i start code in Ruby, but i want to know if in scan i could selection some operations symbol.
For example:
var = ["+100,-20,-15,+30"]
var.scan(/\d+/).map{|i| print i.to_i}
and my answer is:
0312100100102001010010010010
but i need keep ",+-", this si possible with scan?
because my problem is get a finish result in var
var in your question is an array, not a string (the square brackets on each side indicates it is an array). But if you did have a comma-separated string of integers and wanted to add those integers, like your comment suggested, you could do this:
var = "+100,-20,-15,+30"
var.split(",").map(&:to_i).sum(0)
# => 95
Related
I have a string containing 3 or 4 double numbers. what's the best way to extract them in an array of numbers?
First you have to find the numerals. You can use a RegExp pattern for that, say:
var doubleRE = RegExp(r"-?(?:\d*\.)?\d+(?:[eE][+-]?\d+)?");
Then you parse the resulting strings with double.parse. Something like:
var numbers = doubleRE.allMatches(input).map((m) => double.parse(m[0])).toList();
Hello im a newbie in Lua i just want to know if there is a way to get key and value of table not using pairs,ipairs,next or other iterators? thanks in advance.!
I don't believe this is possible, as you've phrased your question in such a way that implies that the key is unknown. The only way to check for a certain value and its corresponding key would be to iterate through the whole table.
However, maybe I misunderstood and you want to get a certain value from a key without iterating through the whole table.
Say you have a table named morse as follows:
morse = { a = ".-"; b = "-..."; } -- And so on
If you wanted to convert a single character to morse you could do as follows:
morse["a"] --Which will return the string ".-"
You can do the opposite, and define a table with all the morse values and their corresponding letters like below. Note the use of square brackets to 'escape' the characters.
morse = { [".-"] = "a"; ["-..."] = "b" }
morse[".-"] -- This will return "a"
Based on your comment, I think you are looking for a string substitution using a mapping table. I think you can use string.gsub here (if your teacher still insists that .gsub is an iterator; you can ask them politely that you are unaware of the method they claim and would be delighted to actually learn about the same):
local str = "sos sos sos"
local morse = {s = "...", o = "---"}
print( str:gsub("%a", morse) )
I am using ruby on rails
I have
article.id = 509969989168Q000475601
I would like the output to be
article.id = 68Q000475601
basically want to get rid of all before it gets to 68Q
the numbers in front of the 68Q can be various length
is there a way to remove up to "68Q"
it will always be 68Q and Q is always the only Letter
is there a way to say remove all characters from 2 digits before "Q"
I'd use:
article.id[/68Q.*/]
Which will return everything from 68Q to the end of the string.
article.id.match(/68Q.+\z/)[0]
You can do this easily with the split method:
'68Q' + article.id.split('68Q')[1]
This splits the string into an array based on the delimiter you give it, then takes the second element of that array. For what it's worth though, #theTinMan's solution is far more elegant.
An interesting Google Spreadsheet problem, I have a language file based on key=value that I have copied into a spreadsheet, eg.
titleMessage=Welcome to My Website
youAreLoggedIn=Hello #{user.name} you are now logged in
facebookPublish=Facebook Publishing
I have managed to split the key / value into two columns, and then translate the value column, and re-join it with the keys and Voila! this gives me a translated language file back
But as you may have spotted there are some variable in there (eg. #{user.name}) which are injected by my application, obviously I dont want to translate them.
So here is my question, given the following cell contents...
Hello #{user.name} you are now logged in
Is there a function that will translate the contents using the TRANSLATE function, but ignore anything inside #{ } (this could be at any point in the sentance)
Any Google Spreadsheet guru's have a solution for me?
Many thanks
If there are at most one occurrence of #{} then you could use the SPLIT function to divide the string into three parts that are arranged as below.
A B C D E
Original =SPLIT(An, "#{}") First piece Tag Rest of string
Translate Keep as is Translate
Put the pieces together with CONCATENATE.
=CONCATINATE(Cn,Dn,En)
I come up with same question.
Assume the escape pattern is #{sth.sth}(in regex as #{[\w.]+}). Replace them with string which Google Translate would view as untranslatable term, like VAR.
After translation, replace the term with original pattern.
Here is how I did this in script editor of spreadsheet:
function myTranslate(text, source_language, target_language) {
if(text.toString()) {
var str = text.toString();
var regex = /#{[\w.]+}/g; // g flag for multiple matches
var replace = 'VAR'; // Replace #{variable} to prevent from translation
var vars = str.match(regex).reverse(); // original patterns
str = str.replace(regex, replace);
str = LanguageApp.translate(str, source_language, target_language);
var ret = '';
for (var idx = str.search(replace); idx; idx = str.search(replace)) {
ret += str.slice(0, idx) + vars.pop();
str = str.slice(idx+replace.length);
}
return ret;
}
return null;
}
You can't just split and concatenate, because different languages use different word order of subject/predicate/object etc., and also because several languages modify nouns with different prefixes/suffixes/spelling changes depending on what they are doing in the sentence. It's all very complicated. Google needs to enable some sort of enclosing parentheses around any term we want to be quoted rather than translated.
I have read a multiline file and converted it to a list with the following code:
Lines = string:tokens(erlang:binary_to_list(Binary), "\n"),
I converted it to a string to do some work on it:
Flat = string:join(Lines, "\r\n"),
I finished working on the string and now I need to convert it back to a multiline list, I tried to repeat the first snippet shown above but that never worked, I tried string:join and that didnt work.. how do i convert it back to a list just like it used to be (although now modified)?
Well that depends on the modifications you made on the flattened string.
string:tokens/2 will always explode a string using the separator you provide. So as long as your transformation preserves a specific string as separator between the individual substrings there should be no problem.
However, if you do something more elaborate and destructive in your transformation then the only way is to iterate on the string manually and construct the individual substrings.
Your first snippet above contains a call to erlang:binary_to_list/1 which first converts a binary to a string (list) which you then split with the call to string:tokens/2 which then join together with string:join/2. The result of doing the tokens then join as you have written it seems to be to convert it from a string containing lines separated by \n into one containing lines separated by \r\n. N.B. that this is a flat list of characters.
Is this what you intended?
What you should do now depends on what you mean by "I need to convert it back to a multiline list". Do you mean everything in a single list of characters (string), or in a nested list of lines where each line is a list of characters (string). I.e. if you ended up with
"here is line 1\r\nhere is line 2\r\nhere is line 3\r\n"
this already is a multiline line list, or do you mean
["here is line 1","here is line 2","here is line 3"]
Note that each "string" is itself a list of characters. What do you intend to do with it afterwards?
You have your terms confused. A string in any language is a sequence of integer values corresponding to a human-readable characters. Whether the representation of the value is a binary or a list does not matter, both are technically strings because of the data they contain.
That being said, you converted a binary string to a list string in your first set of instructions. To convert a list into a binary, you can call erlang:list_to_binary/1, or erlang:iolist_to_binary/1 if your list is not flat. For instance:
BinString = <<"this\nis\na\nstring">>.
ListString = "this\nis\na\nstring" = binary_to_list(BinString).
Words = ["this", "is", "a", "string"] = string:tokens(ListString, "\n").
<<"thisisastring">> = iolist_to_binary(Words).
Rejoined = "this\r\nis\r\na\r\nstring" = string:join(Words, "\r\n").
BinAgain = <<"this\r\nis\r\na\r\nstring">> = list_to_binary(Rejoined).
For your reference, the string module always expects a flat list (e.g., "this is a string", but not ["this", "is", "a", "string"]), except for string:join, which takes a list of flat strings.