I have a string that I would like to print . Is it possible to center it when printing ?
Use the ~< formatting directive. This will return "hello there" centered within 70 columns.
(format nil "~70:#<~A~>" "hello there")
Related
I'm interested in a pure-Lua (i.e., no external Unicode library) solution to extracting the units of a string between certain Unicode control characters and spaces. The code points I would like to use as delimiters are:
0000-0020
007f-00a0
00ad
1680
2000-200a
2028-2029
202f
205f
3000
I know how to access the code points in a string, for example:
> for i,c in utf8.codes("é$ \tπ😃") do print(c) end
233
36
32
9
960
128515
but I am not sure how to "skip" the spaces and tabs and reconstitute the other codepoints into strings themselves. What I would like to do in the example above, is drop the 32 and 9, then perhaps use utf8.char(233, 36) and utf8.char(960, 128515) to somehow get ["é$", "π😃"].
It seems that putting everything into a table of numbers and painstakingly walking through the table with for-loops and if-statements would work, but is there a better way? I looked into string:gmatch but that seems to require making utf8 sequences out of each of the ranges I want, and it's not clear what that pattern would even look like.
Is there a idiomatic way to extract the strings between the spaces? Or must I manually hack tables of code points? gmatch does not look up to the task. Or is it?
would require painstakingly generating the utf8 encodings for all code points at each end of the range.
Yes. But of course not manually.
local function range(from, to)
assert(utf8.codepoint(from) // 64 == utf8.codepoint(to) // 64)
return from:sub(1,-2).."["..from:sub(-1).."-"..to:sub(-1).."]"
end
local function split_unicode(s)
for w in s
:gsub("[\0-\x1F\x7F]", " ")
:gsub("\u{00a0}", " ")
:gsub("\u{00ad}", " ")
:gsub("\u{1680}", " ")
:gsub(range("\u{2000}", "\u{200a}"), " ")
:gsub(range("\u{2028}", "\u{2029}"), " ")
:gsub("\u{202f}", " ")
:gsub("\u{205f}", " ")
:gsub("\u{3000}", " ")
:gmatch"%S+"
do
print(w)
end
end
Test:
split_unicode("#\0#\t#\x1F#\x7F#\u{00a0}#\u{00ad}#\u{1680}#\u{2000}#\u{2005}#\u{200a}#\u{2028}#\u{2029}#\u{202f}#\u{205f}#\u{3000}#")
In restructured text, titles are written with equal number of nonalphanumeric 7-bit ASCII
character as the title text. The underline and overline if both used, should be equal and at least as long as title text. From the official docs:
Titles are underlined (or over-
and underlined) with a printing
nonalphanumeric 7-bit ASCII
character. Recommended choices
are "= - ` : ' " ~ ^ _ * + # < >".
The underline/overline must be at
least as long as the title text.
Example of a title
=========================================================
Main titles are written using equals signs over and under
=========================================================
I want to create a VS Code snippet for this. What I could do was only this,
"Title RST": {
"prefix": "title",
"body": [
"="
"$1"
"=\n"
"$0"
],
"description": "Title for restructured text"
}
Is there a way to know the length of the text that will be typed, and correspondingly insert same number of overline and underline =.
In yasnippet in emacs, they do it as:
${1:$(make-string (string-width yas-text) ?\=)}
${1:Title}
${1:$(make-string (string-width yas-text) ?\=)}
$0
Any help how to implement such snippet in VS code? I looked under snippets in restructured text extension for VS Code here but could not find that suits my needs.
"Title RST": {
"prefix": "title",
"body": [
"${1/./=/g}",
"$1",
"${1/./=/g}",
"$0"
],
"description": "Title for restructured text"
},
The transforms ${1/./=/g} just replace every character in your text $1 with a = in the line above and below your text.
You need commas at the end of your snippet entries and there is no need for the newline as another line in the snippet body is already a newline.
When you type your text hit Tab and the transform will be completed.
You asked if was possible to get the over/underlines to show as =s immediately upon typing your title text. But that isn't possible with vscode snippets, a transform is required and that won't happen until the Tab.
It can be done with HyperSnips version (a little more trouble to set up than plain vscode snippets, but not much):
snippet title "Title" A
``rv = '='.repeat(t[0].length)``
$1
``rv = '='.repeat(t[0].length)``
endsnippet
I have the need to remove leading and trailing spaces around a punctuation character.
For example: Hello , World ... I 'm a newbie iOS Developer.
And I'd like to have: > Hello, World... I'm a newbie iOS Developer.
How can I do this? I tried to get components of the string and enumerate it by sentences. But that is not what I need
Rob's answer is great, but you can trim it down quite a lot by taking advantage of the \p{Po} regular expression class. Getting rid of the spaces around punctuation then becomes a single regular expression replace:
import Foundation
let input = "Hello , World ... I 'm a newbie iOS Developer."
let result = input.replacingOccurrences(of: "\\s*(\\p{Po}\\s?)\\s*",
with: "$1",
options: [.regularExpression])
print(result) // "Hello, World... I'm a newbie iOS Developer."
Rob's answer also tries to trim leading/trailing spaces, but your input doesn't have any of those. If you do care about that you can just call result.trimmingCharacters(in: .whitespacesAndNewlines) on the result.
Here's an explanation for the regular expression. Removing the double-escapes it looks like
\s*(\p{Po}\s?)\s*
This is comprised of the following components:
\s* - Match zero or more whitespace characters (and throw them away)
(…) - Capturing group. Anything inside this group is preserved by the replacement (the $1 in the replacement refers to this group).
\p{Po} - Match a single character in the "Other_Punctuation" unicode category. This includes things like ., ', and …, but excludes things like ( or -.
\s? - Match a single optional whitespace character. This preserves the space after periods (or ellipses).
\s* - Once again, match zero or more whitespace characters (and throw them away). This is what turns your , World into , World.
For Swift 3 or 4 you can use :
let trimmedString = string.trimmingCharacters(in: .whitespaces)
This is a really wonderful problem and a shame that it isn't easier to do in Swift today (someday it will be, but not today).
I kind of hate this code, but I'm getting on a plane for 20 hours, and don't have time to make it nicer. This may at least get you started using NSMutableString. It'd be nice to work in String, and Swift hates regular expressions, so this is kind of hideous, but at least it's a start.
import Foundation
let input = "Hello, World ... I 'm a newbie iOS Developer."
let adjustments = [
(pattern: "\\s*(\\.\\.\\.|\\.|,)\\s*", replacement: "$1 "), // elipsis or period or comma has trailing space
(pattern: "\\s*'\\s*", replacement: "'"), // apostrophe has no extra space
(pattern: "^\\s+|\\s+$", replacement: ""), // remove leading or trailing space
]
let mutableString = NSMutableString(string: input)
for (pattern, replacement) in adjustments {
let re = try! NSRegularExpression(pattern: pattern)
re.replaceMatches(in: mutableString,
options: [],
range: NSRange(location: 0, length: mutableString.length),
withTemplate: replacement)
}
mutableString // "Hello, World... I'm a newbie iOS Developer."
Regular expressions can be very confusing when you first encounter them. A few hints at reading these:
The specific language Foundation uses is described by ICU.
Backslash (\) means "the next character is special" for a regex. But inside a Swift string, backslash means "the next character is special" of the string. So you have to double them all.
\s means "a whitespace character"
\s* means "zero or more whitespace characters"
\s+ means "one or more whitespace characters"
$1 means "the thing we matched in parentheses"
| means "or"
^ means "start of string"
$ means "end of string"
. means "any character" so to mean "an actual dot" you have to type "\\." in a Swift string.
Notice that I check for both "..." and "." in the same regular expression. You kind of have to do something like that, or else the "." will match three times inside the "...". Another approach would be to first replace "..." with "…" (the single ellipsis character, typed on a Mac by pressing Opt-;). Then "…" is a one-character punctuation. (You could also decide to re-expand all ellipsis back to dot-dot-dot at the end of the process.)
Something like this is probably how I'd do it in real life, get it done and shipped, but it may be worth the pain/practice to try to build this as a character-by-character state machine, walking one character at a time, and keeping track of your current state.
You can try something like
string.replacingOccurrences(of: " ,", with: ",") for every punctuation...
Interesting problem; here's my stab at a non-Regex approach:
func correct(input: String) -> String {
typealias Correction = (punctuation: String, replacement: String)
let corrections: [Correction] = [
(punctuation: "...", replacement: "... "),
(punctuation: "'", replacement: "'"),
(punctuation: ",", replacement: ", "),
]
var transformed = input
for correction in corrections {
transformed = transformed
.components(separatedBy: correction.punctuation)
.map({ $0.trimmingCharacters(in: .whitespaces) })
.joined(separator: correction.replacement)
}
return transformed
}
let testInput = "Hello , World ... I 'm a newbie iOS Developer."
let testOutput = correct(input: testInput)
// Hello, World... I'm a newbie iOS Developer.
If you were doing this manually by processing characters arrays, you would merely need to check the previous and next characters around spaces. You can achieve the same result using functional style programming with zip, filter and map:
let testInput = "Hello , World ... I 'm a newbie iOS Developer."
let punctuation = Set(".\',")
let previousNext = zip( [" "] + testInput, String(testInput.dropFirst()) + [" "] )
let filteredChars = zip(Array(previousNext),testInput)
.filter{ $1 != " "
|| !($0.0 != " " && punctuation.contains($0.1))
}
let filteredInput = String(filteredChars.map{$1})
print(testInput) // Hello , World ... I 'm a newbie iOS Developer.
print(filteredInput) // Hello, World... I'm a newbie iOS Developer.
Swift 4, 4.2 and 5
let str = " Akbar Code "
let trimmedString = str.trimmingCharacters(in: .whitespaces)
I am a new to Scala and learning through the language constructs. While using print() with list.foreach() also prints the Nil or "()" in the console. Is this something expected or am I missing some trick here?
Code Snippet:
val oneTwo = "one"::"two"::Nil
println(oneTwo.foreach(s=> print(s+" ")))
o/p: one two ()
You have an extra println.
oneTwo.foreach(s=> print(s+" "))
Prints the contents of the list - "one two".
The println you have outside prints out the return value of the foreach statement, which is Unit (not Nil - that's a completely different beast), represented in scala as ().
To just output the list elements,
oneTwo.foreach(s=> print(s+" "))
would suffice. Now you put another print around that, so you say "and then print whatever oneTwo.foreach(s=> print(s+" ")) evaluates to".
The return type of foreach is Unit, so it'll return the only value of that type, the empty tuple ().
So what you see is the list elements printed by the print in the foreach, and then the outer print prints the result of the foreach. Does that make sense?
In visual lisp, you can use (atoi "123") to convert "123" to 123. It seems there is no "atoi" like function in clisp ?
any suggestion is appreciated !
Now i want to convert '(1 2 3 20 30) to "1 2 3 20 30", then what's the best way to do it ?
parse-interger can convert string to integer, and how to convert integer to string ? Do i need to use format function ?
(map 'list #'(lambda (x) (format nil "~D" x)) '(1 2 3)) => ("1" "2" "3")
But i donot know how to cnovert it to "1 2 3" as haskell does:
concat $ intersperse " " ["1","2","3","4","5"] => "1 2 3 4 5"
Sincerely!
In Common Lisp, you can use the read-from-string function for this purpose:
> (read-from-string "123")
123 ;
3
As you can see, the primary return value is the object read, which in this case happens to be an integer. The second value—the position—is harder to explain, but here it indicates the next would-be character in the string that would need to be read next on a subsequent call to a reading function consuming the same input.
Note that read-from-string is obviously not tailored just for reading integers. For that, you can turn to the parse-integer function. Its interface is similar to read-from-string:
> (parse-integer "123")
123 ;
3
Given that you were asking for an analogue to atoi, the parse-integer function is the more appropriate choice.
Addressing the second part of your question, post-editing, you can interleave (or "intersperse") a string with the format function. This example hard-codes a single space character as the separating string, using the format iteration control directives ~{ (start), ~} (end), and ~^ (terminate if remaining input is empty):
> (format nil "Interleaved: ~{~S~^ ~}." '(1 2 3))
"Interleaved: 1 2 3."
Loosely translated, the format string says,
For each item in the input list (~{), print the item by its normal conversion (~S). If no items remain, stop the iteration (~^). Otherwise, print a space, and then repeat the process with the next item (~}).
If you want to avoid hard-coding the single space there, and accept the separator string as a separately-supplied value, there are a few ways to do that. It's not clear whether you require that much flexibility here.