I have a string s where "substrings" are divided by a pipe. Substrings might or might not contain numbers. And I have a test character string n that contains a number and might or might not contain letters. See example below. Note that spacing can be any
I'm trying to drop all substrings where n is not in a range or is not an exact match. I understand that I need to split by -, convert to numbers, and compare low/high to n converted to numeric. Here's my starting point, but then I got stuck with getting the final good string out of unl_new.
s = "liquid & bar soap 1.0 - 2.0oz | bar 2- 5.0 oz | liquid soap 1-2oz | dish 1.5oz"
n = "1.5oz"
unl = unlist(strsplit(s,"\\|"))
unl_new = (strsplit(unl,"-"))
unl_new = unlist(gsub("[a-zA-Z]","",unl_new))
Desired output:
"liquid & bar soap 1.0 - 2.0oz | liquid soap 1-2oz | dish 1.5oz"
Am I completely on the wrong path? Thanks!
Here an option using r-base ;
## extract the n numeric
nn <- as.numeric(gsub("[^0-9|. ]", "", n))
## keep only numeric and -( for interval)
## and split by |
## for each interval test the condition to create a boolean vector
contains_n <- sapply(strsplit(gsub("[^0-9|. |-]", "", s),'[|]')[[1]],
function(x){
yy <- strsplit(x, "-")[[1]]
yy <- as.numeric(yy[nzchar(yy)])
## the condition
(length(yy)==1 && yy==nn) || length(yy)==2 && nn >= yy[1] && nn <= yy[2]
})
## split again and use the boolean factor to remove the parts
## that don't respect the condition
## paste the result using collapse to get a single character again
paste(strsplit(s,'[|]')[[1]][contains_n],collapse='')
## [1] "liquid & bar soap 1.0 - 2.0oz liquid soap 1-2oz dish 1.5oz"
Don't know if it is general enough, but you might try:
require(stringr)
splitted<-strsplit(s,"\\|")[[1]]
ranges<-lapply(strsplit(
str_extract(splitted,"[0-9\\.]+(\\s*-\\s*[0-9\\.]+|)"),"\\s*-\\s*"),
as.numeric)
tomatch<-as.numeric(str_extract(n,"[0-9\\.]+"))
paste(splitted[
vapply(ranges, function(x) (length(x)==1 && x==tomatch) || (length(x)==2 && findInterval(tomatch,x)==1),TRUE)],
collapse="|")
#[1] "liquid & bar soap 1.0 - 2.0oz | liquid soap 1-2oz | dish 1.5oz"
Here's a method starting from your unl step using stringr:
unl = unlist(strsplit(s,"\\|"))
n2 <- as.numeric(gsub("[[:alpha:]]*", "", n))
num_lst <- str_extract_all(unl, "\\d\\.?\\d*")
indx <- lapply(num_lst, function(x) {
if(length(x) == 1) {isTRUE(all.equal(n2, as.numeric(x)))
} else {n2 >= as.numeric(x[1]) & n2 <= as.numeric(x[2])}})
paste(unl[unlist(indx)], collapse=" | ")
[1] "liquid & bar soap 1.0 - 2.0oz | liquid soap 1-2oz | dish 1.5oz"
I also tested it with other amounts like "2.3oz". With n2 we coerce n to numeric for comparison. The variable num_lst isolates the numbers from the character string.
With indx we apply our comparisions over the string numbers. if there is one number we check if it equals n2. I chose not to use the basic == operator to avoid any rounding issues. Instead isTRUE(all.equal(x, y)) is used.
Finally, the logical index variable indx is used to subset the character string to extract the matches and paste them together with a pipe "|".
Related
I am looking for a shorter and more pretty solution (possibly in tidyverse) to the following problem. I have a data.frame "data":
id string
1 A 1.001 xxx 123.123
2 B 23,45 lorem ipsum
3 C donald trump
4 D ssss 134, 1,45
What I wanted to do is to extract all numbers (no matter if the delimiter is "." or "," -> in this case I assume that string "134, 1,45" can be extracted into two numbers: 134 and 1.45) and create a data.frame "output" looking similar to this:
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
I managed to do this (code below) but the solution is pretty ugly for me also not so efficient (two for-loops). Could someone suggest a better way to do do this (preferably using dplyr)
# data
data <- data.frame(id = c("A", "B", "C", "D"),
string = c("1.001 xxx 123.123",
"23,45 lorem ipsum",
"donald trump",
"ssss 134, 1,45"),
stringsAsFactors = FALSE)
# creating empty data.frame
len <- length(unlist(sapply(data$string, function(x) gregexpr("[0-9]+[,|.]?[0-9]*", x))))
output <- data.frame(id = rep(NA, len), string = rep(NA, len))
# main solution
start = 0
for(i in 1:dim(data)[1]){
tmp_len <- length(unlist(gregexpr("[0-9]+[,|.]?[0-9]*", data$string[i])))
for(j in (start+1):(start+tmp_len)){
output[j,1] <- data$id[i]
output[j,2] <- regmatches(data$string[i], gregexpr("[0-9]+[,|.]?[0-9]*", data$string[i]))[[1]][j-start]
}
start = start + tmp_len
}
# further modifications
output$string <- gsub(",", ".", output$string)
output$string <- as.numeric(ifelse(substring(output$string, nchar(output$string), nchar(output$string)) == ".",
substring(output$string, 1, nchar(output$string) - 1),
output$string))
output
1) Base R This uses relatively simple regular expressions and no packages.
In the first 2 lines of code replace any comma followed by a space with a
space and then replace all remaining commas with a dot. After these two lines s will be: c("1.001 xxx 123.123", "23.45 lorem ipsum", "donald trump", "ssss 134 1.45")
In the next 4 lines of code trim whitespace from beginning and end of each string field and split the string field on whitespace producing a
list. grep out those elements consisting only of digits and dots. (The regular expression ^[0-9.]*$ matches the start of a word followed by zero or more digits or dots followed by the end of the word so only words containing only those characters are matched.) Replace any zero length components with NA. Finally add data$id as the names. After these 4 lines are run the list L will be list(A = c("1.001", "123.123"), B = "23.45", C = NA, D = c("134", "1.45")) .
In the last line of code convert the list L to a data frame with the appropriate names.
s <- gsub(", ", " ", data$string)
s <- gsub(",", ".", s)
L <- strsplit(trimws(s), "\\s+")
L <- lapply(L, grep, pattern = "^[0-9.]*$", value = TRUE)
L <- ifelse(lengths(L), L, NA)
names(L) <- data$id
with(stack(L), data.frame(id = ind, string = values))
giving:
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
2) magrittr This variation of (1) writes it as a magrittr pipeline.
library(magrittr)
data %>%
transform(string = gsub(", ", " ", string)) %>%
transform(string = gsub(",", ".", string)) %>%
transform(string = trimws(string)) %>%
with(setNames(strsplit(string, "\\s+"), id)) %>%
lapply(grep, pattern = "^[0-9.]*$", value = TRUE) %>%
replace(lengths(.) == 0, NA) %>%
stack() %>%
with(data.frame(id = ind, string = values))
3) dplyr/tidyr This is an alternate pipeline solution using dplyr and tidyr. unnest converts to long form, id is made factor so that we can later use complete to recover id's that are removed by subsequent filtering, the filter removes junk rows and complete inserts NA rows for each id that would otherwise not appear.
library(dplyr)
library(tidyr)
data %>%
mutate(string = gsub(", ", " ", string)) %>%
mutate(string = gsub(",", ".", string)) %>%
mutate(string = trimws(string)) %>%
mutate(string = strsplit(string, "\\s+")) %>%
unnest() %>%
mutate(id = factor(id))
filter(grepl("^[0-9.]*$", string)) %>%
complete(id)
4) data.table
library(data.table)
DT <- as.data.table(data)
DT[, string := gsub(", ", " ", string)][,
string := gsub(",", ".", string)][,
string := trimws(string)][,
string := setNames(strsplit(string, "\\s+"), id)][,
list(string = list(grep("^[0-9.]*$", unlist(string), value = TRUE))), by = id][,
list(string = if (length(unlist(string))) unlist(string) else NA_character_), by = id]
DT
Update Removed assumption that junk words do not have digit or dot. Also added (2), (3) and (4) and some improvements.
We can replace the , in between the numbers with . (using gsub), extract the numbers with str_extract_all (from stringr into a list), replace the list elements that have length equal to 0 with NA, set the names of the list with 'id' column, stack to convert the list to data.frame and rename the columns.
library(stringr)
setNames(stack(setNames(lapply(str_extract_all(gsub("(?<=[0-9]),(?=[0-9])", ".",
data$string, perl = TRUE), "[0-9.]+"), function(x)
if(length(x)==0) NA else as.numeric(x)), data$id))[2:1], c("id", "string"))
# id string
#1 A 1.001
#2 A 123.123
#3 B 23.45
#4 C NA
#5 D 134
#6 D 1.45
Same idea as Gabor's. I had hoped to use R's built-in parsing of strings (type.convert, used in read.table) rather than writing custom regex substitutions:
sp = setNames(strsplit(data$string, " "), data$id)
spc = lapply(sp, function(x) {
x = x[grep("[^0-9.,]$", x, invert=TRUE)]
if (!length(x))
NA_real_
else
mapply(type.convert, x, dec=gsub("[^.,]", "", x), USE.NAMES=FALSE)
})
setNames(rev(stack(spc)), names(data))
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
Unfortunately, type.convert is not robust enough to consider both decimal delimiters at once, so we need this mapply malarkey instead of type.convert(x, dec = "[.,]").
Hypothetical dataframe:
strings new column
mesh 1
foo 0
bar 0
tack 1
suture 1
I would like the new column to contain "1" if df$strings contains the strings "mesh", "tack", or "sutur". Otherwise it should display zero in the same row. I tried the following:
df$new_column <- ifelse(grepl("mesh" | "tack" | "sutur",
df$strings, ignore.case = T), "1", "0")
but got this error:
Error in "mesh" | "tack" :
operations are possible only for numeric, logical or complex types
Thanks in advance.
You want to use a single string in grep:
df$new_column <- ifelse(grepl("mesh|tack|sutur", df$strings, ignore.case = T),
"1", "0")
will work, but the following will be faster:
df$new_column <- +(grepl("mesh|tack|sutur", df$strings, ignore.case = T))
This will return a 0 and 1 integer vector
We can also use %in%
df$new_column <- as.integer(df$strings %in% c("mesh", "tack", "sutur"))
I'm formating a data set so each entry has the adegenet format for codominant markers, such as:
Loci1
###/###
208/210
200/204
198/208
where the # represents any digit (the number is a allele size in basepairs). My data has some homozygous entries (all 3 digit integers with no separator) that have the the form of:
Loci1
###
208
198
I intend to paste the 3 digit string to itself with sep='/' to produce the first format. I've tried to use grep to subset these homozygous entries by finding all non ###/### and negating the match using the table matching such as:
a <- grep('\\b\\d{3}?[/]\\d{3}', score$Loci1, value =T ) # Subset all ###/###/
score[!(a %in% 1:nrow(score$Loci1)), ] # works but only on vectors...
After the subset I could paste. The problem arises when I apply this to a data frame. grep seems to treat the data frame as a list (which in part it is) and returns columns that have a match.
So in short how can I go from ### to ###/### in a data frame
self contained example of data:
score2 <- NULL
set.seed(9)
Loci1 <- NULL
Loci2 <- NULL
Loci3 <- NULL
for (i in 1:5) Loci1 <- append(Loci1, paste(sample(seq(from = 230, to=330, by=3), 2, replace = F), collapse = '/'))
for (i in 1:5) Loci2 <- append(Loci2, paste(sample(seq(from = 230, to=330, by=3), 2, replace = F), collapse = '/'))
for (i in 1:5) Loci3 <- append(Loci3, paste(sample(seq(from = 230, to=330, by=3), 2, replace = F), collapse = '/'))
score2 <- data.frame(Loci1, Loci2, Loci3, stringsAsFactors = F)
score2[2,3] <- strsplit(score2[2,3], split = '/')[1]
score2[5,2] <- strsplit(score2[3,3], split = '/')[1]
score2[1,1] <- strsplit(score2[1,1], split = '/')[1]
score2[c(1, 4),c(2,3)] <- NA
score2
You could just replace the 3 digit items with the separator and a copy:
sub("^(...)$", "\\1/\\1", Loci1)
Use lapply with an anonymized function:
data.frame( lapply(score2, function(x) sub("^(...)$", "\\1/\\1", x) ) )
Loci1 Loci2 Loci3
1 251/251 <NA> <NA>
2 251/329 320/257 260/260
3 275/242 278/329 281/320
4 269/266 <NA> <NA>
5 296/326 281/281 326/314
(Not sure what the "paste-part" was supposed to refer to, but I think this was the intent of your question)
If the numeric values could have a varying number of digits then use a pattern argument like "^([0-9]{1,9})$"
An option using grep/paste,
m1 <- as.matrix(score2)
indx <- grep('^...$', m1)
m1[indx] <- paste(m1[indx], m1[indx], sep="/")
as.data.frame(m1)
# Loci1 Loci2 Loci3
#1 251/251 <NA> <NA>
#2 251/329 320/257 260/260
#3 275/242 278/329 281/320
#4 269/266 <NA> <NA>
#5 296/326 281/281 326/314
Or without converting to matrix, this can be done using lapply
score2[] <- lapply(score2, function(x) ifelse(grepl('^...$', x),
paste(x, x, sep="/"),x))
I have big data like this :
> Data[1:7,1]
[1] mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5
[2] mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
[3] mature=hsa-miR-448|mir_Family=mir-448|Gene=OR4F5
[4] mature=hsa-miR-659-3p|mir_Family=-|Gene=OR4F5
[5] mature=hsa-miR-5197-3p|mir_Family=-|Gene=OR4F5
[6] mature=hsa-miR-5093|mir_Family=-|Gene=OR4F5
[7] mature=hsa-miR-650|mir_Family=mir-650|Gene=OR4F5
what I want to do is that, in every row, I want to select the name after word mature= and also the word after Gene= and then pater them together with
paste(a,b, sep="-")
for example, the expected output from first two rows would be like :
hsa-miR-5087-OR4F5
hsa-miR-26a-1-3p-OR4F9
so, the final implementation is like this:
for(i in 1:nrow(Data)){
Data[i,3] <- sub("mature=([^|]*).*Gene=(.*)", "\\1-\\2", Data[i,1])
Name <- strsplit(as.vector(Data[i,2]),"\\|")[[1]][2]
Data[i,4] <- as.numeric(sub("pvalue=","",Name))
print(i)
}
which work well, but it's very slow. the size of Data is very big and it has 200,000,000 rows. this implementation is very slow for that. how can I speed it up ?
If you can guarantee that the format is exactly as you specified, then a regular expression can capture (denoted by the brackets below) everything from the equals sign upto the pipe symbol, and from the Gene= to the end, and paste them together with a minus sign:
sub("mature=([^|]*).*Gene=(.*)", "\\1-\\2", Data[,1])
Another option is to use read.table with = as a separator then pasting the 2 columns:
res = read.table(text=txt,sep='=')
paste(sub('[|].*','',res$V2), ## get rid from last part here
sub('^ +| +$','',res$V4),sep='-') ## remove extra spaces
[1] "hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9" "hsa-miR-448-OR4F5" "hsa-miR-659-3p-OR4F5"
[5] "hsa-miR-5197-3p-OR4F5" "hsa-miR-5093-OR4F5" "hsa-miR-650-OR4F5"
The simple sub solution already given looks quite nice but just in case here are some other approaches:
1) read.pattern Using read.pattern in the gsubfn package we can parse the data into a data.frame. This intermediate form, DF, can then be manipulated in many ways. In this case we use paste in essentially the same way as in the question:
library(gsubfn)
DF <- read.pattern(text = Data[, 1], pattern = "(\\w+)=([^|]*)")
paste(DF$V2, DF$V6, sep = "-")
giving:
[1] "hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9" "hsa-miR-448-OR4F5"
[4] "hsa-miR-659-3p-OR4F5" "hsa-miR-5197-3p-OR4F5" "hsa-miR-5093-OR4F5"
[7] "hsa-miR-650-OR4F5"
The intermediate data frame, DF, that was produced looks like this:
> DF
V1 V2 V3 V4 V5 V6
1 mature hsa-miR-5087 mir_Family - Gene OR4F5
2 mature hsa-miR-26a-1-3p mir_Family mir-26 Gene OR4F9
3 mature hsa-miR-448 mir_Family mir-448 Gene OR4F5
4 mature hsa-miR-659-3p mir_Family - Gene OR4F5
5 mature hsa-miR-5197-3p mir_Family - Gene OR4F5
6 mature hsa-miR-5093 mir_Family - Gene OR4F5
7 mature hsa-miR-650 mir_Family mir-650 Gene OR4F5
Here is a visualization of the regular expression we used:
(\w+)=([^|]*)
Debuggex Demo
1a) names We could make DF look nicer by reading the three columns of data and the three names separately. This also improves the paste statement:
DF <- read.pattern(text = Data[, 1], pattern = "=([^|]*)")
names(DF) <- unlist(read.pattern(text = Data[1,1], pattern = "(\\w+)=", as.is = TRUE))
paste(DF$mature, DF$Gene, sep = "-") # same answer as above
The DF in this section that was produced looks like this. It has 3 instead of 6 columns and remaining columns were used to determine appropriate column names:
> DF
mature mir_Family Gene
1 hsa-miR-5087 - OR4F5
2 hsa-miR-26a-1-3p mir-26 OR4F9
3 hsa-miR-448 mir-448 OR4F5
4 hsa-miR-659-3p - OR4F5
5 hsa-miR-5197-3p - OR4F5
6 hsa-miR-5093 - OR4F5
7 hsa-miR-650 mir-650 OR4F5
2) strapplyc
Another approach using the same package. This extracts the fields coming after a = and not containing a | producing a list. We then sapply over that list pasting the first and third fields together:
sapply(strapplyc(Data[, 1], "=([^|]*)"), function(x) paste(x[1], x[3], sep = "-"))
giving the same result.
Here is a visualization of the regular expression used:
=([^|]*)
Debuggex Demo
Here is one approach:
Data <- readLines(n = 7)
mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5
mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
mature=hsa-miR-448|mir_Family=mir-448|Gene=OR4F5
mature=hsa-miR-659-3p|mir_Family=-|Gene=OR4F5
mature=hsa-miR-5197-3p|mir_Family=-|Gene=OR4F5
mature=hsa-miR-5093|mir_Family=-|Gene=OR4F5
mature=hsa-miR-650|mir_Family=mir-650|Gene=OR4F5
df <- read.table(sep = "|", text = Data, stringsAsFactors = FALSE)
l <- lapply(df, strsplit, "=")
trim <- function(x) gsub("^\\s*|\\s*$", "", x)
paste(trim(sapply(l[[1]], "[", 2)), trim(sapply(l[[3]], "[", 2)), sep = "-")
# [1] "hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9" "hsa-miR-448-OR4F5" "hsa-miR-659-3p-OR4F5" "hsa-miR-5197-3p-OR4F5" "hsa-miR-5093-OR4F5"
# [7] "hsa-miR-650-OR4F5"
Maybe not the more elegant but you can try :
sapply(Data[,1],function(x){
parts<-strsplit(x,"\\|")[[1]]
y<-paste(gsub("(mature=)|(Gene=)","",parts[grepl("mature|Gene",parts)]),collapse="-")
return(y)
})
Example
Data<-data.frame(col1=c("mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5","mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9"),col2=1:2,stringsAsFactors=F)
> Data[,1]
[1] "mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5" "mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9"
> sapply(Data[,1],function(x){
+ parts<-strsplit(x,"\\|")[[1]]
+ y<-paste(gsub("(mature=)|(Gene=)","",parts[grepl("mature|Gene",parts)]),collapse="-")
+ return(y)
+ })
mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5 mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
"hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9"
I have a data.frame with a string column that contains periods e.g "a.b.c.X". I want to split out the string by periods and retain the third segment e.g. "c" in the example given. Here is what I'm doing.
> df = data.frame(v=c("a.b.a.X", "a.b.b.X", "a.b.c.X"), b=seq(1,3))
> df
v b
1 a.b.a.X 1
2 a.b.b.X 2
3 a.b.c.X 3
And what I want is
> df = data.frame(v=c("a.b.a.X", "a.b.b.X", "a.b.c.X"), b=seq(1,3))
> df
v b
1 a 1
2 b 2
3 c 3
I'm attempting to use within, but I'm getting strange results. The value in the first row in the first column is being repeated.
> get = function(x) { unlist(strsplit(x, "\\."))[3] }
> within(df, v <- get(as.character(v)))
v b
1 a 1
2 a 2
3 a 3
What is the best practice for doing this? What am I doing wrong?
Update:
Here is the solution I used from #agstudy's answer:
> df = data.frame(v=c("a.b.a.X", "a.b.b.X", "a.b.c.X"), b=seq(1,3))
> get = function(x) gsub(".*?[.].*?[.](.*?)[.].*", '\\1', x)
> within(df, v <- get(v))
v b
1 a 1
2 b 2
3 c 3
Using some regular expression you can do :
gsub(".*?[.].*?[.](.*?)[.].*", '\\1', df$v)
[1] "a" "b" "c"
Or more concise:
gsub("(.*?[.]){2}(.*?)[.].*", '\\2', v)
The problem is not with within but with your get function. It returns a single character ("a") which gets recycled when added to your data.frame. Your code should look like this:
get.third <- function(x) sapply(strsplit(x, "\\."), `[[`, 3)
within(df, v <- get.third(as.character(v)))
Here is one possible solution:
df[, "v"] <- do.call(rbind, strsplit(as.character(df[, "v"]), "\\."))[, 3]
## > df
## v b
## 1 a 1
## 2 b 2
## 3 c 3
The answer to "what am I doing wrong" is that the bit of code that you thought was extracting the third element of each split string was actually putting all the elements of all your strings in a single vector, and then returning the third element of that:
get = function(x) {
splits = strsplit(x, "\\.")
print("All the elements: ")
print(unlist(splits))
print("The third element:")
print(unlist(splits)[3])
# What you actually wanted:
third_chars = sapply(splits, function (x) x[3])
}
within(df, v2 <- get(as.character(v)))