])
+ con <- if(substring(description, 1L, 7L) == "file://") {
+ file(description, "rb")
+ } else {
+ url(description, "rb")
+ }
+ on.exit(close(con))
+ db <- readRDS(gzcon(con))
+ rownames(db) <- NULL
+
+ db[, c("Package", "Title")]
+ }
R>
R>
R> head(getPackagesWithTitle()) # I shortened one Title here...
Package Title
[1,] "abc" "Tools for Approximate Bayesian Computation (ABC)"
[2,] "abcdeFBA" "ABCDE_FBA: A-Biologist-Can-Do-Everything of Flux ..."
[3,] "abd" "The Analysis of Biological Data"
[4,] "abind" "Combine multi-dimensional arrays"
[5,] "abn" "Data Modelling with Additive Bayesian Networks"
[6,] "AcceptanceSampling" "Creation and evaluation of Acceptance Sampling Plans"
R>
Dirk has provided an answer that is terrific and after finishing my solution and then seeing his I debated for some time posting my solution for fear of looking silly. But I decided to post it anyway for two reasons:
it is informative to beginning scrapers like myself
it took me a while to do and so why not :)
I approached this thinking I'd need to do some web scraping and choose crantastic as the site to scrape from. First I'll provide the code and then two scraping resources that have been very helpful to me as I learn:
library(RCurl)
library(XML)
URL <- "http://cran.r-project.org/web/checks/check_summary.html#summary_by_package"
packs <- na.omit(XML::readHTMLTable(doc = URL, which = 2, header = T,
strip.white = T, as.is = FALSE, sep = ",", na.strings = c("999",
"NA", " "))[, 1])
Trim <- function(x) {
gsub("^\\s+|\\s+$", "", x)
}
packs <- unique(Trim(packs))
u1 <- "http://crantastic.org/packages/"
len.samps <- 10 #for demo purpose; use:
#len.samps <- length(packs) # for all of them
URL2 <- paste0(u1, packs[seq_len(len.samps)])
scraper <- function(urls){ #function to grab description
doc <- htmlTreeParse(urls, useInternalNodes=TRUE)
nodes <- getNodeSet(doc, "//p")[[3]]
return(nodes)
}
info <- sapply(seq_along(URL2), function(i) try(scraper(URL2[i]), TRUE))
info2 <- sapply(info, function(x) { #replace errors with NA
if(class(x)[1] != "XMLInternalElementNode"){
NA
} else {
Trim(gsub("\\s+", " ", xmlValue(x)))
}
}
)
pack_n_desc <- data.frame(package=packs[seq_len(len.samps)],
description=info2) #make a dataframe of it all
Resources:
talkstats.com thread on web scraping (great beginner
examples)
w3schools.com site on html stuff (very
helpful)
I wanted to try to do this using a HTML scraper (rvest) as an exercise, since the available.packages() in OP doesn't contain the package Descriptions.
library('rvest')
url <- 'https://cloud.r-project.org/web/packages/available_packages_by_name.html'
webpage <- read_html(url)
data_html <- html_nodes(webpage,'tr td')
length(data_html)
P1 <- html_nodes(webpage,'td:nth-child(1)') %>% html_text(trim=TRUE) # XML: The Package Name
P2 <- html_nodes(webpage,'td:nth-child(2)') %>% html_text(trim=TRUE) # XML: The Description
P1 <- P1[lengths(P1) > 0 & P1 != ""] # Remove NULL and empty ("") items
length(P1); length(P2);
mdf <- data.frame(P1, P2, row.names=NULL)
colnames(mdf) <- c("PackageName", "Description")
# This is the problem! It lists large sets column-by-column,
# instead of row-by-row. Try with the full list to see what happens.
print(mdf, right=FALSE, row.names=FALSE)
# PackageName Description
# A3 Accurate, Adaptable, and Accessible Error Metrics for Predictive\nModels
# abbyyR Access to Abbyy Optical Character Recognition (OCR) API
# abc Tools for Approximate Bayesian Computation (ABC)
# abc.data Data Only: Tools for Approximate Bayesian Computation (ABC)
# ABC.RAP Array Based CpG Region Analysis Pipeline
# ABCanalysis Computed ABC Analysis
# For small sets we can use either:
# mdf[1:6,] #or# head(mdf, 6)
However, although working quite well for small array/dataframe list (subset), I ran into a display problem with the full list, where the data would be shown either column-by-column or unaligned. I would have been great to have this paged and properly formatted in a new window somehow. I tried using page, but I couldn't get it to work very well.
EDIT:
The recommended method is not the above, but rather using Dirk's suggestion (from the comments below):
db <- tools::CRAN_package_db()
colnames(db)
mdf <- data.frame(db[,1], db[,52])
colnames(mdf) <- c("Package", "Description")
print(mdf, right=FALSE, row.names=FALSE)
However, this still suffers from the display problem mentioned...

Get function's title from documentation

I would like to get the title of a base function (e.g.: rnorm) in one of my scripts. That is included in the documentation, but I have no idea how to "grab" it.
I mean the line given in the RD files as \title{} or the top line in documentation.
Is there any simple way to do this without calling Rd_db function from tools and parse all RD files -- as having a very big overhead for this simple stuff? Other thing: I tried with parse_Rd too, but:
I do not know which Rd file holds my function,
I have no Rd files on my system (just rdb, rdx and rds).
So a function to parse the (offline) documentation would be the best :)
POC demo:
> get.title("rnorm")
[1] "The Normal Distribution"
If you look at the code for help, you see that the function index.search seems to be what is pulling in the location of the help files, and that the default for the associated find.packages() function is NULL. Turns out tha tthere is neither a help fo that function nor is exposed, so I tested the usual suspects for which package it was in (base, tools, utils), and ended up with "utils:
utils:::index.search("+", find.package())
#[1] "/Library/Frameworks/R.framework/Resources/library/base/help/Arithmetic"
So:
ghelp <- utils:::index.search("+", find.package())
gsub("^.+/", "", ghelp)
#[1] "Arithmetic"
ghelp <- utils:::index.search("rnorm", find.package())
gsub("^.+/", "", ghelp)
#[1] "Normal"
What you are asking for is \title{Title}, but here I have shown you how to find the specific Rd file to parse and is sounds as though you already know how to do that.
EDIT: #Hadley has provided a method for getting all of the help text, once you know the package name, so applying that to the index.search() value above:
target <- gsub("^.+/library/(.+)/help.+$", "\\1", utils:::index.search("rnorm",
find.package()))
doc.txt <- pkg_topic(target, "rnorm") # assuming both of Hadley's functions are here
print(doc.txt[[1]][[1]][1])
#[1] "The Normal Distribution"
It's not completely obvious what you want, but the code below will get the Rd data structure corresponding to the the topic you're interested in - you can then manipulate that to extract whatever you want.
There may be simpler ways, but unfortunately very little of the needed coded is exported and documented. I really wish there was a base help package.
pkg_topic <- function(package, topic, file = NULL) {
# Find "file" name given topic name/alias
if (is.null(file)) {
topics <- pkg_topics_index(package)
topic_page <- subset(topics, alias == topic, select = file)$file
if(length(topic_page) < 1)
topic_page <- subset(topics, file == topic, select = file)$file
stopifnot(length(topic_page) >= 1)
file <- topic_page[1]
}
rdb_path <- file.path(system.file("help", package = package), package)
tools:::fetchRdDB(rdb_path, file)
}
pkg_topics_index <- function(package) {
help_path <- system.file("help", package = package)
file_path <- file.path(help_path, "AnIndex")
if (length(readLines(file_path, n = 1)) < 1) {
return(NULL)
}
topics <- read.table(file_path, sep = "\t",
stringsAsFactors = FALSE, comment.char = "", quote = "", header = FALSE)
names(topics) <- c("alias", "file")
topics[complete.cases(topics), ]
}

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