Count repeated characters in a string - r

Question

How do find the length of highest repeated character in a string
col1 repeated letter repeated number
apples333 2 3
summer13 2 0
talk77 0 2
Aa6668 2 3
I can use lengths(regmatches(str, gregexpr("a",str) or str_count(str,"a") but the idea is to automatically check which is the highest repeating char/number and return count.

Answer 1

Using rle and rawConversion functions:
d <- data.frame(col1 = c("apples333", "summer13", "talk77", "Aa6668"))
foo <- function(x, p){
r <- rle(charToRaw(tolower(x)))
res <- max(r$lengths[ grepl(p, rawToChar(r$values, multiple = TRUE)) ])
if(res == 1) res <- 0
res
}
d$repLetter <- sapply(d$col1, foo, p = "[a-z]")
d$repNumber <- sapply(d$col1, foo, p = "[0-9]")
d
# col1 repLetter repNumber
# 1 apples333 2 3
# 2 summer13 2 0
# 3 talk77 0 2
# 4 Aa6668 2 3

Answer 2

There is probably an elegant regex-based solution for this (obviously I am not a big regex-er). The following is based on determining the run length of a vector using the base-rle() function, i.e. counting the repetition of elements.
As a strategy, we develop a function to work on a single string input providing the different portions and associated occurrences/counts. Then, to operate over several input strings, we apply (loop) a function to each element of the input vector.
single loop
Let's see how rle() works:
x <- "abba" # a test string - who does not know ABBA
x_split <- strsplit(x, "") %>% unlist # split the string, unlist to coerce vector
x_rle <- rle(x_split) # apply rle()
# now let's check what we have
x_rle
Run Length Encoding
lengths: int [1:3] 1 2 1
values : chr [1:3] "a" "b" "a"
rle() returns a list. As you want to filter, etc. on your results, it might be easier to turn this into a data frame. We also store the actual input.
With a view to apply this to other strings (e.g. loop over input vector), we wrap this into a function call:
library(dplyr)
check_rle_char_num <- function(x){
# split the string and count occurrences
x_split <- strsplit(x, "") %>% unlist()
x_rle <- rle(x_split).
# turn it into a tibble
df <- with(x_rle, tibble(values, lengths)) %>%
# ----------- store the input string and check for chars/numerics
mutate( input = x
, is_num = grepl(pattern = "[0-9]", values) # logical check for numbers
) %>%
# ----------- order output tibble
select(input, everything())
}
check that it works:
> ( check_rle_char_num("Appllles44777") )
# A tibble: 7 x 4
input values lengths is_num
<chr> <chr> <int> <lgl>
1 Appllles44777 A 1 FALSE
2 Appllles44777 p 2 FALSE
3 Appllles44777 l 3 FALSE
4 Appllles44777 e 1 FALSE
5 Appllles44777 s 1 FALSE
6 Appllles44777 4 2 TRUE
7 Appllles44777 7 3 TRUE
We have all the pieces on which you can filter, select, etc. your desired output.
loop over multiple input strings
We use tidyverse's {purrr} package for this.
# multiple input strings
my_strings <- c("apples333", "summer13","talk77","Aa6668","Appllles44777")
# loop over my_strings
library(purrr)
test <- my_strings %>%
map_dfr(.f = ~ check_rle_char_num(.x)) # map_dfr returns a data frame
test
# A tibble: 29 x 4
input values lengths is_num
<chr> <chr> <int> <lgl>
1 apples333 a 1 FALSE
2 apples333 p 2 FALSE
3 apples333 l 1 FALSE
4 apples333 e 1 FALSE
5 apples333 s 1 FALSE
6 apples333 3 3 TRUE
7 summer13 s 1 FALSE
8 summer13 u 1 FALSE
9 summer13 m 2 FALSE
10 summer13 e 1 FALSE
final push, filter, and reshape a nice output tibble
# per problem statement - filter for maximum and min 2 counts (i.e. > 1)
result <- test %>%
group_by(input, is_num) %>%
filter(lengths == max(lengths), lengths > 1)
> result
# A tibble: 7 x 4
# Groups: input, is_num [7]
input values lengths is_num
<chr> <chr> <int> <lgl>
1 apples333 p 2 FALSE
2 apples333 3 3 TRUE
3 summer13 m 2 FALSE
4 talk77 7 2 TRUE
5 Aa6668 6 3 TRUE
6 Appllles44777 l 3 FALSE
7 Appllles44777 7 3 TRUE
Emulating a bit your results listed in the problem statement, one can reshuffle the columns and provide "nice" column names:
library(tidyr) # for reshuffling
result %>%
tidyr::pivot_wider( names_from = is_num
, values_from = c(values, lengths)
) %>%
#---------- we spread the tibble, "spread" column-names combine previous colnames and TRUE/FALSE - mind that TRUE were numbers
rename( char = values_FALSE
, char_count = lengths_FALSE
, nums = values_TRUE
, nums_count = lengths_TRUE) %>%
#---------- changing order of columns for nice output
select(input, starts_with("char"), starts_with("num"))
# A tibble: 5 x 5
# Groups: input [5]
input char char_count nums nums_count
<chr> <chr> <int> <chr> <int>
1 apples333 p 2 3 3
2 summer13 m 2 NA NA
3 talk77 NA NA 7 2
4 Aa6668 NA NA 6 3
5 Appllles44777 l 3 7 3
final notes
The solution presented
does the filtering on the result data frame (after the loop). If there are no other operations on your data, you can lift this into the function.
does not clean the NAs in the final output. If you need zeros for no letter or no number, you can replace the NAs.
keeps characters and numbers in a single data frame. Obviously, you can split them. One could combine both again based on a join() or bind_cols() on the input-variable. This saves the pivot-wider bit.
does not care for "ties", i.e. you have a sequence of multiple characters and/or numbers with the same count. You may have to handle this.
Last but not least: simplify the code, if none of the columns/variables kept in the tibble help for your problem.

Answer 3

Solution
You can go with this:
library(stringr)
max_freq <- Vectorize(function(x) max(tabulate(factor(x))))
df$repeated_letter <- max_freq(str_extract_all(str_to_lower(df$col1), "[:alpha:]"))
df$repeated_letter <- max_freq(str_extract_all(str_to_lower(df$col1), "[:digit:]"))
df
#> col1 repeated_letter repeated_number
#> 1 apples333 2 3
#> 2 summer13 2 1
#> 3 talk77 1 2
#> 4 Aa6668 2 3
#> 5 Appllles44777 3 3
Explanation
Following a breakdown of the solution step by step with some explanations:
# take your column
df$col1 |>
# set to lower so A and a is the same character
str_to_lower() |>
# extract only letters or digits as list of vectors
str_extract_all("[:alpha:]") |>
# get frequency table for each vector
lapply(factor) |> lapply(tabulate) |>
# extract the count of most repeated letter for each table and return a vector
sapply(max)
#> [1] 2 2 1 2 3
Data
Where df is:
df <- data.frame(col1 = c("apples333", "summer13", "talk77", "Aa6668", "Appllles44777"))
Warnings
When there are no repeated characters, 1 will be returned, which is actually a more consistent answer, since the most repeated character will be repeated once. If you prefer zero, you can replace all ones with zeros.
In case of no characters or no numbers, -Inf will be returned. If you want a different result (like zero) you can replace it. In your example, it was not specified an occurrence like that.

Answer 4

Though late to the party but this method might still be of interest:
library(tidyr)
library(stringr)
library(dplyr)
d %>%
# count the number of character repetitions:
mutate(
# for letters:
dup_w = lapply(str_extract_all(col1, "(?i)([a-z])\\1+"), nchar),
# for numbers:
dup_n = lapply(str_extract_all(col1, "([0-9])\\1+"), nchar)) %>%
# throw all repetition counts into a single column:
pivot_longer(c(dup_w, dup_n)) %>%
# show items in list:
unnest(cols = value) %>%
# group:
group_by(col1, name) %>%
# reduce dataframe to maximum values per group:
filter(value == max(value)) %>%
# widen the dataframe back to original format:
pivot_wider(names_from = name, values_from = value)
# A tibble: 5 x 3
# Groups: col1 [5]
col1 dup_w dup_n
<chr> <int> <int>
1 11applesssss333 5 3
2 summer13 2 NA
3 talk77 NA 2
4 Aa6668 2 3
5 Appllles44777 3 3
Data (with lots ore repetitions to make things clearer):
d <- data.frame(col1 = c("11applesssss333", "summer13", "talk77",
"Aa6668", "Appllles44777"))