I am using Amibroker ver5.97
Suppose I have a stock symbol DAS.G and EFS.XF. I would like to remove the substring after the dot (including dot itself) so that the output becomes DAS and EFS respectively.
How can this be done in AFL?
Try this Explore:
Substring = StrFind(Name(),".");
Length = StrLen(Name());
Position = Length - (Length-Substring) - 1;
if ( Substring )
{
Ticker=StrLeft( Name(), Position);
}
else
{
Ticker=Name();
}
Filter = 1;
AddTextColumn(Ticker,"Ticker");
Related
I want to obtain the index chars or words in a string
for example
tv.text=" hey how are you, are you okay"
val res=tv.text.indexOf('h')
(have any way for put string instead of char?
output res=0
index of return only first char with h but in my tv text I have more h chars
can we return all h chars indexs
You can use filter function to get all the string indices with the required character.
val text = " hey how are you, are you okay"
val charToSearch = 'h'
val occurrences = text.indices.filter { text[it] == charToSearch }
println(occurences)
Try it yourself
And, if you want to search for strings instead of a single character, you can do this:
text.indices.filter { text.startsWith(stringToSearch, it) }
The following should work (you try to find an index if you found one in the previous iteration and you begin the folloing iteration from the previously found character instance plus 1, so that you don't find the same again and again):
fun main() {
val word = " hey how are you, are you okay"
val character = 'h'
var index: Int = word.indexOf(character)
while (index >= 0) {
println(index)
index = word.indexOf(character, index + 1)
}
}
If you want to store the indexes for later usage you can also do the following:
fun main() {
val word = " hey how are you, are you okay"
val character = 'h'
val indexes = mutableListOf<Int>()
var index: Int = word.indexOf(character)
while (index >= 0) {
index = word.indexOf(character, index + 1)
indexes.add(index)
}
println(indexes)
}
If you just want all the indices matching a char you can do this:
text.indices.filter { text[it] == 'h' }
Finding string matches is trickier, you could use Kotlin's regionMatches function to check if the part of the string starting at index matches what you're looking for:
val findMe = "you"
text.indices.filter { i ->
text.regionMatches(i, findMe, 0, findMe.length)
}
You could use a regular expression too, so long as you're careful about validating the search pattern:
Regex(findMe).findAll(text)
.map { it.range.first() } // getting the first index of each matching range
.toList()
Ex 10.0 = 10
10.3 = 10.3
10.7 = 10. 7
Looking for a convenient way from Kotlin Standard library
You can use the following function:
fun removeTrailingZeros(num: String): String {
if(!num.contains('.')) // Return the original number if it doesn't contain decimal
return num
return num
.dropLastWhile { it == '0' } // Remove trailing zero
.dropLastWhile { it == '.' } // Remove decimal in case it's the last character in the resultant string
}
You can verify the code here
You can try this:
double number = 23.471;
if (number % 1 != 0)
{
//round off here
System.out.print ("Decimal");
}
else
{
System.out.print ("Integer");
}
If you want to get a string, the easiest way is to work with a string like num.toString().replace(".0",""). For numbers conversion does not make sense since the resulting type is different for different inputs.
Try to understand running time of below algorithm for problem; Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results
This is a simple BFS solution that generates all possible strings by removing "(" or ")".
public List<String> removeInvalidParentheses(String s) {
List<String> ret = new LinkedList<>();
Set<String> visited = new HashSet<>();
Queue<String> queue = new LinkedList<>();
queue.add(s);
while (!queue.isEmpty()) {
String current = queue.poll();
if (isValidParentheses(current)) {
ret.add(current);
}
if (ret.size() > 0) continue;
for (int i = 0; i < current.length(); i++) {
if (current.charAt(i) == '(' || current.charAt(i) == ')') {
String next = current.substring(0, i) + current.substring(i + 1);
if (!visited.contains(next)) {
visited.add(next);
queue.offer(next);
}
}
}
}
return ret;
}
public boolean isValidParentheses(String current) {
int open = 0;
int close = 0;
for (char c : current.toCharArray()) {
if (c == '(') open++;
else if (c == ')') close++;
if (close > open) return false;
}
return open == close;
}
It starts with generate n possible strings and next level it generate all strings with size n-1 length, and n-2 length, etc .. for )()( example
)()( len n
()( ))( ()( )() n-1
() (( () n-2
each level it checks all possible strings with n-level length.
given this - I was having hard to time figure out how to finalize the running time of this algorithm. How do I generalize this algorithm and analyze the complexity?
For the worst case, lets try with input as ((((. As per the logic above, it will push (((( in the queue, checking that this is invalid. So it would generate 4 more possible substrings of length 3 pushing them inside the queue. Again, on processing that queue elements, it would again generate more strings of length 2 for each substring of length 3, then for two and then end. We are assuming that T(1) = 1.
If you try to make a recurrence relation for the above logic, it would be
T(n) = nT(n-1) + 1, which can be written as
= `n((n-1)T(n-2) + 1) + 1` and so on.
On solving it completely, we would get T(n) = n! + n(n-1)/2 + 1 which would be O(n!).
Hence, I think the time complexity would be of order O(n!)
For more details on how to solve the recurrence relation T(n) = nT(n-1) + 1, please refer:
this post on its solution
I got a question in vb.net
can I identify each of the character in string
for an example
i got a string of "Hello!. Good Afternoon!"
from this string can i trim away the period symbol?
Thank you
You should look at the methods of the String class, as they support different forms of string manipulation.
At its simplest, the Replace() method can be used to replace all occurrences of a period character with an empty string.
Alternatively, you can use the IndexOf() method to locate a specific string (e.g. the period) and the Remove() method to remove that character.
According to my 8-ball Magic, you actually want to :
Remove concecutive punctuation from a string:
With a Regex we are going to find all punctuation in the string.
The index of Match will be into an int[].
We will go iterate throught the array to find if the index is concecutive to the last punctuation index.
We will delete all the punctuation starting by the last one. Because starting with the 1rst will modify the index.
Code:
string Input = "....Thalassius! vero ea--*/-*/-- tempestate+- fectus";
string Output = Input;
var regex = new Regex(#"[^\w\s]|_"); // *1.
var matches = regex.Matches(Input) ;
var MatchesIndex = matches .Cast<Match>()
.Select(match => match.Index)
.ToArray(); // *2.
int last = 0;
List<int> toDelete = new List<int>();
for (int i = 0; i < MatchesIndex.Length; i++) // *3.
{
if ( MatchesIndex[i] == last + 1)
toDelete.Add(MatchesIndex[i]);
last = MatchesIndex[i];
}
foreach (int i in toDelete.OrderByDescending(x => x)) // *4.
Output = Output.Remove(i, 1);
Console.WriteLine("Input : " + Input);
Console.WriteLine("Output : " + Output);
C# Snippet
You can learn more about the regex used, thanks to #John Kugelman.
I want to keep white space when I call text attribute of token, is there any way to do it?
Here is the situation:
We have the following code
IF L > 40 THEN;
ELSE
IF A = 20 THEN
PUT "HELLO";
In this case, I want to transform it into:
if (!(L>40){
if (A=20)
put "hello";
}
The rule in Antlr is that:
stmt_if_block: IF expr
THEN x=stmt
(ELSE y=stmt)?
{
if ($x.text.equalsIgnoreCase(";"))
{
WriteLn("if(!(" + $expr.text +")){");
WriteLn($stmt.text);
Writeln("}");
}
}
But the result looks like:
if(!(L>40))
{
ifA=20put"hello";
}
The reason is that the white space in $stmt was removed. I was wondering if there is anyway to keep these white space
Thank you so much
Update: If I add
SPACE: [ ] -> channel(HIDDEN);
The space will be preserved, and the result would look like below, many spaces between tokens:
IF SUBSTR(WNAME3,M-1,1) = ')' THEN M = L; ELSE M = L - 1;
This is the C# extension method I use for exactly this purpose:
public static string GetFullText(this ParserRuleContext context)
{
if (context.Start == null || context.Stop == null || context.Start.StartIndex < 0 || context.Stop.StopIndex < 0)
return context.GetText(); // Fallback
return context.Start.InputStream.GetText(Interval.Of(context.Start.StartIndex, context.Stop.StopIndex));
}
Since you're using java, you'll have to translate it, but it should be straightforward - the API is the same.
Explanation: Get the first token, get the last token, and get the text from the input stream between the first char of the first token and the last char of the last token.
#Lucas solution, but in java in case you have troubles in translating:
private String getFullText(ParserRuleContext context) {
if (context.start == null || context.stop == null || context.start.getStartIndex() < 0 || context.stop.getStopIndex() < 0)
return context.getText();
return context.start.getInputStream().getText(Interval.of(context.start.getStartIndex(), context.stop.getStopIndex()));
}
Looks like InputStream is not always updated after removeLastChild/addChild operations. This solution helped me for one grammar, but it doesn't work for another.
Works for this grammar.
Doesn't work for modern groovy grammar (for some reason inputStream.getText contains old text).
I am trying to implement function name replacement like this:
enterPostfixExpression(ctx: PostfixExpressionContext) {
// Get identifierContext from ctx
...
const token = CommonTokenFactory.DEFAULT.createSimple(GroovyParser.Identifier, 'someNewFnName');
const node = new TerminalNode(token);
identifierContext.removeLastChild();
identifierContext.addChild(node);
UPD: I used visitor pattern for the first implementation