I have a list of IP address pairs separated by "::".
ip_pairs <- c("104.124.199.136::192.168.1.67", "104.124.199.136::192.168.137.174", "192.168.1.67::104.124.199.136", "192.168.137.174::104.124.199.136")
As you can see, the third and fourth elements of the vector are the same as the first two, but reversed (my actual problem is to find all unique pairings of IPs, so the solution would drop the pair B::A if A::B is already present. This could be solved using stringr or regex, I'm guessing.
One option:
library(stringr)
split_function = function(x) {
x = sort(x)
paste(x, collapse="::")
}
pairs = str_split(ip_pairs, "::")
unique(sapply(pairs, split_function))
[1] "104.124.199.136::192.168.1.67" "104.124.199.136::192.168.137.174"
Use read.table to create a two column data frame from the pairs, sort each row and find the duplicates using duplicated. Then extract out the non-duplicates. No packages are used.
DF <- read.table(text = ip_pairs, sep = ":")[-2]
ip_pairs[! duplicated(t(apply(DF, 1, sort)))]
## [1] "192.168.1.67::104.124.199.136" "192.168.137.174::104.124.199.136"
Related
Example:
df <- data.frame(Name = c("J*120_234_458_28", "Z*23_205_a834_306", "H*_39_004_204_99_04902"))
I would like to be able to select everything before the third underscore for each row in the dataframe. I understand how to split the string apart:
df$New <- sapply(strsplit((df$Name),"_"), `[`)
But this places a list in each row. I've thus far been unable to figure out how to use sapply to unlist() each row of df$New select the first N elements of the list to paste/collapse them back together. Because the length of each subelement can be distinct, and the number of subelements can also be distinct, I haven't been able to figure out an alternative way of getting this info.
We specify the 'n', after splitting the character column by '_', extract the n-1 first components
n <- 4
lapply(strsplit(as.character(df$Name), "_"), `[`, seq_len(n - 1))
If we need to paste it together, can use anonymous function call (function(x)) after looping over the list with lapply/sapply, get the first n elements with head and paste them together`
sapply(strsplit(as.character(df$Name), "_"), function(x)
paste(head(x, n - 1), collapse="_"))
#[1] "J*120_234_458" "Z*23_205_a834" "H*_39_004"
Or use regex method
sub("^([^_]+_[^_]+_[^_]+)_.*", "\\1", df$Name)
#[1] "J*120_234_458" "Z*23_205_a834" "H*_39_004"
Or if the 'n' is really large, then
pat <- sprintf("^([^_]+){%d}[^_]+).*", n-1)
sub(pat, "\\1", df$Name)
Or
sub("^(([^_]+_){2}[^_]+)_.*", "\\1", df$Name)
#[1] "J*120_234_458" "Z*23_205_a834" "H*_39_004"
I would like to add a number indicating the x^th occurrence of a word in a vector. (So this question is different from Make a column with duplicated values unique in a dataframe , because I have a simple vector and try to avoid the overhead of casting it to a data.frame).
E.g. for the vector:
book, ship, umbrella, book, ship, ship
the output would be:
book, ship, umbrella, book2, ship2, ship3
I have solved this myself by transposing the vector to a dataframe and next using the grouping function. That feels like using a sledgehammer to crack nuts:
# add consecutive number for equal string
words <- c("book", "ship", "umbrella", "book", "ship", "ship")
# transpose word vector to data.frame for grouping
df <- data.frame(words = words)
df <- df %>% group_by(words) %>% mutate(seqN = row_number())
# combine columns and remove '1' for first occurrence
wordsVec <- paste0(df$words, df$seqN)
gsub("1", "", wordsVec)
# [1] "book" "ship" "umbrella" "book2" "ship2" "ship3"
Is there a more clean solution, e.g. using the stringr package?
You can still utilize row_number() from dplyr but you don't need to convert to data frame, i.e.
sub('1$', '', ave(words, words, FUN = function(i) paste0(i, row_number(i))))
#[1] "book" "ship" "umbrella" "book2" "ship2" "ship3"
Another option is to use make.unique along with gsubfn to increment your values by 1, i.e.
library(gsubfn)
gsubfn("\\d+", function(x) as.numeric(x) + 1, make.unique(words))
#[1] "book" "ship" "umbrella" "book.2" "ship.2" "ship.3"
I have a character vector in which long names are used, which will consist of several words connected by delimiters in the form of a dot.
x <- c("Duschekia.fruticosa..Rupr...Pouzar",
"Betula.nana.L.",
"Salix.glauca.L.",
"Salix.jenisseensis..F..Schmidt..Flod.",
"Vaccinium.minus..Lodd...Worosch")
The length of the names is different. But only the first two words of the entire name are important.
My goal is to get names up to 7 symbols: 3 initial symbols from the first two words and a separator in the form of a "dot" between them.
Very close to my request are these examples, but I do not know how to apply these code variations to my case.
R How to remove characters from long column names in a data frame and
how to append names to " column names" of the output data frame in R?
What should I do to get exit names to look like this?
x <- c("Dus.fru",
"Bet.nan",
"Sal.gla",
"Sal.jen",
"Vac.min")
Any help would be appreciated.
You can do the following:
gsub("(\\w{1,3})[^\\.]*\\.(\\w{1,3}).*", "\\1.\\2", x)
# [1] "Dus.fru" "Bet.nan" "Sal.gla" "Sal.jen" "Vac.min"
First we match up to 3 characters (\\w{1,3}), then ignore anything which is not a dot [^\\.]*, match a dot \\. and then again up to 3 characters (\\w{1,3}). Finally anything, that comes after that .*. We then only use the things in the brackets and separate them with a dot \\1.\\2.
Split on dot, substring 3 characters, then paste back together:
sapply(strsplit(x, ".", fixed = TRUE), function(i){
paste(substr(i[ 1 ], 1, 3), substr(i[ 2], 1, 3), sep = ".")
})
# [1] "Dus.fru" "Bet.nan" "Sal.gla" "Sal.jen" "Vac.min"
Here a less elegant solution than kath's, but a bit more easy to read, if you are not an expert in regex.
# Your data
x <- c("Duschekia.fruticosa..Rupr...Pouzar",
"Betula.nana.L.",
"Salix.glauca.L.",
"Salix.jenisseensis..F..Schmidt..Flod.",
"Vaccinium.minus..Lodd...Worosch")
# A function that takes three characters from first two words and merges them
cleaner_fun <- function(ugly_string) {
words <- strsplit(ugly_string, "\\.")[[1]]
short_words <- substr(words, 1, 3)
new_name <- paste(short_words[1:2], collapse = ".")
return(new_name)
}
# Testing function
sapply(x, cleaner_fun)
[1]"Dus.fru" "Bet.nan" "Sal.gla" "Sal.jen" "Vac.min"
I have a long character vector of protein names which I want to reduce.
I want to remove from the vector all entries that are == "5-FCL-like_protein" and all entries that start with "CBSS-"
For the first problem, I can just use %in%
remove <- c("5-FCL-like_protein")
vec[! vec %in% remove]
But how can I include the entries that start with "CBSS-" as well?
Thank you.
You can use two conditions in your subset. The first one is very similar to your %in% except I use == instead just because of personal preference. If you have multiple strings you want to exclude you can go back to %in%. The second one uses grepl to match "CBSS-" at the beginning of the string.
vec <- c("Protein1","Protein2", "CBSS-Protein 2", "5-FCL-like_protein")
vec[!vec == "5-FCL-like_protein" & !grepl("^CBSS-", vec)]
#[1] "Protein1" "Protein2"
Or we can use this within grep
grep("^(CBSS|5-FCL-like_protein$)", vec, value = TRUE, invert = TRUE)
#[1] "Protein1" "Protein2"
data
vec <- c("Protein1","Protein2", "CBSS-Protein 2", "5-FCL-like_protein")
I' ve got a list of names.
l1 <- rep(paste("Session", 1:6, sep=""), each=4)
l2 <- rep(paste("ID", 1:4, sep=""), 6)
list <- paste(l1, l2, sep="")
With real data the list is far more complicated ;)
How do create a new list from this list, that includes only those items from Session 1-4?
In dplyr there is the >>select(contains("Session1"|"Session2"))<< which is used to select variables in data.frames.
I am looking for something similar to use for lists.
Is this what you want ?
list[grepl("Session(1|2|3|4)ID", list)]
[1] "Session1ID1" "Session1ID2" "Session1ID3" "Session1ID4" "Session2ID1" "Session2ID2" "Session2ID3" "Session2ID4"
[9] "Session3ID1" "Session3ID2" "Session3ID3" "Session3ID4" "Session4ID1" "Session4ID2" "Session4ID3" "Session4ID4"
Here is another option with regex lookarounds to match "Session" followed by numbers 1-4 and not followed by any number ((?![0-9]))
grep("Session[1-4](?![0-9])", c(list, "Session10ID"), value = TRUE, perl = TRUE)
#[1] "Session1ID1" "Session1ID2" "Session1ID3" "Session1ID4" "Session2ID1" "Session2ID2" "Session2ID3" "Session2ID4" "Session3ID1" "Session3ID2"
#[11] "Session3ID3" "Session3ID4" "Session4ID1" "Session4ID2" "Session4ID3" "Session4ID4"