Given a String: 3 Design Features, I'm trying to check if the term preceeding "Design Features" is a number or not using the below. (The number can exist as 2 or 2.)
score=0;
str = <P>3 Design Features</P>
regexp_number = "/^[0-9]+./";
if(str_detect(y,regexp_number) ==TRUE)
{
score=score++;
}
This returns 0. What am I doing wrong here? Hoping someone can point out?
Thanks in advance.
-Simak
Your regex is wrong. It says it must contain a . to match, rather than optionally contain 0 or 1 ..
Change it to
regexp_number = "/^[0-9]+.?/";
w <- "aghj 3 Design Features kjkl"
x <- "aghj 3. Design Features kjkl"
y <- "aghj c Design Features kjkl"
z <- "4 aghj c gn Features kjkl"
fun <- function(x) grepl("[[:digit:]]",
regmatches(x,
regexpr(".\\.?(?= Design Features)",x,perl = TRUE)))
fun(w)
[1] TRUE
fun(x)
[1] TRUE
fun(y)
[1] FALSE
fun(z)
[1] logical(0)
Related
I have a dataframe dat like this
P pedigree cas
1 M rs2745406 T
2 M rs6939431 A
3 M SNP_DPB1_33156641 G
4 M SNP_DPB1_33156664_G P
5 M SNP_DPB1_33156664_A A
6 M SNP_DPB1_33156664_T A
I want to exclude all rows where the pedigree column starts with SNP_ and ends with either G, C, T, or A (_[GCTA]). In this case, this would be rows 4,5,6.
How can I achieve this in R? I have tried
multisnp <- which(grepl("^SNP_*_[GCTA]$", dat$pedigree)=="TRUE")
new_dat <- dat[-multisnp,]
My multisnp vector is empty, but I can't figure out how to fix it so that it matches the pattern I want. I think it is my wildcard * usage that is wrong.
You can use the following with .*? (match everything in non greedy way):
multisnp <- which(grepl("^SNP_.*?_[GCTA]$", dat$pedigree))
^^^
You can subset dat like this
new_dat <- dat[!grepl("^SNP_.*_[GCTA]$", dat$pedigree), ]
Regarding the code that you've tried, I'm not sure that grepl("^SNP_*_[GCTA]$") will complete without an error since you aren't passing in an x vector to grepl. See ?grepl for more info.
I have big data like this :
> Data[1:7,1]
[1] mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5
[2] mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
[3] mature=hsa-miR-448|mir_Family=mir-448|Gene=OR4F5
[4] mature=hsa-miR-659-3p|mir_Family=-|Gene=OR4F5
[5] mature=hsa-miR-5197-3p|mir_Family=-|Gene=OR4F5
[6] mature=hsa-miR-5093|mir_Family=-|Gene=OR4F5
[7] mature=hsa-miR-650|mir_Family=mir-650|Gene=OR4F5
what I want to do is that, in every row, I want to select the name after word mature= and also the word after Gene= and then pater them together with
paste(a,b, sep="-")
for example, the expected output from first two rows would be like :
hsa-miR-5087-OR4F5
hsa-miR-26a-1-3p-OR4F9
so, the final implementation is like this:
for(i in 1:nrow(Data)){
Data[i,3] <- sub("mature=([^|]*).*Gene=(.*)", "\\1-\\2", Data[i,1])
Name <- strsplit(as.vector(Data[i,2]),"\\|")[[1]][2]
Data[i,4] <- as.numeric(sub("pvalue=","",Name))
print(i)
}
which work well, but it's very slow. the size of Data is very big and it has 200,000,000 rows. this implementation is very slow for that. how can I speed it up ?
If you can guarantee that the format is exactly as you specified, then a regular expression can capture (denoted by the brackets below) everything from the equals sign upto the pipe symbol, and from the Gene= to the end, and paste them together with a minus sign:
sub("mature=([^|]*).*Gene=(.*)", "\\1-\\2", Data[,1])
Another option is to use read.table with = as a separator then pasting the 2 columns:
res = read.table(text=txt,sep='=')
paste(sub('[|].*','',res$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 the following data.table:
id fShort
1 432-12 1245
2 3242-12 453543
3 324-32 45543
4 322-34 45343
5 2324-34 13543
DT <- data.table(
id=c("432-12", "3242-12", "324-32", "322-34", "2324-34"),
fShort=c("1245", "453543", "45543", "45343", "13543"))
and the following list:
filenames <- list("3242-124342345.png", "432-124343.png", "135-13434.jpeg")
I would like to create a new column "fComplete" that includes the complete filename from the list. For this the values of column "id" need to be matched with the filename-list. If the filename starts with the "id" string, the complete filename should be returned. I use the following regex
t <- grep("432-12","432-124343.png",value=T)
that return the correct filename.
This is how the final table should look like:
id fShort fComplete
1 432-12 1245 432-124343.png
2 3242-12 453543 3242-124342345.png
3 324-32 45543 NA
4 322-34 45343 NA
5 2324-34 13543 NA
DT2 <- data.table(
id=c("432-12", "3242-12", "324-32", "322-34", "2324-34"),
fshort=c("1245", "453543", "45543", "45343", "13543"),
fComplete = c("432-124343.png", "3242-124342345.png", NA, NA, NA))
I tried using apply and data.table approaches but I always get warnings like
argument 'pattern' has length > 1 and only the first element will be used
What is a simple approach to accomplish this?
Here's a data.table solution:
DT[ , fComplete := lapply(id, function(x) {
m <- grep(x, filenames, value = TRUE)
if (!length(m)) NA else m})]
id fShort fComplete
1: 432-12 1245 432-124343.png
2: 3242-12 453543 3242-124342345.png
3: 324-32 45543 NA
4: 322-34 45343 NA
5: 2324-34 13543 NA
In my experience with similar functions, sometimes the regex functions return a list, so you have to consider that in the apply - I usually do an example manually
Also apply will not always in y experience on its own return something that always works into a data.frame,sometimes I had to use lap ply, and or unlist and data.frame to modify it
Here is an answer - I am not familiar with data.tables and I was having issues with the filenames being in a list, but with some transformations this works. I worked it out by seeing what apply was outputting and adding the [1] to get the piece I needed
DT <- data.frame(
id=c("432-12", "3242-12", "324-32", "322-34", "2324-34"),
fShort=c("1245", "453543", "45543", "45343", "13543"))
filenames <- list("3242-124342345.png", "432-124343.png", "135-13434.jpeg")
filenames1 <- unlist(filenames)
x<-apply(DT[1],1,function(x) grep(x,filenames1)[1])
DT$fielname <- filenames1[x]
Is there a way to make the R code below run quicker (i.e. vectorized to avoid use of for loops)?
My example contains two data frames. First is dimension n1*p. One of the p columns contains names. Second data frame is a column vector (n2*1). It contains names as well. I want to keep all rows of the first data frame, where some part of the name in the column vector of the second data frame appears in the corresponding first data frame. Sorry for the brutal explanation.
Example (Data frame 1):
x y
Doggy 1
Hello 2
Hi Dog 3
Zebra 4
Example (Data frame 2)
z
Hello
Dog
So in the above example I want to keep rows 1,2,3 but NOT 4. Since "Dog" appears in "Doggy" and "Hi Dog". And "Hello" appears in "Hello". Exclude row four since no part of "Hello" or "Dog" appears in "Zebra".
Below is my R code to do this...runs fine. However, for my real task. Data frame 1 has 1 million rows and data frame 2 has 50 items to match on. So runs pretty slow. Any suggestion on how to speed this up are appreciated.
x <- c("Doggy", "Hello", "Hi Dog", "Zebra")
y <- 1:4
dat <- as.data.frame(cbind(x,y))
names(dat) <- c("x","y")
z <- as.data.frame(c("Hello", "Dog"))
names(z) <- c("z")
dat$flag <- NA
for(j in 1:length(z$z)){
for(i in 1:dim(dat)[1]){
if ( is.na(dat$flag[i])==TRUE ) {
dat$flag[i] <- length(grep(paste(z[j,1]), dat[i,1], perl=TRUE, value=TRUE))
} else {
if (dat$flag[i]==0) {
dat$flag[i] <- length(grep(paste(z[j,1]), dat[i,1], perl=TRUE, value=TRUE))
} else {
if (dat$flag[i]==1) {
dat$flag[i]==1
}
}
}
}
}
dat1 <- subset(dat, flag==1)
dat1
Try this:
dat[grep(paste(z$z, collapse = "|"), dat$x), ]
or
subset(dat, grepl(paste(z$z, collapse = "|"), x))
This question inspired a boolean text search function (%bs%) in the qdap package and thus I thought I'd share the approach to this question:
library(qdap)
dat[dat$x %bs% paste(z$z, collapse = "OR"), ]
In this case no less typing but if multiple or/and statements are involved this may be a useful approach.
In the process of (mostly) answering this question, I stumbled across something that I feel like I really should already have seen before. Let's say you've got a list:
l <- list(a = 1:3, b = letters[1:3], c = runif(3))
Attempting to coerce l to various types returns an error:
> as.numeric(l)
Error: (list) object cannot be coerced to type 'double'
> as.logical(l)
Error: (list) object cannot be coerced to type 'logical'
However, I'm apparently allowed to coerce a list to character, I just wasn't expecting this result:
> as.character(l)
[1] "1:3"
[2] "c(\"a\", \"b\", \"c\")"
[3] "c(0.874045701464638, 0.0843329173512757, 0.809434881201014)"
Rather, if I'm allowed to coerce lists to character, I would have thought I'd see behavior more like this:
> as.character(unlist(l))
[1] "1" "2" "3" "a" "b"
[6] "c" "0.874045701464638" "0.0843329173512757" "0.809434881201014"
Note that how I specify the list elements originally affects the output of as.character:
l <- list(a = c(1,2,3), b = letters[1:3], c = runif(3))
> as.character(l)
[1] "c(1, 2, 3)"
[2] "c(\"a\", \"b\", \"c\")"
[3] "c(0.344991483259946, 0.0492411875165999, 0.625746068544686)"
I have two questions:
How is as.character dredging up the information from my original creation of the list l in order to spit out 1:3 versus c(1,2,3).
In what circumstances would I want to do this, exactly? When would I want to call as.character() on a list and get output of this form?
For non-trivial lists, as.character uses deparse to generate the strings.
Only if the vector is integer and 1,2,3,...,n - then it deparses as 1:n.
c(1,2,3) is double whereas 1:3 is integer...
No idea :-)
...but look at deparse if you want to understand as.character here:
deparse(c(1L, 2L, 3L)) # 1:3
deparse(c(3L, 2L, 1L)) # c(3L, 2L, 1L)
deparse(c(1, 2, 3)) # c(1, 2, 3)
The help file does say
For lists it deparses the elements individually, except that it extracts the first element of length-one character vectors.
I'd seen this before in trying to answer a question [not online] about grep. Consider:
> x <- list(letters[1:10],letters[10:19])
> grep("c",x)
[1] 1 2
grep uses as.character on x, with the result that, since both have c( in them, both components match. That took a while to figure out.
On "Why does it do this?", I'd guess that one of the members of R core wanted it to do this.