Convert bitstring to tuple - binary

I'm trying to find out how to convert an Erlang bitstring to a tuple, but so far without any luck.
What I want is to get from for example <<"{1,2}">> the tuple {1,2}.

You can use the modules erl_scan and erl_parse, as in this answer. Since erl_scan:string requires a string, not a binary, you have to convert the value with binary_to_list first:
> {ok, Scanned, _} = erl_scan:string(binary_to_list(<<"{1,2}">>)).
{ok,[{'{',1},{integer,1,1},{',',1},{integer,1,2},{'}',1}],1}
Then, you'd use erl_parse:parse_term to get the actual term. However, this function expects the term to end with a dot, so we have to add it explicitly:
> {ok, Parsed} = erl_parse:parse_term(Scanned ++ [{dot,0}]).
{ok,{1,2}}
Now the variable Parsed contains the result:
> Parsed.
{1,2}

You can use binary functions and erlang:list_to_tuple/1
1> B = <<"{1,2}">>.
<<"{1,2}">>
2> list_to_tuple([list_to_integer(binary_to_list(X)) || X <- binary:split(binary:part(B, 1, byte_size(B)-2), <<",">>, [global])]).
{1,2}

Related

Spliting string of arithmetic operations in lua

The following is string of arithmetic operations in lua.
local str ='x+abc*def+y^z+10'
Can this string be splitted so that individual variables or numbers will appear? For example, say string str is splitted into table s. Then the output will be
s[1] = x
s[2] = abc
s[3] = def
s[4] = y
s[5] = z
s[6] = 10
The splitting is to be done with operators +,-,*,\,^,%
Try also this simpler pattern:
local str ='x+(abc*def)+y^z+10'
for w in str:gmatch("%w+") do
print(w)
end
You can use string.gmatch to iterate over your string.
Feel free to add other operators to the pattern.
Refer to https://www.lua.org/manual/5.3/manual.html#6.4.1
local str ='x+abc*def+y^z+10'
local s = {}
for operand in str:gmatch('[^%+%*%^]+') do
table.insert(s, operand)
end
You can use string.gmatch to do what your looking for. you would use the pattern %+%-%*%^/
local str ='x+abc*def+y^z+10'
local s = {}
for value in str:gmatch("[%+%-%*%^/]*(%w*)[%+%-%*%^/]*") do
s[#s + 1] = value
end
print(unpack(s))
Also not that if you need \ as operator as shown in your question it would need to be escaped using an additional \.
Resourse for leaning more about lua patterns: understanding_lua_patterns

How to parse a string to key value pair using regex?

What is the best way to parse the string into key value pair using regex?
Sample input:
application="fre" category="MessagingEvent" messagingEventType="MessageReceived"
Expected output:
application "fre"
Category "MessagingEvent"
messagingEventType "MessageReceived"
We already tried the following regex and its working.
application=(?<application>(...)*) *category=(?<Category>\S*) *messagingEventType=(?<messagingEventType>\S*)
But we want a generic regex which will parse the sample input to the expected output as key value pair?
Any idea or solution will be helpful.
input = 'application="fre" category="MessagingEvent" messagingEventType="MessageReceived"'
puts input.
scan(/(\w+)="([^"]+)"/). # scan for KV-pairs
map{ |k, v| %Q|#{k.ljust(30,' ')}"#{v}"| }. # adjust as you requested
join($/) # join with platform-dependent line delimiters
#⇒ application "fre"
# category "MessagingEvent"
# messagingEventType "MessageReceived"
Instead of using regex, it can be done by spliting and storing the string in hash like below:
input = 'application="fre" category="MessagingEvent" messagingEventType="MessageReceived"'
res = {}
input.split.each { |str| a,b = str.split('='); res[a] = b}
puts res
==> {"application"=>"\"fre\"", "category"=>"\"MessagingEvent\"", "messagingEventType"=>"\"MessageReceived\""}

Parse a MySQL insert statement with multiple rows [duplicate]

I need a regular expression to select all the text between two outer brackets.
Example:
START_TEXT(text here(possible text)text(possible text(more text)))END_TXT
^ ^
Result:
(text here(possible text)text(possible text(more text)))
I want to add this answer for quickreference. Feel free to update.
.NET Regex using balancing groups:
\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\)
Where c is used as the depth counter.
Demo at Regexstorm.com
Stack Overflow: Using RegEx to balance match parenthesis
Wes' Puzzling Blog: Matching Balanced Constructs with .NET Regular Expressions
Greg Reinacker's Weblog: Nested Constructs in Regular Expressions
PCRE using a recursive pattern:
\((?:[^)(]+|(?R))*+\)
Demo at regex101; Or without alternation:
\((?:[^)(]*(?R)?)*+\)
Demo at regex101; Or unrolled for performance:
\([^)(]*+(?:(?R)[^)(]*)*+\)
Demo at regex101; The pattern is pasted at (?R) which represents (?0).
Perl, PHP, Notepad++, R: perl=TRUE, Python: PyPI regex module with (?V1) for Perl behaviour.
(the new version of PyPI regex package already defaults to this → DEFAULT_VERSION = VERSION1)
Ruby using subexpression calls:
With Ruby 2.0 \g<0> can be used to call full pattern.
\((?>[^)(]+|\g<0>)*\)
Demo at Rubular; Ruby 1.9 only supports capturing group recursion:
(\((?>[^)(]+|\g<1>)*\))
Demo at Rubular  (atomic grouping since Ruby 1.9.3)
JavaScript  API :: XRegExp.matchRecursive
XRegExp.matchRecursive(str, '\\(', '\\)', 'g');
Java: An interesting idea using forward references by #jaytea.
Without recursion up to 3 levels of nesting:
(JS, Java and other regex flavors)
To prevent runaway if unbalanced, with * on innermost [)(] only.
\((?:[^)(]|\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\))*\)
Demo at regex101; Or unrolled for better performance (preferred).
\([^)(]*(?:\([^)(]*(?:\([^)(]*(?:\([^)(]*\)[^)(]*)*\)[^)(]*)*\)[^)(]*)*\)
Demo at regex101; Deeper nesting needs to be added as required.
Reference - What does this regex mean?
RexEgg.com - Recursive Regular Expressions
Regular-Expressions.info - Regular Expression Recursion
Mastering Regular Expressions - Jeffrey E.F. Friedl 1 2 3 4
Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.
But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.
You can use regex recursion:
\(([^()]|(?R))*\)
[^\(]*(\(.*\))[^\)]*
[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.
This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.
Regular expressions can not do this.
Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.
In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:
0 1 1 0
-> S1 -> S2 -> S2 -> S2 ->S1
In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.
In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.
However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.
(?<=\().*(?=\))
If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible(*).
This regex just returns the text between the first opening and the last closing parentheses in your string.
(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.
It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.
You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.
Angle brackets <> were used because they do not require escaping.
The regular expression looks like this:
<
[^<>]*
(
(
(?<Open><)
[^<>]*
)+
(
(?<Close-Open>>)
[^<>]*
)+
)*
(?(Open)(?!))
>
I was also stuck in this situation when dealing with nested patterns and regular-expressions is the right tool to solve such problems.
/(\((?>[^()]+|(?1))*\))/
This is the definitive regex:
\(
(?<arguments>
(
([^\(\)']*) |
(\([^\(\)']*\)) |
'(.*?)'
)*
)
\)
Example:
input: ( arg1, arg2, arg3, (arg4), '(pip' )
output: arg1, arg2, arg3, (arg4), '(pip'
note that the '(pip' is correctly managed as string.
(tried in regulator: http://sourceforge.net/projects/regulator/)
I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing
balanced.matches({
source: source,
open: '(',
close: ')'
});
You can even do replacements:
balanced.replacements({
source: source,
open: '(',
close: ')',
replace: function (source, head, tail) {
return head + source + tail;
}
});
Here's a more complex and interactive example JSFiddle.
Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.
Lua
Use %b() (%b{} / %b[] for curly braces / square brackets):
for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)
Raku (former Perl6):
Non-overlapping multiple balanced parentheses matches:
my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* }
say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/;
# => (「(a(b)c)」 「((d)f(g))」)
Overlapping multiple balanced parentheses matches:
say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/;
# => (「(a(b)c)」 「(b)」 「((d)f(g))」 「(d)」 「(g)」)
See demo.
Python re non-regex solution
See poke's answer for How to get an expression between balanced parentheses.
Java customizable non-regex solution
Here is a customizable solution allowing single character literal delimiters in Java:
public static List<String> getBalancedSubstrings(String s, Character markStart,
Character markEnd, Boolean includeMarkers)
{
List<String> subTreeList = new ArrayList<String>();
int level = 0;
int lastOpenDelimiter = -1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == markStart) {
level++;
if (level == 1) {
lastOpenDelimiter = (includeMarkers ? i : i + 1);
}
}
else if (c == markEnd) {
if (level == 1) {
subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i)));
}
if (level > 0) level--;
}
}
return subTreeList;
}
}
Sample usage:
String s = "some text(text here(possible text)text(possible text(more text)))end text";
List<String> balanced = getBalancedSubstrings(s, '(', ')', true);
System.out.println("Balanced substrings:\n" + balanced);
// => [(text here(possible text)text(possible text(more text)))]
The regular expression using Ruby (version 1.9.3 or above):
/(?<match>\((?:\g<match>|[^()]++)*\))/
Demo on rubular
The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.
If you need to match matching nested brackets, then you need something more than regular expressions. - see #dehmann
If it's just first open to last close see #Zach
Decide what you want to happen with:
abc ( 123 ( foobar ) def ) xyz ) ghij
You need to decide what your code needs to match in this case.
"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.
This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns. This is where the re package greatly
assists in parsing.
"""
import re
# The pattern below recognises a sequence consisting of:
# 1. Any characters not in the set of open/close strings.
# 2. One of the open/close strings.
# 3. The remainder of the string.
#
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included. However quotes are not ignored inside
# quotes. More logic is needed for that....
pat = re.compile("""
( .*? )
( \( | \) | \[ | \] | \{ | \} | \< | \> |
\' | \" | BEGIN | END | $ )
( .* )
""", re.X)
# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.
matching = { "(" : ")",
"[" : "]",
"{" : "}",
"<" : ">",
'"' : '"',
"'" : "'",
"BEGIN" : "END" }
# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.
def matchnested(s, term=""):
lst = []
while True:
m = pat.match(s)
if m.group(1) != "":
lst.append(m.group(1))
if m.group(2) == term:
return lst, m.group(3)
if m.group(2) in matching:
item, s = matchnested(m.group(3), matching[m.group(2)])
lst.append(m.group(2))
lst.append(item)
lst.append(matching[m.group(2)])
else:
raise ValueError("After <<%s %s>> expected %s not %s" %
(lst, s, term, m.group(2)))
# Unit test.
if __name__ == "__main__":
for s in ("simple string",
""" "double quote" """,
""" 'single quote' """,
"one'two'three'four'five'six'seven",
"one(two(three(four)five)six)seven",
"one(two(three)four)five(six(seven)eight)nine",
"one(two)three[four]five{six}seven<eight>nine",
"one(two[three{four<five>six}seven]eight)nine",
"oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
"ERROR testing ((( mismatched ))] parens"):
print "\ninput", s
try:
lst, s = matchnested(s)
print "output", lst
except ValueError as e:
print str(e)
print "done"
You need the first and last parentheses. Use something like this:
str.indexOf('('); - it will give you first occurrence
str.lastIndexOf(')'); - last one
So you need a string between,
String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')');
because js regex doesn't support recursive match, i can't make balanced parentheses matching work.
so this is a simple javascript for loop version that make "method(arg)" string into array
push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)
const parser = str => {
let ops = []
let method, arg
let isMethod = true
let open = []
for (const char of str) {
// skip whitespace
if (char === ' ') continue
// append method or arg string
if (char !== '(' && char !== ')') {
if (isMethod) {
(method ? (method += char) : (method = char))
} else {
(arg ? (arg += char) : (arg = char))
}
}
if (char === '(') {
// nested parenthesis should be a part of arg
if (!isMethod) arg += char
isMethod = false
open.push(char)
} else if (char === ')') {
open.pop()
// check end of arg
if (open.length < 1) {
isMethod = true
ops.push({ method, arg })
method = arg = undefined
} else {
arg += char
}
}
}
return ops
}
// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)
console.log(test)
the result is like
[ { method: 'push', arg: 'number' },
{ method: 'map', arg: 'test(a(a()))' },
{ method: 'bass', arg: 'wow,abc' } ]
[ { method: '$$', arg: 'groups' },
{ method: 'filter',
arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
{ method: 'pickBy', arg: '_id,type' },
{ method: 'map', arg: 'test()' },
{ method: 'as', arg: 'groups' } ]
While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.
Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.
Read more # here
I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:
def extract_code(data):
""" returns an array of code snippets from a string (data)"""
start_pos = None
end_pos = None
count_open = 0
count_close = 0
code_snippets = []
for i,v in enumerate(data):
if v =='{':
count_open+=1
if not start_pos:
start_pos= i
if v=='}':
count_close +=1
if count_open == count_close and not end_pos:
end_pos = i+1
if start_pos and end_pos:
code_snippets.append((start_pos,end_pos))
start_pos = None
end_pos = None
return code_snippets
I used this to extract code snippets from a text file.
This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:
Parse parmeters from function string (with nested structures) in javascript
Match structures like:
matches brackets, square brackets, parentheses, single and double quotes
Here you can see generated regexp in action
/**
* get param content of function string.
* only params string should be provided without parentheses
* WORK even if some/all params are not set
* #return [param1, param2, param3]
*/
exports.getParamsSAFE = (str, nbParams = 3) => {
const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/;
const params = [];
while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine
str = str.replace(nextParamReg, (full, p1) => {
params.push(p1);
return '';
});
}
return params;
};
This might help to match balanced parenthesis.
\s*\w+[(][^+]*[)]\s*
This one also worked
re.findall(r'\(.+\)', s)

Json Files parsing

So I am trying to open some json files to look for a publication year and sort them accordingly. But before doing this, I decided to experiment on a single file. I am having trouble though, because although I can get the files and the strings, when I try to print one word, it starts printinf the characters.
For example:
print data2[1] #prints
THE BRIDES ORNAMENTS, Viz. Fiue MEDITATIONS, Morall and Diuine. #results
but now
print data2[1][0] #should print THE
T #prints T
This is my code right now:
json_data =open(path)
data = json.load(json_data)
i=0
data2 = []
for x in range(0,len(data)):
data2.append(data[x]['section'])
if len(data[x]['content']) > 0:
for i in range(0,len(data[x]['content'])):
data2.append(data[x]['content'][i])
I probably need to look at your json file to be absolutely sure, but it seems to me that the data2 list is a list of strings. Thus, data2[1] is a string. When you do data2[1][0], the expected result is what you are getting - the character at the 0th index in the string.
>>> data2[1]
'THE BRIDES ORNAMENTS, Viz. Fiue MEDITATIONS, Morall and Diuine.'
>>> data2[1][0]
'T'
To get the first word, naively, you can split the string by spaces
>>> data2[1].split()
['THE', 'BRIDES', 'ORNAMENTS,', 'Viz.', 'Fiue', 'MEDITATIONS,', 'Morall', 'and', 'Diuine.']
>>> data2[1].split()[0]
'THE'
However, this will cause issues with punctuation, so you probably need to tokenize the text. This link should help - http://www.nltk.org/_modules/nltk/tokenize.html

Converting epgsql results to JSON

I am a total beginner with Erlang and functional programming in general. For fun, to get me started, I am converting an existing Ruby Sinatra REST(ish) API that queries PostgreSQL and returns JSON.
On the Erlang side I am using Cowboy, Epgsql and Jiffy as the JSON library.
Epgsql returns results in the following format:
{ok, [{column,<<"column_name">>,int4,4,-1,0}], [{<<"value">>}]}
But Jiffy expects the following format when encoding to JSON:
{[{<<"column_name">>,<<"value">>}]}
The following code works to convert epgsql output into suitable input for jiffy:
Assuming Data is the Epgsql output and Key is the name of the JSON object being created:
{_, C, R} = Data,
Columns = [X || {_, X, _, _, _, _} <- C,
Rows = tuple_to_list(hd(R)),
Result = {[{atom_to_binary(Key, utf8), {lists:zip(Columns, Rows)}}]}.
However, I am wondering if this is efficient Erlang?
I've looked into the documentation for Epgsql and Jiffy and can't see any more obvious ways to perform the conversion.
Thank you.
Yes, need parse it.
For example function parse result
parse_result({error, #error{ code = <<"23505">>, extra = Extra }}) ->
{match, [Column]} =
re:run(proplists:get_value(detail, Extra),
"Key \\(([^\\)]+)\\)", [{capture, all_but_first, binary}]),
throw({error, {non_unique, Column}});
parse_result({error, #error{ message = Msg }}) ->
throw({error, Msg});
parse_result({ok, Cols, Rows}) ->
to_map(Cols, Rows);
parse_result({ok, Counts, Cols, Rows}) ->
{ok, Counts, to_map(Cols, Rows)};
parse_result(Result) ->
Result.
And function convert result to map
to_map(Cols, Rows) ->
[ maps:from_list(lists:zipwith(fun(#column{name = N}, V) -> {N, V} end,
Cols, tuple_to_list(Row))) || Row <- Rows ].
And encode it to json. You can change my code and make output as proplist.