I want to remove blanks, periods and hyphens from a string and enclose each element of the resulting string by inverted commas. Furthermore, I'd like to ensure that all letters are uppercase.
I know how to remove a list of special characters but I cannot add enclosing inverted commas due to my lack of experience with regular expressions or other string manipulation functions (e.g., stringr functions).
How can I convert a string such as
test1 <- "A.1, b-1, C" # start string
test2 <- gsub("[ .-]", "", test1) # remove period and hyphen
to generate the string 'A1','B1','C'?
We can use strsplit on , followed by zero or more spaces after removing the .- with gsub and converting the case to upper case
strsplit(gsub("[.-]", "", toupper(test1)), ",\\s*")[[1]]
#[1] "A1" "B1" "C"
If we need a single string, after removing the ., and -, capture the word (\\w+) and replace it by wrapping the ' around the backreference (\\1) of the captured group
gsub('(\\w+)', "'\\1'", gsub("[.-]+", "", toupper(test1)))
#[1] "'A1', 'B1', 'C'"
Use
test1 <- "A.1, b-1, C"
test2 <- gsub("[^,\\w]+", "", test1, perl=TRUE)
test2 <- paste0("'", gsub(",", "','", toupper(test2)), "'")
Remove all non-word characters other than commas with gsub("[^,\\w]+", "", test1, perl=TRUE) and then replace all commas with commas inside quotes and wrap with quotes using paste0("'", gsub(",", "','", toupper(test2)), "'").
Related
I have a character string of names which look like
"_6302_I-PAL_SPSY_000237_001"
I need to remove the first occurred underscore, so that it will be as
"6302_I-PAL_SPSY_000237_001"
I aware of gsub but it removes all of underscores. Thank you for any suggestions.
gsub function do the same, to remove starting of the string symbol ^ used
x <- "_6302_I-PAL_SPSY_000237_001"
x <- gsub("^\\_","",x)
[1] "6302_I-PAL_SPSY_000237_001"
We can use sub with pattern as _ and replacement as blanks (""). This will remove the first occurrence of '_'.
sub("_", "", str1)
#[1] "6302_I-PAL_SPSY_000237_001"
NOTE: This will remove the first occurence of _ and it will not limit based on the position i.e. at the start of the string.
For example, suppose we have string
str2 <- "6302_I-PAL_SPSY_000237_001"
sub("_", "", str2)
#[1] "6302I-PAL_SPSY_000237_001"
As the example have _ in the beginning, another option is substring
substring(str1, 2)
#[1] "6302_I-PAL_SPSY_000237_001"
data
str1 <- "_6302_I-PAL_SPSY_000237_001"
This can be done with base R's trimws() too
string1<-"_6302_I-PAL_SPSY_000237_001"
trimws(string1, which='left', whitespace = '_')
[1] "6302_I-PAL_SPSY_000237_001"
In case we have multiple words with leading underscores, we may have to include a word boundary (\\b) in our regex, and use either gsub or stringr::string_remove:
string2<-paste(string1, string1)
string2
[1] "_6302_I-PAL_SPSY_000237_001 _6302_I-PAL_SPSY_000237_001"
library(stringr)
str_remove_all(string2, "\\b_")
> str_remove_all(string2, "\\b_")
[1] "6302_I-PAL_SPSY_000237_001 6302_I-PAL_SPSY_000237_001"
I have a string as"dfgdf" sa"2323":
a <- "as\"dfgdf\" sa\"2323\""
The delimiter (same for the start and the end) here is ". So what I want is to get a string were everything is deleted within delimiter but not delimiter itself. So the end result string should look like as"" sa""
You could match " and forget what is matched using \K
Then use a negated character class matching any char except " or a whitespace character and use lookarounds to assert " to the right.
Use perl=TRUE to enable Perl-like regular expressions.
a <- "as\"dfgdf\" sa\"2323\""
gsub('"\\K[^"\\s]+(?=")', "", a, perl=TRUE)
Output
[1] "as\"\" sa\"\""
R demo
Here is another base R option using paste0 + strsplit
s <- paste0(paste0(unlist(strsplit(a, '"\\w+"')), '""'), collapse = "")
which gives
> s
[1] "as\"\" sa\"\""
> cat(s)
as"" sa""
Here is one option with a regex lookaround to match a word (\\w+) that succeeds a double quote and precedes one as pattern and is replaced by blank ("")
cat(gsub('(?<=")\\w+(?=")', "", a, perl = TRUE), "\n")
#as"" sa""
Or without regex lookaround
cat(gsub('"\\w+"', '""', a), "\n")
#as"" sa""
I also found a way with stringr library:
library(stringr)
a <- "as\"dfgdf\" sa\"2323\""
result <- str_replace_all(a, "\".*?\"", "\"\"")
cat(result)
I am looking to manipulate/cut a character string from right to left until a particular character is met
I want to take this:
a <- "L1.L2.L3.L4.L5"
And output this: a <- "L5"
I have specifically worded this problem as needing to cut the string from right to left because the strings can be of variable length and the output string can be of variable length as well
For example the code needs to work on:
b <- "L1.L555"
c <- "L1.L2.L3.L4.L5.L6.LLLL"
We can use sub to match characters (.*) until a . (. is a metacharacter for any character. So we escape (\\) to evaluate it literally) and replace it with blank ("")
sub(".*\\.", "", a)
#[1] "L5"
sub(".*\\.", "", b)
#[1] "L555"
sub(".*\\.", "", c)
#[1] "LLLL"
Or using trimws
trimws(a, whitespace = ".*\\.")
#[1] "L5"
I want to write a regex in R to remove all words of a string containing numbers.
For example:
first_text = "a2c if3 clean 001mn10 string asw21"
second_text = "clean string
Try with gsub
trimws(gsub("\\w*[0-9]+\\w*\\s*", "", first_text))
#[1] "clean string"
It is easier to select words with no numbers than to select and delete words with numbers:
> library(stringr)
> str1 <- "a2c if3 clean 001mn10 string asw21"
> paste(unlist(str_extract_all(str1, "(\\b[^\\s\\d]+\\b)")), collapse = " ")
[1] "clean string"
Note:
Backslashes have to be escaped in R to work properly, hence double backslashes
\b is word boundary
\s is white space
\d is digit character
a caret (^) inside square brackets is a negater: find characters that do not match ...
"+" after the character group inside [] means "1 or more" occurrences of those (non white space and non digit) characters
Just another alternative using gsub
trimws(gsub("[^\\s]*[0-9][^\\s]*", "", first_text, perl=T))
#[1] "clean string"
A bit longer than some of the answers but very tractable is to first convert the string to a vector of words, then check word by word if there are any numbers and use standard R subsetting.
first_text_vec <- strsplit(first_text, " ")[[1]]
first_text_vec
[1] "a2c" "if3" "clean" "001mn10" "string" "asw21"
paste(first_text_vec[!grepl("[0-9]", first_text_vec)], collapse = " ")
[1] "clean string"
I have this kind of char vector:
"MODIS.evi.2013116.yL2.BOKU.tif"
The number in the middle of the vector is gonna change. And the evi word will change to ndvi some times.
I want to use substr (or other function, maybe) to sub-string the vector after the second point: ., ie, just take the 2013116.yL2.BOKU.tif, even when the string is MODIS.evi.2013116.yL2.BOKU.tif or MODIS.ndvi.2013116.yL2.BOKU.tif.
We can use sub to match two instance of one or more characters that are not a . followed by a . from the start (^) of the string and replace it with blank ("")
sub("^([^.]+\\.){2}", "", str1)
#[1] "2013116.yL2.BOKU.tif" "2013116.yL2.BOKU.tif"
If the pattern to keep always start with numbers, then the above can be simplified to match only one or more non-numeric characters and replace it with blank from the start (^) of the string
sub("^\\D+", "", str1)
#[1] "2013116.yL2.BOKU.tif" "2013116.yL2.BOKU.tif"
data
str1 <- c("MODIS.evi.2013116.yL2.BOKU.tif", "MODIS.ndvi.2013116.yL2.BOKU.tif")
This deletes all leading non-digit characters in s :
sub("^\\D*", "", s)
If s is as in the Note at the end then the result of running the above is:
[1] "2013116.yL2.BOKU.tif" "2013116.yL2.BOKU.tif"
Note:
s <- c("MODIS.evi.2013116.yL2.BOKU.tif", "MODIS.ndvi.2013116.yL2.BOKU.tif")
l = c("MODIS.evi.2013116.yL2.BOKU.tif","MODIS.ndvi.2013116.yL2.BOKU.tif")
sapply(l, function(x) strsplit(x, "vi.", fixed = T)[[1]][2])