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Excel BI’s Excel Challenge #308 — solved in R

### Defining the Puzzle:

The newest puzzle by ExcelBI is still focused on world of words. We need to determine if words are Portmanteau, which means that they are complex structure taking parts of root words which gave them new meaning: like breakfast + lunch = brunch.

List Portmanteau words.
A Portmanteau word is made by starting few alphabets from Word1 and starting or ending few alphabets from Word2.
Ex. Biopic is made from Biography and picture.

Today we have 3 source columns and 1 solution column. In first part we have complex word and two columns for its potential partial words. We have to check if using column 2 and 3 we can construct column 1. Let’s read the data.

```library(tidyverse)
library(data.table)
library(stringi)

input = read_excel(“Portmanteau Words.xlsx”, range =”A1:C10")
test = read_excel(“Portmanteau Words.xlsx”, range =”D1:D6")```

### Approach 1: Tidyverse with purrr

```detect_portmanteau <- function(portmanteau, word1, word2) {
indices <- seq(1, str_length(portmanteau) — 1)
portmanteau_checks <- map_lgl(indices, function(i) {
pattern1 <- str_c(‘^’, str_sub(portmanteau, 1, i))
pattern2 <- str_c(‘^’, str_sub(portmanteau, i + 1, -1), ‘|’, str_sub(portmanteau, i + 1, -1), ‘\$’)

match_word1 <- str_detect(word1, regex(pattern1, ignore_case = TRUE))
match_word2 <- str_detect(word2, regex(pattern2, ignore_case = TRUE))

return(match_word1 && match_word2)
})

is_portmanteau <- any(portmanteau_checks)

return(is_portmanteau)
}

result <- input %>%
mutate(is_portmanteau = pmap_lgl(list(Word, Word1, Word2), detect_portmanteau)) %>%
filter(is_portmanteau) %>%
select(Word)```

### Approach 2: Base R

```detect_portmanteau_base <- function(portmanteau, word1, word2) {
indices <- seq(1, nchar(portmanteau) — 1)
is_portmanteau <- FALSE

for (i in indices) {
pattern1 <- paste0(‘^’, substring(portmanteau, 1, i))
pattern2 <- paste0(‘^’, substring(portmanteau, i + 1, nchar(portmanteau)), ‘|’, substring(portmanteau, i + 1, nchar(portmanteau)), ‘\$’)

match_word1 <- grepl(pattern1, word1, ignore.case = TRUE)
match_word2 <- grepl(pattern2, word2, ignore.case = TRUE)

if (match_word1 && match_word2) {
is_portmanteau <- TRUE
break
}
}
return(is_portmanteau)
}

result_base <- data.frame(Word = character(), stringsAsFactors = FALSE)

for (row in 1:nrow(input)) {
is_portmanteau <- with(input[row, ], detect_portmanteau_base(Word, Word1, Word2))
if (is_portmanteau) {
result_base <- rbind(result_base, input[row, "Word", drop = FALSE])
}
}```

### Approach 3: Data.table

Data.table syntax doesn’t really affect function construction, that’s why I only call tidyverse function from above using dt syntax.

```df <- setDT(input)

input_dt <- as.data.table(input)

input_dt[, is_portmanteau := mapply(detect_portmanteau, Word, Word1, Word2)]
result_dt <- input_dt[is_portmanteau == TRUE, .(Word)]```

### Validating Our Solutions:

```identical(test\$`Answer Expected`, result\$Word)
#> [1] TRUE

#> [1] TRUE

#> [1] TRUE```

If you like my publications or have your own ways to solve those puzzles in R, Python or whatever tool you choose, let me know.

Portmanteau Words was originally published in Numbers around us on Medium, where people are continuing the conversation by highlighting and responding to this story.