Sorry, I ran into another question about using PetitParser. I've figured out my recursive issues, but now I have a problem with parentheses. If I need to be able to parse the following two expressions:
'(use = "official").empty()'
'(( 5 + 5 ) * 5) + 5'
I've tried doing something like the following:
final expression = (char('(') & any().starGreedy(char(')')).flatten() & char(')')).map((value) => ParenthesesParser(value));
But that doesnt' work on the first expression.
If I try this:
final expression = (char('(') & any().starLazy(char(')')).flatten() & char(')')).map((value) => ParenthesesParser(value));
It doesn't work on the second expression. Any suggestions on how to parse both?
I think neither of the parsers does what you want: The first parser, the greedy one with starGreedy, will consume up to the last closing parenthesis. The second parser, the lazy one with starLazy, will consume up to the first closing parenthesis.
To parse a balanced parenthesis you need recursion, so that each opening parenthesis is followed by a matching closing one:
final inner = undefined();
final parser = char('(') & inner.star().flatten() & char(')');
inner.set(parser | pattern('^)'));
In the snippet above, the inner parser is recursively trying to either parse another parenthesis pair, or otherwise it simply consumes any character that is not a closing parenthesis.
Related
How to properly format a math expression before passing it to the math_expressions package in flutter?
Context
I'm using math_expressions package but there are two cases I found when it throws an error:
A. Missing an asterisk before a parenthesis.
B. Missing parenthesis within the expression.
E.g.
// Throws error
final expression = "8(3+1)"; // A
final expression = "8(3+1"; // B
// Executes correctly
final expression = "8*(3+1)";
final Parser parser = Parser();
Expression exp = parser.parse(expression);
ContextModel cm = ContextModel();
final double result = exp.evaluate(EvaluationType.REAL, cm);
I'm aware of the syntactic requirement of the package so I'd like to properly format the expression before passing it to the parser since I cannot guarantee user input will comply to the requirement mentioned before.
What I've got so far
A. Missing an asterisk before a parenthesis:
I read about the replaceAllMapped method but I don't really know how to start from here in order to add the missing asterisks when needed.
B. Missing parenthesis within the expression. (solved)
Hypothesis
A. Missing an asterisk before a parenthesis:
I think the way is to create an array of digits, search for coincidences of a digit + parenthesis and then replace it with the addition of an asterisk like this: digit + "*" + parenthesis
Any ideas on how to solve this appropriately?
My code has a ton of occurrences of something like:
idof(some_object)
I want to replace them with:
some_object["id"]
It sounds simple:
sed -i 's/idof(\([^)]\+\))/\1["id"]/g' source.py
The problem is that some_object might be something like idof(get_some_object()), or idof(my_class().get_some_object()), in which case, instead of getting what I want (get_some_object()["id"] or my_class().get_some_object()["id"]), I get get_some_object(["id"]) or my_class(["id"].get_some_object()).
Is there a way to have sed match closing bracket, so that it internally keeps track of any opening/closing brackets inside my (), and ignores those?
It needs to keep everything that's between those brackets: idof(ANYTHING) becomes ANYTHING["id"].
Using sed
$ sed -E 's/idof\(([[:alpha:][:punct:]]*)\)/\1["id"]/g' input_file
Using ERE, exclude idof and the first opening parenthesis.
As a literal closing parenthesis is also excluded, everything in-between the capture parenthesis including additional parenthesis will be captured.
[[:alpha:]] will match all alphabetic characters including upper and lower case while [[:punct:]] will capture punctuation characters including ().-{} and more.
The g option will make the substitution as many times as the pattern is found.
Theoretically, you can write a regex that will handle all combinations of idof(....) up to some limit of nested () calls inside ..... Such regex would have to list with all possible combinations of calls, like idof(one(two(three))) or idof(one(two(three)four(five)) you can match with an appropriate regex like idof([^()]*([^()]*([^()]*)[^()]*)[^()]*) or idof([^()]*([^()]*([^()]*)[^()]*([^()]*)[^()]*) respectively.
The following regex handles only some cases, but shows the complexity and general path. Writing a regex to handle all possible cases to "eat" everything in front of the trailing ) is left to OP as an exercise why it's better to use something else. Note that handling string literals ")" becomes increasingly complex.
The following Bash code:
sed '
: begin
# No idof? Just print the line!
/^\(.*\)idof(\([^)]*)\)/!n
# Note: regex is greedy - we start from the back!
# Note: using newline as a stack separator.
s//\1\n\2/
# hold the front
{ h ; x ; s/\n.*// ; x ; s/[^\n]*\n// ; }
: handle_brackets
# Eat everything before final ) up to some number of nested ((())) calls.
# Insert more jokes here.
: eat_brackets
/^[^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)/{
s//&\n/
# Hold the front.
{ H ; x ; s/\n\([^\n]*\)\n.*/\1/ ; x ; s/[^\n]*\n// ; }
b eat_brackets
}
/^\([^()]*\))/!{
s/^/ERROR: eating brackets did not work: /
q1
}
# Add the id after trailing ) and remove it.
s//\1["id"]/
# Join with hold space and clear the hold space for next round
{ H ; s/.*// ; x ; s/\n//g ; }
# Restart for another idof if in input.
b begin
' <<EOF
before idof(some_object) after
before idof(get_some_object()) after
before idof(my_class().get_some_object()) after
before idof(one(two(three)four)five) after
before idof(one(two(three)four)five) between idof(one(two(three)four)five) after
before idof( one(two(three)four)five one(two(three)four)five ) after
before idof(one(two(three(four)five)six(seven(eight)nine)ten) between idof(one(two(three(four)five)six(seven(eight)nine)ten) after
EOF
Will output:
before some_object["id"] after
before get_some_object()["id"] after
before my_class().get_some_object()["id"] after
before one(two(three)four)five["id"] after
before one(two(three)four)five["id"] between one(two(three)four)five["id"] after
before one(two(three)four)five one(two(three)four)five ["id"] after
ERROR: eating brackets did not work: one(two(three(four)five)six(seven(eight)nine)ten) after
The last line is not handled correctly, because (()()) case is not correctly handled. One would have to write a regex to match it.
I'm using AutoHotkey for this as the code is the most understandable to me. So I have a document with numbers and text, for example like this
120344 text text text
234000 text text
and the desired output is
12:03:44 text text text
23:40:00 text text
I'm sure StrReplace can be used to insert the colons in, but I'm not sure how to specify the position of the colons or ask AHK to 'find' specific strings of 6 digit numbers. Before, I would have highlighted the text I want to apply StrReplace to and then press a hotkey, but I was wondering if there is a more efficient way to do this that doesn't need my interaction. Even just pointing to the relevant functions I would need to look into to do this would be helpful! Thanks so much, I'm still very new to programming.
hfontanez's answer was very helpful in figuring out that for this problem, I had to use a loop and substring function. I'm sure there are much less messy ways to write this code, but this is the final version of what worked for my purposes:
Loop, read, C:\[location of input file]
{
{ If A_LoopReadLine = ;
Continue ; this part is to ignore the blank lines in the file
}
{
one := A_LoopReadLine
x := SubStr(one, 1, 2)
y := SubStr(one, 3, 2)
z := SubStr(one, 5)
two := x . ":" . y . ":" . z
FileAppend, %two%`r`n, C:\[location of output file]
}
}
return
Assuming that the "timestamp" component is always 6 characters long and always at the beginning of the string, this solution should work just fine.
String test = "012345 test test test";
test = test.substring(0, 2) + ":" + test.substring(2, 4) + ":" + test.substring(4, test.length());
This outputs 01:23:45 test test test
Why? Because you are temporarily creating a String object that it's two characters long and then you insert the colon before taking the next pair. Lastly, you append the rest of the String and assign it to whichever String variable you want. Remember, the substring method doesn't modify the String object you are calling the method on. This method returns a "new" String object. Therefore, the variable test is unmodified until the assignment operation kicks in at the end.
Alternatively, you can use a StringBuilder and append each component like this:
StringBuilder sbuff = new StringBuilder();
sbuff.append(test.substring(0,2));
sbuff.append(":");
sbuff.append(test.substring(2,4));
sbuff.append(":");
sbuff.append(test.substring(4,test.length()));
test = sbuff.toString();
You could also use a "fancy" loop to do this, but I think for something this simple, looping is just overkill. Oh, I almost forgot, this should work with both of your test strings because after the last colon insert, the code takes the substring from index position 4 all the way to the end of the string indiscriminately.
diary_file = tempname();
diary(diary_file);
myFun();
diary('off');
output = fileread(diary_file);
I would like to search a string from output, but also to ignore spaces and upper/lower cases. Here is an example for what's in output:
the test : passed
number : 4
found = 'thetest:passed'
a = strfind(output,found )
How could I ignore spaces and cases from output?
Assuming you are not too worried about accidentally matching something like: 'thetEst:passed' here is what you can do:
Remove all spaces and only compare lower case
found = 'With spaces'
found = lower(found(found ~= ' '))
This will return
found =
withspaces
Of course you would also need to do this with each line of output.
Another way:
regexpi(output(~isspace(output)), found, 'match')
if output is a single string, or
regexpi(regexprep(output,'\s',''), found, 'match')
for the more general case (either class(output) == 'cell' or 'char').
Advantages:
Fast.
robust (ALL whitespace (not just spaces) is removed)
more flexible (you can return starting/ending indices of the match, tokenize, etc.)
will return original case of the match in output
Disadvantages:
more typing
less obvious (more documentation required)
will return original case of the match in output (yes, there's two sides to that coin)
That last point in both lists is easily forced to lower or uppercase using lower() or upper(), but if you want same-case, it's a bit more involved:
C = regexpi(output(~isspace(output)), found, 'match');
if ~isempty(C)
C = found; end
for single string, or
C = regexpi(regexprep(output, '\s', ''), found, 'match')
C(~cellfun('isempty', C)) = {found}
for the more general case.
You can use lower to convert everything to lowercase to solve your case problem. However ignoring whitespace like you want is a little trickier. It looks like you want to keep some spaces but not all, in which case you should split the string by whitespace and compare substrings piecemeal.
I'd advertise using regex, e.g. like this:
a = regexpi(output, 'the\s*test\s*:\s*passed');
If you don't care about the position where the match occurs but only if there's a match at all, removing all whitespaces would be a brute force, and somewhat nasty, possibility:
a = strfind(strrrep(output, ' ',''), found);
I have the following code:
object testLines extends App {
val items = Array("""a-b-c d-e-f""","""a-b-c th-i-t""")
val lines = items.map(_.replaceAll("-", "")split("\t"))
print(lines.map(_.mkString(",")).mkString("\n"))
}
By mistake i did not put a dot between replaceAll and split but it worked.
By contrary when putting a dot between replaceAll and split i got an error
identifier expected but ';' found.
Implicit conversions found: items =>
What is going on?
Why does it work without a dot but is not working with a dot.
Update:
It works also with dot. The error message is a bug in the scala ide. The first part of the question is still valid
Thanks,
David
You have just discovered that Operators are methods. x.split(y) can also be written x split y in cases where the method is operator-like and it looks nicer. However there is nothing stopping you putting either side in parentheses like x split (y), (x) split y, or even (x) split (y) which may be necessary (and is a good idea for readability even if not strictly necessary) if you are passing in a more complex expression than a simple variable or constant and need parentheses to override the precedence.
With the example code you've written, it's not a bad idea to do the whole thing in operator style for clarity, using parentheses only where the syntax requires and/or they make groupings more obvious. I'd probably have written it more like this:
object testLines extends App {
val items = Array("a-b-c d-e-f", "a-b-c th-i-t")
val lines = items map (_ replaceAll ("-", "") split "\t")
print(lines map (_ mkString ",") mkString "\n")
}