I'm trying to extract certain records from a dataframe with grepl.
This is based on the comparison between two columns Result and Names. This variable is build like this "WordNumber" but for the same word I have multiple numbers (more than 30), so when I use the grepl expression to get for instance Word1 I get also results that I would like to avoid, like Word12.
Any ideas on how to fix this?
Names <- c("Word1")
colnames(Names) <- name
Results <- c("Word1", "Word11", "Word12", "Word15")
Records <- c("ThisIsTheResultIWant", "notThis", "notThis", "notThis")
Relationships <- data.frame(Results, Records)
Relationships <- subset(Relationships, grepl(paste(Names$name, collapse = "|"), Relationships$Results))
This doesn't work, if I use fixed = TRUE than it doesn't return any result at all (which is weird). I have also tried concatenating the name part with other numbers like this, but with no success:
Relationships <- subset(Relationships, grepl(paste(paste(Names$name, '3', sep = ""), collapse = "|"), Relationships$Results))
Since I'm concatenating I'm not really sure of how to use the \b to enforce a full match.
Any suggestions?
In addition to #Richard's solution, there are multiple ways to enforce a full match.
\b
"\b" is an anchor to identify word before/after pattern
> grepl("\\bWord1\\b",c("Word1","Word2","Word12"))
[1] TRUE FALSE FALSE
\< & \>
"\<" is an escape sequence for the beginning of a word, and ">" is used for end
> grepl("\\<Word1\\>",c("Word1","Word2","Word12"))
[1] TRUE FALSE FALSE
Use ^ to match the start of the string and $ to match the end of the string
Names <-c('^Word1$')
Or, to apply to the entire names vector
Names <-paste0('^',Names,'$')
I think this is just:
Relationships[Relationships$Results==Names,]
If you end up doing ^Word1$ you're just doing a straight subset.
If you have multiple names, then instead use:
Relationships[Relationships$Results %in% Names,]
Related
I'm trying to use filter(grepl()) to match some words in my column. Let's suppose I want to extract the word "Guartelá". In my column, i have variations such as "guartela" "guartelá" and "Guartela". To match upper/lowercase words I'm using (?i). However, I haven't found a good way to match accent/no-accent (i.e., "guartelá" and "guartela").
I know that I can simply substitute á by a, but is there a way to assign the accent-insensitive in the code? It can be base R/tidyverse/any, I don't mind.
Here's how my curent code line is:
cobras <- final %>% filter(grepl("(?i)guartelá", NAME)
| grepl("(?i)guartelá", locality))
Cheers
you can use stri_trans_general fron stringi to remove all accents:
unaccent_chars= stringi::stri_trans_general(c("guartelá","with_é","with_â","with_ô") ,"Latin-ASCII")
unaccent_chars
# [1] "guartela" "with_e" "with_a" "with_o"
# grepl(paste(unaccent_chars,collapse = "|"), string)
You can pass options in OR statements using [ to capture different combinations
> string <- c("Guartelá", "Guartela", "guartela", "guartelá", "any")
> grepl("[Gg]uartel[aá]", string)
[1] TRUE TRUE TRUE TRUE FALSE
Another option using str_detect():
library(tidyverse)
tibble(name = c("guartela","guartelá", "Guartela", "Other")) |>
filter(str_detect(name, "guartela|guartelá|Guartela"))
I am trying to get the host of an IP address from a list of strings.
ips <- c('140.112.204.42', '132.212.14.139', '31.2.47.93', '7.112.221.238')
I want to get the first 2 digits from the ips. output:
ips <- c('140.112', '132.212', '31.2', '7.112')
This is the code that I wrote to convert them:
cat(unlist(strsplit(ips, "\\.", fixed = FALSE))[1:2], sep = ".")
When I check the type of individual ips in the end I get something like this:
140.112 NULL
Not sure what I am doing wrong. If you have some other ideas completely different from this that is completely fine too.
With sub:
ips <- c('140.112.204.42', '132.212.14.139', '31.2.47.93', '7.112.221.238')
sub('\\.\\d+\\.\\d+$', '', ips)
# [1] "140.112" "132.212" "31.2" "7.112"
With str_extract from stringr:
library(stringr)
str_extract(ips, '^\\d+\\.\\d+')
# [1] "140.112" "132.212" "31.2" "7.112"
With strsplit + sapply:
sapply(strsplit(ips, '\\.'), function(x) paste(x[1:2], collapse = '.'))
# [1] "140.112" "132.212" "31.2" "7.112"
With read.table + apply:
apply(read.table(textConnection(ips), sep='.')[1:2], 1, paste, collapse = '.')
#[1] "140.112" "132.212" "31.2" "7.112"
Notes:
sub('\\.\\d+\\.\\d+$', '', ips):
i. \\.\\d+\\.\\d+$ matches a literal dot, a digit one or more times, a literal dot again, and a digit one or more times at the end of the string
ii. sub removes the above match from the string
str_extract(ips, '^\\d+\\.\\d+'):
i. ^\\d+\\.\\d+ matches a digit one or more times, a literal dot and a digit one or more times in the beginning of the string
ii. str_extract extracts the above match from the string
sapply(strsplit(ips, '\\.'), function(x) paste(x[1:2], collapse = '.')):
i. strsplit(ips, '\\.') splits each ip using a literal dot as the delimiter. This returns a list of vectors after the split
ii. With sapply, paste(x[1:2], collapse = '.') is applied to every element of the list, thus taking only the first two numbers from each vector, and collapsing them with a dot as the separator. sapply then coerces the list to a vector, thus returning a vector of the desired ips.
apply(read.table(textConnection(ips), sep='.')[1:2], 1, paste, collapse = '.'):
i. read.table(textConnection(ips), sep='.')[1:2] treats ips as text input and reads it in with dot as a delimiter. Only taking the first two columns.
ii. apply enables paste to be operated on each row, and collapses with a dot.
Could you please try following.
gsub("([0-9]+.[0-9]+)(.*)","\\1",ips)
Explanation: Using gsub function and putting regex there to match digits then DOT then digits in memory's 1st place holder and keeping .* everything after it in 2nd place holder of memory. Then substituting these with \\1 with first regex's value which will be first 2 fields.
One solution is the following:
vapply(strsplit(ips, ".", fixed = TRUE),
function(x) paste(x[1:2], collapse = "."),
character(1L))
vapply applies function(x) to each element of the output of strsplit
strsplit produces a list where each element of the list is the components of the IP addresses separated by "."; setting fixed = TRUE requests to split using the exact value of the splitting string (i.e., "."), not using regex
function(x) takes the first two elements (x[1:2]) of each item coming out of strsplit and pastes them together, seperated by "."
character(1L) tells vapply that each element of the output (i.e., returned from function(x) should be a string of length 1.
Edit: #useR posted this solution right before me (using sapply).
substr is vectorised on the stop argument, so you can use this with a vector of positions before the second dot. regexpr gives the positions of the first match, so if you sub out the first one you can match on the second - which will be conveniently one before it's true position as needed (since you removed the first one).
substr(ips,1,regexpr("\\.",sub("\\.","",ips)))
[1] "140.112" "132.212" "31.2" "7.112"
We can convert the ip addresses to numeric_version class and then format using this base R one-liner that employs no regular expressions:
format(numeric_version(ips)[, 1:2])
[1] "140.112" "132.212" "31.2" "7.112"
Suppose I have the following two strings and want to use grep to see which match:
business_metric_one
business_metric_one_dk
business_metric_one_none
business_metric_two
business_metric_two_dk
business_metric_two_none
And so on for various other metrics. I want to only match the first one of each group (business_metric_one and business_metric_two and so on). They are not in an ordered list so I can't index and have to use grep. At first I thought to do:
.*metric.*[^_dk|^_none]$
But this doesn't seem to work. Any ideas?
You need to use a PCRE pattern to filter the character vector:
x <- c("business_metric_one","business_metric_one_dk","business_metric_one_none","business_metric_two","business_metric_two_dk","business_metric_two_none")
grep("metric(?!.*_(?:dk|none))", x, value=TRUE, perl=TRUE)
## => [1] "business_metric_one" "business_metric_two"
See the R demo
The metric(?!.*(?:_dk|_none)) pattern matches
metric - a metric substring
(?!.*_(?:dk|none)) - that is not followed with any 0+ chars other than line break chars followed with _ and then either dk or none.
See the regex demo.
NOTE: if you need to match only such values that contain metric and do not end with _dk or _none, use a variation, metric.*$(?<!_dk|_none) where the (?<!_dk|_none) negative lookbehind fails the match if the string ends with either _dk or _none.
You can also do something like this:
grep("^([[:alpha:]]+_){2}[[:alpha:]]+$", string, value = TRUE)
# [1] "business_metric_one" "business_metric_two"
or use grepl to match dk and none, then negate the logical when you're indexing the original string:
string[!grepl("(dk|none)", string)]
# [1] "business_metric_one" "business_metric_two"
more concisely:
string[!grepl("business_metric_[[:alpha:]]+_(dk|none)", string)]
# [1] "business_metric_one" "business_metric_two"
Data:
string = c("business_metric_one","business_metric_one_dk","business_metric_one_none","business_metric_two","business_metric_two_dk","business_metric_two_none")
I have the next vector of strings
[1] "/players/playerpage.htm?ilkidn=BRYANPHI01"
[2] "/players/playerpage.htm?ilkidhh=WILLIROB027"
[3] "/players/playerpage.htm?ilkid=THOMPWIL01"
I am looking for a way to retrieve the part of the string that is placed after the equal sign meaning I would like to get a vector like this
[1] "BRYANPHI01"
[2] "WILLIROB027"
[3] "THOMPWIL01"
I tried using substr but for it to work I have to know exactly where the equal sign is placed in the string and where the part i want to retrieve ends
We can use sub to match the zero or more characters that are not a = ([^=]*) followed by a = and replace it with ''.
sub("[^=]*=", "", str1)
#[1] "BRYANPHI01" "WILLIROB027" "THOMPWIL01"
data
str1 <- c("/players/playerpage.htm?ilkidn=BRYANPHI01",
"/players/playerpage.htm?ilkidhh=WILLIROB027",
"/players/playerpage.htm?ilkid=THOMPWIL01")
Using stringr,
library(stringr)
word(str1, 2, sep = '=')
#[1] "BRYANPHI01" "WILLIROB027" "THOMPWIL01"
Using strsplit,
strsplit(str1, "=")[[1]][2]
# [1] "BRYANPHI01"
With Sotos comment to get results as vector:
sapply(str1, function(x){
strsplit(x, "=")[[1]][2]
})
Another solution based on regex, but extracting instead of substituting, which may be more efficient.
I use the stringi package which provides a more powerful regex engine than base R (in particular, supporting look-behind).
str1 <- c("/players/playerpage.htm?ilkidn=BRYANPHI01",
"/players/playerpage.htm?ilkidhh=WILLIROB027",
"/players/playerpage.htm?ilkid=THOMPWIL01")
stri_extract_all_regex(str1, pattern="(?<==).+$", simplify=T)
(?<==) is a look-behind: regex will match only if preceded by an equal sign, but the equal sign will not be part of the match.
.+$ matches everything until the end. You could replace the dot with a more precise symbol if you are confident about the format of what you match. For example, '\w' matches any alphanumeric character, so you could use "(?<==)\\w+$" (the \ must be escaped so you end up with \\w).
I am using grepl() in R to match patterns to a string.
I need to match multiple strings to a common string and return TRUE if they all match.
For example:
a <- 'DEARBORN TRUCK INCDBA'
b <- 'DEARBORN TRUCK INC DBA'
I want to see if all words in variable b are also in variable a.
I can't just use grepl(b, a) because the patterns (spaces) aren't the same.
It seems like it should be something like this:
grepl('DEARBORN&TRUCK&INC&DBA', a)
or
grepl('DEARBORN+TRUCK+INC+DBA', a)
but neither work. I need to compare each individual word in b to a. In this case, since all the words exist in a, it should return TRUE.
Thanks!
Use strsplit to split b into words and then use sapply to perform a grepl on each such word. The result will be a logical vector and if its all TRUE then return TRUE:
all(sapply(strsplit(b, " ")[[1]], grepl, a))
giving:
[1] TRUE
Note: If you are only looking to determine if a and b are the same aside from spaces then remove the spaces from both and compare what is left:
gsub(" ", "", a) == gsub(" ", "", b)