Remove quotes from a character vector in R
RQuotesR Problem Overview
Suppose you have a character vector:
char <- c("one", "two", "three")
When you make reference to an index value, you get the following:
> char[1]
[1] "one"
How can you strip off the quote marks from the return value to get the following?
[1] one
R Solutions
Solution 1 - R
Just try noquote(a)
> noquote("a")
[1] a
Solution 2 - R
as.name(char[1]) will work, although I'm not sure why you'd ever really want to do this -- the quotes won't get carried over in a paste for example:
> paste("I am counting to", char[1], char[2], char[3])
[1] "I am counting to one two three"
Solution 3 - R
There are no quotes in the return value, only in the default output from print() when you display the value. Try
> print(char[1], quote=FALSE)
[1] one
or
> cat(char[1], "\n")
one
to see the value without quotes.
Solution 4 - R
You are confusing quantmod's 'symbol' (a term relating to a code for some financial thingamuwot) with R's 'symbol', which is a 'type' in R.
You've said:
> I have a character vector of stock > symbols that I pass to > quantmod::getSymbols() and the > function returns the symbol to the > environment without the quotes
Well almost. What it does is create objects with those names in the specified environment. What I think you want to do is to get things out of an environment by name. And for that you need 'get'. Here's how, example code, working in the default environment:
> getSymbols('F',src='yahoo',return.class='ts') [1] "F"
so you have a vector of characters of the things you want:
> z="F"
> z
[1] "F"
and then the magic:
> summary(get(z))
F.Open F.High F.Low F.Close
Min. : 1.310 Min. : 1.550 Min. : 1.010 Min. : 1.260
1st Qu.: 5.895 1st Qu.: 6.020 1st Qu.: 5.705 1st Qu.: 5.885
Median : 7.950 Median : 8.030 Median : 7.800 Median : 7.920
Mean : 8.358 Mean : 8.495 Mean : 8.178 Mean : 8.332
3rd Qu.:11.210 3rd Qu.:11.400 3rd Qu.:11.000 3rd Qu.:11.180
Max. :18.810 Max. :18.970 Max. :18.610 Max. :18.790
and if you don't believe me:
> identical(F,get(z))
[1] TRUE
Solution 5 - R
If:
> char<-c("one", "two", "three")
You can:
> print(char[1],quote = FALSE)
Your result should be:
[1] one
Solution 6 - R
I'm just guessing, is this in the ball park of what you're trying to achieve?
> a <- "a"
> a
[1] "a" # quote yes
> as.factor(a)
[1] a #quote no
Solution 7 - R
Easiest way is :
> a = "some string"
> write(a, stdout()) # Can specify stderr() also.
some string
Gives you the option to print to stderr if you're doing some error handling printing.
Solution 8 - R
I think I was trying something very similar to the original poster. the get() worked for me, although the name inside the chart was not inherited. Here is the code that worked for me.
#install it if you dont have it
library(quantmod)
# a list of stock tickers
myStocks <- c("INTC", "AAPL", "GOOG", "LTD")
# get some stock prices from default service
getSymbols(myStocks)
# to pause in between plots
par(ask=TRUE)
# plot all symbols
for (i in 1:length(myStocks)) {
chartSeries(get(myStocks[i]), subset="last 26 weeks")
}
Solution 9 - R
Here is one combining noquote and paste:
noquote(paste("Argument is of length zero",sQuote("!"),"and",dQuote("double")))
#[1] Argument is of length zero ‘!’ and “double”
Solution 10 - R
Try this: (even [1] will be removed)
> cat(noquote("love"))
love
else just use noquote
> noquote("love")
[1] love
Solution 11 - R
nump function :)
> nump <- function(x) print(formatC(x, format="fg", big.mark=","), quote=FALSE)
correct answer: > x <- 1234567890123456
> nump(x)
[1] 1,234,567,890,123,456