Split lua string into characters - lua

I only found this related to what I am looking for: Split string by count of characters but it is not useful for what I mean.
I have a string variable, which is an ammount of 3 numbers (can be from 000 to 999). I need to separate each of the numbers (characters) and get them into a table.
I am programming for a game mod which uses lua, and it has some extra functions. If you could help me to make it using: http://wiki.multitheftauto.com/wiki/Split would be amazing, but any other way is ok too.
Thanks in advance

Corrected to what the OP wanted to ask:
To just split a 3-digit number in 3 numbers, that's even easier:
s='429'
c1,c2,c3=s:match('(%d)(%d)(%d)')
t={tonumber(c1),tonumber(c2),tonumber(c3)}
The answer to "How do I split a long string composed of 3 digit numbers":
This is trivial. You might take a look at the gmatch function in the reference manual:
s="123456789"
res={}
for num in s:gmatch('%d%d%d') do
res[#res+1]=tonumber(num)
end
or if you don't like looping:
res={}
s:gsub('%d%d%d',function(n)res[#res+1]=tonumber(n)end)

I was looking for something like this, but avoiding looping - and hopefully having it as one-liner. Eventually, I found this example from lua-users wiki: Split Join:
fields = {str:match((str:gsub("[^"..sep.."]*"..sep, "([^"..sep.."]*)"..sep)))}
... which is exactly the kind of syntax I'd like - one liner, returns a table - except, I don't really understand what is going on :/ Still, after some poking about, I managed to find the right syntax to split into characters with this idiom, which apparently is:
fields = { str:match( (str:gsub(".", "(.)")) ) }
I guess, what happens is that gsub basically puts parenthesis '(.)' around each character '.' - so that match would consider those as a separate match unit, and "extract" them as separate units as well... But I still don't get why is there extra pair of parenthesis around the str:gsub(".", "(.)") piece.
I tested this with Lua5.1:
str = "a - b - c"
fields = { str:match( (str:gsub(".", "(.)")) ) }
print(table_print(fields))
... where table_print is from lua-users wiki: Table Serialization; and this code prints:
"a"
" "
"-"
" "
"b"
" "
"-"
" "
"c"

Related

How to remove from string before __

I am building a Rails 5.2 app.
In this app I got outputs from different suppliers (I am building a webshop).
The name of the shipping provider is in this format:
dhl_freight__233433
It could also be in this format:
postal__US-320202
How can I remove all that is before (and including) the __ so all that remains are the things after the ___ like for example 233433.
Perhaps some sort of RegEx.
A very simple approach would be to use String#split and then pick the second part that is the last part in this example:
"dhl_freight__233433".split('__').last
#=> "233433"
"postal__US-320202".split('__').last
#=> "US-320202"
You can use a very simple Regexp and a ask the resulting MatchData for the post_match part:
p "dhl_freight__233433".match(/__/).post_match
# another (magic) way to acces the post_match part:
p $'
Postscript: Learnt something from this question myself: you don't even have to use a RegExp for this to work. Just "asddfg__qwer".match("__").post_match does the trick (it does the conversion to regexp for you)
r = /[^_]+\z/
"dhl_freight__233433"[r] #=> "233433"
"postal__US-320202"[r] #=> "US-320202"
The regular expression matches one or more characters other than an underscore, followed by the end of the string (\z). The ^ at the beginning of the character class reads, "other than any of the characters that follow".
See String#[].
This assumes that the last underscore is preceded by an underscore. If the last underscore is not preceded by an underscore, in which case there should be no match, add a positive lookbehind:
r = /(?<=__[^_]+\z/
This requires the match to be preceded by two underscores.
There are many ruby ways to extract numbers from string. I hope you're trying to fetch numbers out of a string. Here are some of the ways to do so.
Ref- http://www.ruby-forum.com/topic/125709
line.delete("^0-9")
line.scan(/\d/).join('')
line.tr("^0-9", '')
In the above delete is the fastest to trim numbers out of strings.
All of above extracts numbers from string and joins them. If a string is like this "String-with-67829___numbers-09764" outut would be like this "6782909764"
In case if you want the numbers split like this ["67829", "09764"]
line.split(/[^\d]/).reject { |c| c.empty? }
Hope these answers help you! Happy coding :-)

Match a word or whitespaces in Lua

(Sorry for my broken English)
What I'm trying to do is matching a word (with or without numbers and special characters) or whitespace characters (whitespaces, tabs, optional new lines) in a string in Lua.
For example:
local my_string = "foo bar"
my_string:match(regex) --> should return 'foo', ' ', 'bar'
my_string = " 123!#." -- note: three whitespaces before '123!#.'
my_string:match(regex) --> should return ' ', ' ', ' ', '123!#.'
Where regex is the Lua regular expression pattern I'm asking for.
Of course I've done some research on Google, but I couldn't find anything useful. What I've got so far is [%s%S]+ and [%s+%S+] but it doesn't seem to work.
Any solution using the standart library, e.g. string.find, string.gmatch etc. is OK.
Match returns either captures or the whole match, your patterns do not define those. [%s%S]+ matches "(space or not space) multiple times more than once", basically - everything. [%s+%S+] is plain wrong, the character class [ ] is a set of single character members, it does not treat sequences of characters in any other way ("[cat]" matches "c" or "a"), nor it cares about +. The [%s+%S+] is probably "(a space or plus or not space or plus) single character"
The first example 'foo', ' ', 'bar' could be solved by:
regex="(%S+)(%s)(%S+)"
If you want a variable number of captures you are going to need the gmatch iterator:
local capt={}
for q,w,e in my_string:gmatch("(%s*)(%S+)(%s*)") do
if q and #q>0 then
table.insert(capt,q)
end
table.insert(capt,w)
if e and #e>0 then
table.insert(capt,e)
end
end
This will not however detect the leading spaces or discern between a single space and several, you'll need to add those checks to the match result processing.
Lua standard patterns are simplistic, if you are going to need more intricate matching, you might want to have a look at lua lpeg library.

Lua String Manipulation (Find Words Before & After)

I'm fairly new to this forum. I am having trouble with manipulating the correct string to achieve this.
Basically, what I'm trying to do is receive an input string like this example:
str = "Say hello to=Stack overflow, Say goodbye to=other resources"
for question, answer in pairs(string.gmatch(s, "(%w+)=(%w+)"))
print(question, answer)
end
I want it to return: question = "Say hello to" and answer = "Stack overflow, question = "Say goodbye to" and so on and so forth. but instead, it picks up the word just before the equal sign and the word just after. I've even tried the * quantifier, and it does the same exact thing.
I've also tried this pattern
[%w%s]*=[%w%s]
I just want to be able to sort this string into a key-value table where the key is all words before each = and the value is all words after that equal but before the comma.
Does anyone have a suggestion?
You can use something like this:
local str = "Say hello to=Stack overflow, Say goodbye to=other resources"
for question, answer in string.gmatch(str..",", "([^=]+)=([^,]+),%s*") do
print(question, answer)
end
"([^=]+)=([^,]+),%s*" means the following: anything except = ([^=]) repeated 1 or more times (+) followed by = and then anything except ',', followed by comma and optional whitespaces (to avoid including them in the next question). I also added comma to the string, so it parses the last pair as well.
To elaborate a bit further per request in the comments: in the expression [^=]+, [=] designates a set with one allowed character (=) and [^=] negates that, so it's a set with any character allowed except = and + allows the set to be repeated 1 or more times.
As #lhf suggested you can use a simpler expression: (.-)=(.-),%s*, which means: take all characters until the first = (- makes matching non-greedy) and then take all characters until the first ,.

Lua Pattern Matching, get character before match

Currently I have code that looks like this:
somestring = "param=valueZ&456"
local stringToPrint = (somestring):gsub("(param=)[^&]+", "%1hello", 1)
StringToPrint will look like this:
param=hello&456
I have replaced all of the characters before the & with the string "hello". This is where my question becomes a little strange and specific.
I want my string to appear as: param=helloZ&456. In other words, I want to preserve the character right before the & when replacing the string valueZ with hello to make it helloZ instead. How can this be done?
I suggest:
somestring:gsub("param=[^&]*([^&])", "param=hello%1", 1)
See the Lua demo
Here, the pattern matches:
param= - literal substring param=
[^&]* - 0 or more chars other than & as many as possible
([^&]) - Group 1 capturing a symbol other than & (here, backtracking will occur, as the previous pattern grabs all such chars other than & and then the engine will take a step back and place the last char from that chunk into Group 1).
There are probably other ways to do this, but here is one:
somestring = "param=valueZ&456"
local stringToPrint = (somestring):gsub("(param=).-([^&]&)", "%1hello%2", 1)
print(stringToPrint)
The thing here is that I match the shortest string that ends with a character that is not & and a character that is &. Then I add the two ending characters to the replaced part.

Funny CSV format help

I've been given a large file with a funny CSV format to parse into a database.
The separator character is a semicolon (;). If one of the fields contains a semicolon it is "escaped" by wrapping it in doublequotes, like this ";".
I have been assured that there will never be two adjacent fields with trailing/ leading doublequotes, so this format should technically be ok.
Now, for parsing it in VBScript I was thinking of
Replacing each instance of ";" with a GUID,
Splitting the line into an array by semicolon,
Running back through the array, replacing the GUIDs with ";"
It seems to be the quickest way. Is there a better way? I guess I could use substrings but this method seems to be acceptable...
Your method sounds fine with the caveat that there's absolutely no possibility that your GUID will occur in the text itself.
On approach I've used for this type of data before is to just split on the semi-colons regardless then, if two adjacent fields end and start with a quote, combine them.
For example:
Pax;is;a;good;guy";" so;says;his;wife.
becomes:
0 Pax
1 is
2 a
3 good
4 guy"
5 " so
6 says
7 his
8 wife.
Then, when you discover that fields 4 and 5 end and start (respectively) with a quote, you combine them by replacing the field 4 closing quote with a semicolon and removing the field 5 opening quote (and joining them of course).
0 Pax
1 is
2 a
3 good
4 guy; so
5 says
6 his
7 wife.
In pseudo-code, given:
input: A string, first character is input[0]; last
character is input[length]. Further, assume one dummy
character, input[length+1]. It can be anything except
; and ". This string is one line of the "CSV" file.
length: positive integer, number of characters in input
Do this:
set start = 0
if input[0] = ';':
you have a blank field in the beginning; do whatever with it
set start = 2
endif
for each c between 1 and length:
next iteration unless string[c] = ';'
if input[c-1] ≠ '"' or input[c+1] ≠ '"': // test for escape sequence ";"
found field consting of half-open range [start,c); do whatever
with it. Note that in the case of empty fields, start≥c, leaving
an empty range
set start = c+1
endif
end foreach
Untested, of course. Debugging code like this is always fun….
The special case of input[0] is to make sure we don't ever look at input[-1]. If you can make input[-1] safe, then you can get rid of that special case. You can also put a dummy character in input[0] and then start your data—and your parsing—from input[1].
One option would be to find instances of the regex:
[^"];[^"]
and then break the string apart with substring:
List<string> ret = new List<string>();
Regex r = new Regex(#"[^""];[^""]");
Match m;
while((m = r.Match(line)).Success)
{
ret.Add(line.Substring(0,m.Index + 1);
line = line.Substring(m.Index + 2);
}
(Sorry about the C#, I don't known VBScript)
Using quotes is normal for .csv files. If you have quotes in the field then you may see opening and closing and the embedded quote all strung together two or three in a row.
If you're using SQL Server you could try using T-SQL to handle everything for you.
SELECT * INTO MyTable FROM OPENDATASOURCE('Microsoft.JET.OLEDB.4.0',
'Data Source=F:\MyDirectory;Extended Properties="text;HDR=No"')...
[MyCsvFile#csv]
That will create and populate "MyTable". Read more on this subject here on SO.
I would recommend using RegEx to break up the strings.
Find every ';' that is not a part of
";" and change it to something else
that does not appear in your fields.
Then go through and replace ";" with ;
Now you have your fields with the correct data.
Most importers can swap out separator characters pretty easily.
This is basically your GUID idea. Just make sure the GUID is unique to your file before you start and you will be fine. I tend to start using 'Z'. After enough 'Z's, you will be unique (sometimes as few as 1-3 will do).
Jacob

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