Count patterns and differentiate them

Question

I'd like to count a defined pattern (here: 'Y') in a string for each row of a dataframe. Ideally, I'd like to get a number of occurrences in V3 and length in V4.

Input:

V1  V2
A   XXYYYYY
B   XXYYXX
C   XYXXYX
D   XYYXYX

Output:

V1       V2 V3   V4
 A  XXYYYYY  1    5
 B   XXYYXX  1    2
 C   XYXXYX  2  1,1
 D   XYYXYX  2  2,1

I tried different modifications of the function below, with no success.

dict <- setNames(nm=c("Y"))
seqs <- df$V2
sapply(dict, str_count, string=seqs)

Thanks in advance!

Answer 1

another base R solution but using regexpr:

df <- data.frame(
  V1 = c("A", "B", "C", "D"),
  V2 = c("XXYYYYY", "XXYYXX" , "XYXXYX", "XYYXYX")
)

extract match.length attribute of the regexpr output, then count length of each attribute (which tells you how many matches there are):

r <- gregexpr("Y+", df$V2)
len <- lapply(r, FUN = function(x) as.array((attributes(x)[[1]])))
df$V3 <- lengths(len)
df$V4 <- len

df
#V1      V2 V3   V4
#1  A XXYYYYY  1    5
#2  B  XXYYXX  1    2
#3  C  XYXXYX  2 1, 1
#4  D  XYYXYX  2 2, 1

if you have an old version of R that doesn't have lengths yet you can use df$V3 <- sapply(len, length) instead. and if you need a more generic function to do the same for any vector x and pattern a:

foo <- function(x, a){
  ans <- data.frame(x)
  r <- gregexpr(a, x)
  len <- lapply(r, FUN = function(z) as.array((attributes(z)[[1]])))
  ans$quantity <- lengths(len)
  ans$lengths <- len
  ans
}

try foo(df$V2, 'Y+').

Answer 2

Here is a stringr solution:

df <- data.frame(
  V1 = c("A", "B", "C", "D"),
  V2 = c("XXYYYYY", "XXYYXX" , "XYXXYX", "XYYXYX")
  )

df$V3 <- str_count(df$V2, "Y+")

df$V4 <- lapply(str_locate_all(df$V2, "Y+"), function(x) {
    paste(x[, 2] - x[, 1] + 1, collapse = ",")
  })