Determine if a string has all unique characters: Difference between revisions
Determine if a string has all unique characters (view source)
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"1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ" (36): '0' (0x30) duplicates at 9, 24
</pre>
=={{header|R}}==
Most of this is adapted from [[Determine if a string has all the same characters#R]].
<lang r>isAllUnique<-function(string)
{
strLength<-nchar(string)
if(length(strLength)>1)
{
#R has a distinction between the length of a string and that of a character vector. It is a common source
#of problems when coming from another language. We will try to avoid the topic here.
#For our purposes, let us only say that there is a good reason why we have made
#isAllUnique(c("foo", "bar") immediately throw an error.
stop("This task is intended for character vectors with lengths of at most 1.")
}
else if(length(strLength)==0)
{
cat("Examining a character vector of length 0.",
"It is therefore made entirely of unique characters.\n")
TRUE
}
else if(strLength==0)
{
cat("Examining a character vector of length 1, containing an empty string.",
"It is therefore made entirely of unique characters.\n")
TRUE
}
else if(strLength==1)
{
cat("Examining the string", paste0(sQuote(string), ","),
"which is of length", paste0(strLength, "."),
"It is therefore made entirely of unique characters.\n")
TRUE
}
else
{
cat("Examining the string", paste0(sQuote(string), ","),
"which is of length", paste0(strLength, ":"), "\n")
#strsplit outputs a list. Its first element is the vector of characters that we desire.
characters<-strsplit(string, "")[[1]]
#Our use of match is using R's vector recycling rules. Element i is being checked
#against every other.
indexesOfDuplicates<-sapply(seq_len(strLength), function(i) match(TRUE, characters[i]==characters[-i], nomatch = -1)) + 1
firstDuplicateElementIndex<-indexesOfDuplicates[indexesOfDuplicates != 0][1]
if(is.na(firstDuplicateElementIndex))
{
cat("It has no duplicates. It is therefore made entirely of unique characters.\n")
TRUE
}
else
{
cat("It has duplicates. ")
firstDuplicatedCharacter<-characters[firstDuplicateElementIndex]
cat(sQuote(firstDuplicatedCharacter), "is the first duplicated character. It has hex value",
sprintf("0x%X", as.integer(charToRaw(firstDuplicatedCharacter))),
"and is at index", paste0(firstDuplicateElementIndex, "."),
"\nThis is a duplicate of the character at index",
paste0(match(firstDuplicateElementIndex, indexesOfDuplicates), "."), "\n")
FALSE
}
}
}
#Tests:
cat("Test: A string of length 0 (an empty string):\n")
cat("Test 1 of 2: An empty character vector:\n")
print(isAllUnique(character(0)))
cat("Test 2 of 2: A character vector containing the empty string:\n")
print(isAllUnique(""))
cat("Test: A string of length 1 which contains .:\n")
print(isAllUnique("."))
cat("Test: A string of length 6 which contains abcABC:\n")
print(isAllUnique("abcABC"))
cat("Test: A string of length 7 which contains XYZ ZYX:\n")
print(isAllUnique("XYZ ZYX"))
cat("Test: A string of length 36 doesn't contain the letter 'oh':\n")
print(isAllUnique("1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ"))</lang>
{{out}}
<pre>Test: A string of length 0 (an empty string):
Test 1 of 2: An empty character vector:
Examining a character vector of length 0. It is therefore made entirely of unique characters.
[1] TRUE
Test 2 of 2: A character vector containing the empty string:
Examining a character vector of length 1, containing an empty string. It is therefore made entirely of unique characters.
[1] TRUE
Test: A string of length 1 which contains .:
Examining the string ‘.’, which is of length 1. It is therefore made entirely of unique characters.
[1] TRUE
Test: A string of length 6 which contains abcABC:
Examining the string ‘abcABC’, which is of length 6:
It has no duplicates. It is therefore made entirely of unique characters.
[1] TRUE
Test: A string of length 7 which contains XYZ ZYX:
Examining the string ‘XYZ ZYX’, which is of length 7:
It has duplicates. ‘X’ is the first duplicated character. It has hex value 0x58 and is at index 7.
This is a duplicate of the character at index 1.
[1] FALSE
Test: A string of length 36 doesn't contain the letter 'oh':
Examining the string ‘1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ’, which is of length 36:
It has duplicates. ‘0’ is the first duplicated character. It has hex value 0x30 and is at index 25.
This is a duplicate of the character at index 10.
[1] FALSE</pre>
=={{header|Raku}}==
(formerly Perl 6)
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