I see numerous backticks in your code, not quotes. "`" and "'" are not the
same. Backticks are not string delimiters.
As for valid names: look at the help page for make.names().
HTH,
Boris
On Feb 22, 2016, at 1:32 PM, Burhan ul haq wrote:
> Hi,
>
> # 1) I have read in a CSV file
>
> df =
Hi,
# 1) I have read in a CSV file
df = read.csv(file="GiftCards - v1.csv",stringsAsFactors=FALSE)
head(df)
str(df)
# 2) converted to a tbl_df
df2 = tbl_df(df)
# 3) fixed the names to remove leading "X" character
n = names(df2)
n2 = gsub(pattern="^\\w","\\1",n)
names(df2) = n2
# 4) somehow the
r-h...@stat.math.ethz.ch
Cc:
Sent: Wednesday, May 1, 2013 4:37 AM
Subject: [R] grep help (character ommission)
Hello,
Banging my head against a wall here ... can anyone light the way to a
pattern modification that would make the following TRUE?
identical(
grep(
"^Intensity\\s[^HL]&qu
Hello,
The following pattern seems to do it.
grep("^Intensity$|^Intensity\\s[^HL]",
c("Intensity","Intensity L", "Intensity H", "Intensity Rep1"))
Hope this helps,
Rui Barradas
Em 01-05-2013 09:37, Johannes Graumann escreveu:
Hello,
Banging my head against a wall here ... can anyone li
try this:
> identical(
+ grep(
+ "^Intensity *[HL]",
+ c("Intensity","Intensity L", "Intensity H", "Intensity Rep1"),
+ invert = TRUE),
+ as.integer(c(1,4)))
[1] TRUE
>
On Wed, May 1, 2013 at 4:37 AM, Johannes Graumann
wrote:
> Hello,
>
> Banging my head against a wall here ...
Hello,
Banging my head against a wall here ... can anyone light the way to a
pattern modification that would make the following TRUE?
identical(
grep(
"^Intensity\\s[^HL]",
c("Intensity","Intensity L", "Intensity H", "Intensity Rep1")),
as.integer(c(1,4)))
Thank you for your time.
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