I am trying following code:
open Str
let ss = (Str.first_chars "testing" 3);;
print_endline ("The first 3 chars of 'testing' are: "^ss);;
if (ss == "tes")
then print_endline "These are equal to 'tes'"
else print_endline "These are NOT equal to 'tes'"
However, I am getting these are NOT equal:
$ ocaml str.cma testing2.ml
The first 3 chars of 'testing' are: tes
These are NOT equal to 'tes'
Why first 3 characters pulled by Str.first_chars from "testing" not equal to "tes"?
Also, I had to use ;; to make this code work (combinations of in and ; which I tried did not work). What is the best way to put these 3 statements together?
The (==) function is the physical equality operator. If you want to test whether two objects have the same contents, then you should use the structural equality operator which has one equal sign (=).
What is the best way to put these 3 statements together?
There are no statements in OCaml. Only expressions, all returning values. It is like a mathematical formula, where you have numbers, operators, and functions and you combine them together into bigger formulae, e.g., sin (2 * pi). The closest thing to the statement is an expression which has side effects and returns a value of type unit. But this is still expression.
Here is an example, how you can build your expression, which will first bind the returned substring to the ss variable, and then compute in order two expressions: an unconditional print, and a conditional print. Altogether, this will be one expression evaluating to the unit value.
open Str
let () =
let ss = Str.first_chars "testing" 3 in
print_endline ("The first 3 chars of 'testing' are: " ^ ss);
if ss = "tes"
then print_endline "These are equal to 'tes'"
else print_endline "These are NOT equal to 'tes'"
and here is how it works
$ ocaml str.cma test.ml
The first 3 chars of 'testing' are: tes
These are equal to 'tes'
Related
For two given strings, is there a pythonic way to count how many consecutive characters of both strings (starting at postion 0 of the strings) are identical?
For example in aaa_Hello and aa_World the "leading matching characters" are aa, having a length of 2. In another and example there are no leading matching characters, which would give a length of 0.
I have written a function to achive this, which uses a for loop and thus seems very unpythonic to me:
def matchlen(string0, string1): # Note: does not work if a string is ''
for counter in range(min(len(string0), len(string1))):
# run until there is a mismatch between the characters in the strings
if string0[counter] != string1[counter]:
# in this case the function terminates
return(counter)
return(counter+1)
matchlen(string0='aaa_Hello', string1='aa_World') # returns 2
matchlen(string0='another', string1='example') # returns 0
You could use zip and enumerate:
def matchlen(str1, str2):
i = -1 # needed if you don't enter the loop (an empty string)
for i, (char1, char2) in enumerate(zip(str1, str2)):
if char1 != char2:
return i
return i+1
An unexpected function in os.path, commonprefix, can help (because it is not limited to file paths, any strings work). It can also take in more than 2 input strings.
Return the longest path prefix (taken character-by-character) that is a prefix of all paths in list. If list is empty, return the empty string ('').
from os.path import commonprefix
print(len(commonprefix(["aaa_Hello","aa_World"])))
output:
2
from itertools import takewhile
common_prefix_length = sum(
1 for _ in takewhile(lambda x: x[0]==x[1], zip(string0, string1)))
zip will pair up letters from the two strings; takewhile will yield them as long as they're equal; and sum will see how many there are.
As bobble bubble says, this indeed does exactly the same thing as your loopy thing. Its sole pro (and also its sole con) is that it is a one-liner. Take it as you will.
I need to create a regular expression (for program in haskell) that will catch the strings containing "X" and ".", assuming that there are 4 "X" and only one ".". It cannot catch any string with other X-to-dot relations.
I have thought about something like
[X\.]{5}
But it catches also "XXXXX" or ".....", so it isn't what I need.
That's called permutation parsing, and while "pure" regular expressions can't parse permutations it's possible if your regex engine supports lookahead. (See this answer for an example.)
However I find the regex in the linked answer difficult to understand. It's cleaner in my opinion to use a library designed for permutation parsing, such as megaparsec.
You use the Text.Megaparsec.Perm module by building a PermParser in a quasi-Applicative style using the <||> operator, then converting it into a regular MonadParsec action using makePermParser.
So here's a parser which recognises any combination of four Xs and one .:
import Control.Applicative
import Data.Ord
import Data.List
import Text.Megaparsec
import Text.Megaparsec.Perm
fourXoneDot :: Parsec Dec String String
fourXoneDot = makePermParser $ mkFive <$$> x <||> x <||> x <||> x <||> dot
where mkFive a b c d e = [a, b, c, d, e]
x = char 'X'
dot = char '.'
I'm applying the mkFive function, which just stuffs its arguments into a five-element list, to four instances of the x parser and one dot, combined with <||>.
ghci> parse fourXoneDot "" "XXXX."
Right "XXXX."
ghci> parse fourXoneDot "" "XX.XX"
Right "XXXX."
ghci> parse fourXoneDot "" "XX.X"
Left {- ... -}
This parser always returns "XXXX." because that's the order I combined the parsers in: I'm mapping mkFive over the five parsers and it doesn't reorder its arguments. If you want the permutation parser to return its input string exactly, the trick is to track the current position within the component parsers, and then sort the output.
fourXoneDotSorted :: Parsec Dec String String
fourXoneDotSorted = makePermParser $ mkFive <$$> x <||> x <||> x <||> x <||> dot
where mkFive a b c d e = map snd $ sortBy (comparing fst) [a, b, c, d, e]
x = withPos (char 'X')
dot = withPos (char '.')
withPos = liftA2 (,) getPosition
ghci> parse fourXoneDotSorted "" "XX.XX"
Right "XX.XX"
As the megaparsec docs note, the implementation of the Text.Megaparsec.Perm module is based on Parsing Permutation Phrases; the idea is described in detail in the paper and the accompanying slides.
The other answers look quite complicated to me, given that there are only five strings in this language. Here's a perfectly fine and very readable regex for this:
\.XXXX|X\.XXX|XX\.XX|XXX\.X|XXXX\.
Are you attached to regex, or did you just end up at regex because this was a question you didn't want to try answering with applicative parsers?
Here's the simplest possible attoparsec implementation I can think of:
parseDotXs :: Parser ()
parseDotXs = do
dotXs <- count 5 (satisfy (inClass ".X"))
let (dots,xS) = span (=='.') . sort $ dotXs
if (length dots == 1) && (length xS == 4) then do
return ()
else do
fail "Mismatch between dots and Xs"
You may need to adjust slightly depending on your input type.
There are tons of fancy ways to do stuff in applicative parsing land, but there is no rule saying you can't just do things the rock-stupid simple way.
Try the following regex :
(?<=^| )(?=[^. ]*\.)(?=(?:[^X ]*X){4}).{5}(?=$| )
Demo here
If you have one word per string, you can simplify the regex by this one :
^(?=[^. \n]*\.)(?=(?:[^X \n]*X){4}).{5}$
Demo here
Given an input string map three types of possible sequences of numbers contained in the string to a single number and leave the other elements of the string unchanged:
Single number should be mapped to the char 1: "help3me" -> "help1me"
Two numbers in a row should be mapped to the char 2: "help18me" -> "help2me"
Three or more numbers in a row should be mapped to 3: "test3432help234312me" -> "test3help3me"
Our input strings can contain any number of 1,2,3+ length sequences of digits so that a valid input example is "help3490897test73me23435please5"
What is an effective solution for the above problem in Scala does it just involve enumerating through the three possible cases as a regex ?
Use regular expression and method replaceAllIn. The second argument is the function that takes Match object and transforms it to its length.
val str = "help3me34"
val expr = "(\\d+)".r
expr.replaceAllIn(str, x => (x.group(0).length min 3).toString)
res2: String = help1me2
I need to validate a string against a character vector pattern. My current code is:
trim <- function (x) gsub("^\\s+|\\s+$", "", x)
# valid pattern is lowercase alphabet, '.', '!', and '?' AND
# the string length should be >= than 2
my.pattern = c(letters, '!', '.', '?')
check.pattern = function(word, min.size = 2)
{
word = trim(word)
chars = strsplit(word, NULL)[[1]]
all(chars %in% my.pattern) && (length(chars) >= min.size)
}
Example:
w.valid = 'special!'
w.invalid = 'test-me'
check.pattern(w.valid) #TRUE
check.pattern(w.invalid) #FALSE
This is VERY SLOW i guess...is there a faster way to do this? Regex maybe?
Thanks!
PS: Thanks everyone for the great answers. My objective was to build a 29 x 29 matrix,
where the row names and column names are the allowed characters. Then i iterate over each word of a huge text file and build a 'letter precedence' matrix. For example, consider the word 'special', starting from the first char:
row s, col p -> increment 1
row p, col e -> increment 1
row e, col c -> increment 1
... and so on.
The bottleneck of my code was the vector allocation, i was 'appending' instead of pre-allocate the final vector, so the code was taking 30 minutes to execute, instead of 20 seconds!
There are some built-in functions that can clean up your code. And I think you're not leveraging the full power of regular expressions.
The blaring issue here is strsplit. Comparing the equality of things character-by-character is inefficient when you have regular expressions. The pattern here uses the square bracket notation to filter for the characters you want. * is for any number of repeats (including zero), while the ^ and $ symbols represent the beginning and end of the line so that there is nothing else there. nchar(word) is the same as length(chars). Changing && to & makes the function vectorized so you can input a vector of strings and get a logical vector as output.
check.pattern.2 = function(word, min.size = 2)
{
word = trim(word)
grepl(paste0("^[a-z!.?]*$"),word) & nchar(word) >= min.size
}
check.pattern.2(c(" d ","!hello ","nA!"," asdf.!"," d d "))
#[1] FALSE TRUE FALSE TRUE FALSE
Next, using curly braces for number of repetitions and some paste0, the pattern can use your min.size:
check.pattern.3 = function(word, min.size = 2)
{
word = trim(word)
grepl(paste0("^[a-z!.?]{",min.size,",}$"),word)
}
check.pattern.3(c(" d ","!hello ","nA!"," asdf.!"," d d "))
#[1] FALSE TRUE FALSE TRUE FALSE
Finally, you can internalize the regex from trim:
check.pattern.4 = function(word, min.size = 2)
{
grepl(paste0("^\\s*[a-z!.?]{",min.size,",}\\s*$"),word)
}
check.pattern.4(c(" d ","!hello ","nA!"," asdf.!"," d d "))
#[1] FALSE TRUE FALSE TRUE FALSE
If I understand the pattern you are desiring correctly, you would want a regex of a similar format to:
^\\s*[a-z!\\.\\?]{MIN,MAX}\\s*$
Where MIN is replaced with the minimum length of the string, and MAX is replaced with the maximum length of the string. If there is no maximum length, then MAX and the comma can be omitted. Likewise, if there is neither maximum nor minimum everything within the {} including the braces themselves can be replaced with a * which signifies the preceding item will be matched zero or more times; this is equivalent to {0}.
This ensures that the regex only matches strings where every character after any leading and trailing whitespace is from the set of
* a lower case letter
* a bang (exclamation point)
* a question mark
Note that this has been written in Perl style regex as it is what I am more familiar with; most of my research was at this wiki for R text processing.
The reason for the slowness of your function is the extra overhead of splitting the string into a number of smaller strings. This is a lot of overhead in comparison to a regex (or even a manual iteration over the string, comparing each character until the end is reached or an invalid character is found). Also remember that this algorithm ENSURES a O(n) performance rate, as the split causes n strings to be generated. This means that even FAILING strings must do at least n actions to reject the string.
Hopefully this clarifies why you were having performance issues.
Given an input string such as " word1 word2 word3 word4 ", what would be the best approach to split this as an array of strings in Go? Note that there can be any number of spaces or unicode-spacing characters between each word.
In Java I would just use someString.trim().split("\\s+").
(Note: possible duplicate Split string using regular expression in Go doesn't give any good quality answer. Please provide an actual example, not just a link to the regexp or strings packages reference.)
The strings package has a Fields method.
someString := "one two three four "
words := strings.Fields(someString)
fmt.Println(words, len(words)) // [one two three four] 4
DEMO: http://play.golang.org/p/et97S90cIH
From the docs:
Fields splits the string s around each instance of one or more consecutive white space characters, as defined by unicode.IsSpace, returning a slice of substrings of s or an empty slice if s contains only white space.
If you're using tip: regexp.Split
func (re *Regexp) Split(s string, n int) []string
Split slices s into substrings separated by the expression and returns
a slice of the substrings between those expression matches.
The slice returned by this method consists of all the substrings
of s not contained in the slice returned by FindAllString. When called
on an expression that contains no metacharacters, it is equivalent to strings.SplitN.
Example:
s := regexp.MustCompile("a*").Split("abaabaccadaaae", 5)
// s: ["", "b", "b", "c", "cadaaae"]
The count determines the number of substrings to return:
n > 0: at most n substrings; the last substring will be the unsplit remainder.
n == 0: the result is nil (zero substrings)
n < 0: all substrings
I came up with the following, but that seems a bit too verbose:
import "regexp"
r := regexp.MustCompile("[^\\s]+")
r.FindAllString(" word1 word2 word3 word4 ", -1)
which will evaluate to:
[]string{"word1", "word2", "word3", "word4"}
Is there a more compact or more idiomatic expression?
You can use package strings function split
strings.Split(someString, " ")
strings.Split