Afternoon,
I have an odd algorithm. I would like to populate a string of code dynamically based on some user entry.
I have a multi-dimensional array with data in it and a multi-line input text field.
What I want is for a user to be able to enter some text
example:
00
01 - 02 - 03
comments: 12
my code would identify the numbers an treat everything else as text.
Thus, if my array is data[x][#], the # will correspond to their entry.
I would get
algorithm_string = data[x][0] + "\n" + data[x][1] + " - " + data[x][2] + " - " + data[x][3] + "\n" + "comments: " + data[x][12]
So the algorithm would construct the above, and then I could run through the code.
for(var x:int = 0; x < data.length; x++){
some_object._display_text.text = algorithm_string;
}
Ok so I want to first say that relying on a user to put in the entry exactly the way you want is probably not a good idea. They WILL make mistakes and your code WILL eventually not work as expected. I would recommend using 5 inputs restricted to numeric input, and labeling each field with which number should go in it.
However, you can accomplish what you are trying to do above like this:
var parts:Array = myInput.text.split(" ");
for (var i:int=0; i<parts.length, i++){
if(!isNaN(parseInt(parts[i]))){
// you have a number here.
data[x].push(parts[i]);
} else {
//this was not a number so ignore it
}
}
Again let me state I think you should refactor how you get the numbers, but that code will grab the numbers out and put them in the 0,1,2,3,and 4 indexes of your data[x], but relies on the user perfectly inputting the text every time.
Good luck! (refactor) :)
Related
I have tried to use say-as interpret-as to make Alexa speak number in digits
Example - 9822 must not read in words instead '9,8,2,2'
One of the two ways I have tried is as follows:
this.emit(':tell',"Hi "+clientname+" your "+theIntentConfirmationStatus+" ticket is sent to "+ "<say-as interpret-as='digits'>" + clientno + "</say-as>",'backup');
The other one is this:
this.response.speak("Hi "+clientname+" your "+theIntentConfirmationStatus+" ticket is sent to "+ "<say-as interpret-as='digits'>" + clientno + "</say-as>");
Both are not working but working on a separate fresh function.
Actually your code SHOULD work.
Maybe you can try in test simulator and send us the code your script produces? Or the logs?
I've tried the following:
<speak>
1. The numbers are: <say-as interpret-as="digits">5498</say-as>.
2. The numbers are: <say-as interpret-as="spell-out">5498</say-as>.
3. The numbers are: <say-as interpret-as="characters">5498</say-as>.
4. The numbers are: <prosody rate="x-slow"><say-as interpret-as="digits">5498</say-as></prosody>.
5. The number is: 5498.
</speak>
Digits, Spell-out and Characters all have the effect you want.
If you want to Alexa to say it extra slow, use the prosody in #4.
Try using examples #2 or #3, maybe this works out?
Otherwise the example from Amod will work too.
You can split number into individual digits using sample function ( please test it for your possible inputs-its not tested for all input). You can search for similar function on stackoverflow
function getNumber(tablenumber) {
var number = (""+tablenumber).split("");
var arrayLength = number.length;
var tmp =" ";
for (var i = 0; i < arrayLength; i++) {
var tmp = tmp + myStringArray[i] + ", <break time=\"0.4s\"/> ";
}
return tmp;
}
In your main function... call this
var finalresult = getNumber(clientno);
this.emit(':tell',"Hi "+clientname+" your "+theIntentConfirmationStatus+" ticket is sent to "+ finalresult ,'backup');
Edited: Yep, nightflash's answer is great.
You could also break the numbers up yourself if you need other formatting, such as emphasizing particular digits, add pauses, etc. You would need to use your Lambda code to convert the numeric string to multiple digits separated by spaces and any other formatting you need.
Here's an example based on the answers in this post:
var inputNumber = 12354987;
var output = '';
var sNumber = inputNumber.toString();
for (var i = 0, len = sNumber.length; i < len; i += 1) {
// just adding spaces here, but could be SSML attributes, etc.
output = output + sNumber.charAt(i) + ' ';
}
console.log(output);
This code could be refactored and done many other ways, but I think this is about the easiest to understand.
Most Frequent Character
Design a program that prompts the user to enter a string, and displays the character that appears most frequently in the string.
It is a homework question, but my teacher wasn't helpful and its driving me crazy i can't figure this out.
Thank You in advance.
This is what i have so far!
Declare String str
Declare Integer maxChar
Declare Integer index
Set maxChar = 0
Display “Enter anything you want.”
Input str
For index = 0 To length(str) – 1
If str[index] =
And now im stuck. I dont think its right and i dont know where to go with it!
It seems to me that the way you want to do it is:
"Go through every character in the string and remember the character we've seen most times".
However, that won't work. If we only remember the count for a single character, like "the character we've seen most times is 'a' with 5 occurrences", we can't know if perhaps the character in the 2nd place doesn't jump ahead.
So, what you have to do is this:
Go through every character of the string.
For every character, increase the occurrence count for that character. Yes, you have to save this count for every single character you encounter. Simple variables like string or int are not going to be enough here.
When you're done, you're left with a bunch of data looking like "a"=5, "b"=2, "e"=7,... you have to go though that and find the highest number (I'm sure you can find examples for finding the highest number in a sequence), then return the letter which this corresponds to.
Not a complete answer, I know, but that's all I'm going to say.
If you're stuck, I suggest getting a pen and a piece of paper and trying to calculate it manually. Try to think - how would you do it without a computer? If your answer is "look and see", what if the text is 10 pages? I know it can be pretty confusing, but the point of all this is to get you used to a different way of thinking. If you figure this one out, the next time will be easier because the basic principles are always the same.
This is the code I have created to count all occurences in a string.
String abc = "aabcabccc";
char[] x = abc.toCharArray();
String _array = "";
for(int i = 0; i < x.length; i++) //copy distinct data to a new string
{
if(_array.indexOf(x[i]) == -1)
_array = _array+x[i];
}
char[] y = _array.toCharArray();
int[] count1 = new int[_array.length()];
for(int j = 0; j<x.length;j++) //count occurences
{
count1[new String(String.valueOf(y)).indexOf(x[j])]++;
}
for(int i = 0; i<y.length;i++) //display
{
System.out.println(y[i] + " = " + count1[i]);
}
Right now i have to write a code that will print out "*" for each collectedDots, however if it doesn't collect any collectedDots==0 then it print out "". Using too many if statements look messy and i was wandering how you would implement the forloop in this case.
As a general principle the kind of rearrangement you've done here is good. You have found a way to express the rule in a general way rather than as a sequence of special cases. This is much easier to reason about and to check, and it's obviously extensible to cases where you have more than 3 dots.
You probably have made an error in confusing your target number and the iteration value, I assume that collectedDots contains the number of dots you have (as per your if statement) and so you need to introduce a variable to count up to that value
for (int i =0; i <= collectedDots; i++)
{
stars = "*";
System.out.print(stars);
}
Ok, so you already have a variable called collectedDots that is a number which tells you how many stars to print?
So your loop would be something like
for every collected dot
print *
But you can't just print it out, you need to return a string that will be printed out. So it's more like
for every collected dot
add a * to our string
return the string
They key difference between this and your attempt so far is that you were assigning a star to be your string each time through the loop, then at the end of it, you return that string–no matter how many times you assign a star to the string, the string will always just be one star.
You also need a separate variable to keep track of your loop, this should do the trick:
String stars = "";
for(int i = 0; i < collectedDots; i++)
{
stars = stars + "*";
}
return stars;
You are almost correct. Just need to change range limit of looping. Looping initial value is set to 1. So whenever you have collectedDots = 0, it will not go in loop and will return "", as stars is intialized with "" before loop.
String stars = "";
for (int i =1; i <= collectedDots; i++)
{
stars = "*";
System.out.print(stars);
}
return stars;
A bit of background before I astart: I am by no means an InDesign expert, but I've been tasked with pulling some text information out of our company's InDesign files to do a bit of analytics. Each InDesign file that I'm working with has multiple pages, and each of those pages use Stories from the Master page (Forgive me if I'm using the wrong terminology, I'm working on a limited understanding of the InDesign object model). Using the default Export All Text script as a starting point, I was able to expand that functionality to dump out other useful information; such as the geometric bounds. I've posted some of the code below:
myFileName = "StoryID" + myID + myExtension;
myFilePath = myFolder + "/" + myFileName;
myFile = new File(myFilePath);
myFile.encoding='text';
myFile.open('w');
containers = myStory.textContainers;
if (containers[0].overridden) {
for (i = 0; i < myStory.words.length; i++) {
myFile.write(myStory.words[i].contents);
myFile.write(" ");
}
myFile.write("\n");
}
else {
myFile.write(myStory.contents + '\n');
}
var geometry = containers[0].geometricBounds
myFile.write("++" + containers[0].parent.name + '\n');
myFile.write("$$" + geometry[0] + "\t" + geometry[1] + "\t" + geometry[2] + "\t" + geometry[3] + "\n");
pageitems = myStory.allPageItems;
for (j = 0; j < pageitems.length; j++){
myFile.write("--" + pageitems[0]);
}
myFile.close();
The question I have is, how do I determine all the pages that a Story appears in? For example, I lets say I have 4 pages: {Master, Region 1, Region 2, Region 3}. Region 1 - 3 all inherit a Story block from Master. When I run this script I only get one instance of the Story block, the instance belonging to Master. However, I would be interested to know that the Story block in question belongs to all 4 pages. I've tried accessing the allPageItems property but I keep getting 0 results.
Any help would be appreciated. I'm working on InDesign CS 6 if that helps
When you're accessing allPageItems of the Story element, you're accessing all of the page items that are anchored within that story, which I don't think is what you want.
When you access the allPageItems property you get 0 results because your Story doesn't contain any textFrames. The Story element is contained by textContainers not textFrames. I know it's confusing!
If you want to determine whether a story is on a particular page, you can access the textContainers’ parentPage property.
A Story flows through multiple textContainers. For example:
link link
TextContainer --> TextContainer2 --> TextContainer3
|_____________________________________________________________________|
|
Story
A Story can also contain textFrames, rectangles, words, etc. For example:
Story
___________________________________|__________________________________
| |
Word --> Word2 --> TextFrame --> Word3 --> Word4 --> Word5
This is a bit of a side project I have taken on to solve a no-fix issue for work. Our system outputs a code to represent a combination of things on another thing. Some example codes are:
9-9-0-4-4-5-4-0-2-0-0-0-2-0-0-0-0-0-2-1-2-1-2-2-2-4
9-5-0-7-4-3-5-7-4-0-5-1-4-2-1-5-5-4-6-3-7-9-72
9-15-0-9-1-6-2-1-2-0-0-1-6-0-7
The max number in one of the slots I've seen so far is about 150 but they will likely go higher.
When the system was designed there was no requirement for what this code would look like. But now the client wants to be able to type it in by hand from a sheet of paper, something the code above isn't suited for. We've said we won't do anything about it, but it seems like a fun challenge to take on.
My question is where is a good place to start loss-less compressing this code? Obvious solutions such as store this code with a shorter key are not an option; our database is read only. I need to build a two way method to make this code more human friendly.
1) I agree that you definately need a checksum - data entry errors are very common, unless you have really well trained staff and independent duplicate keying with automatic crosss-checking.
2) I suggest http://en.wikipedia.org/wiki/Huffman_coding to turn your list of numbers into a stream of bits. To get the probabilities required for this, you need a decent sized sample of real data, so you can make a count, setting Ni to the number of times number i appears in the data. Then I suggest setting Pi = (Ni + 1) / (Sum_i (Ni + 1)) - which smooths the probabilities a bit. Also, with this method, if you see e.g. numbers 0-150 you could add a bit of slack by entering numbers 151-255 and setting them to Ni = 0. Another way round rare large numbers would be to add some sort of escape sequence.
3) Finding a way for people to type the resulting sequence of bits is really an applied psychology problem but here are some suggestions of ideas to pinch.
3a) Software licences - just encode six bits per character in some 64-character alphabet, but group characters in a way that makes it easier for people to keep place e.g. BC017-06777-14871-160C4
3b) UK car license plates. Use a change of alphabet to show people how to group characters e.g. ABCD0123EFGH4567IJKL...
3c) A really large alphabet - get yourself a list of 2^n words for some decent sized n and encode n bits as a word e.g. GREEN ENCHANTED LOGICIAN... -
i worried about this problem a while back. it turns out that you can't do much better than base64 - trying to squeeze a few more bits per character isn't really worth the effort (once you get into "strange" numbers of bits encoding and decoding becomes more complex). but at the same time, you end up with something that's likely to have errors when entered (confusing a 0 with an O etc). one option is to choose a modified set of characters and letters (so it's still base 64, but, say, you substitute ">" for "0". another is to add a checksum. again, for simplicity of implementation, i felt the checksum approach was better.
unfortunately i never got any further - things changed direction - so i can't offer code or a particular checksum choice.
ps i realised there's a missing step i didn't explain: i was going to compress the text into some binary form before encoding (using some standard compression algorithm). so to summarize: compress, add checksum, base64 encode; base 64 decode, check checksum, decompress.
This is similar to what I have used in the past. There are certainly better ways of doing this, but I used this method because it was easy to mirror in Transact-SQL which was a requirement at the time. You could certainly modify this to incorporate Huffman encoding if the distribution of your id's is non-random, but it's probably unnecessary.
You didn't specify language, so this is in c#, but it should be very easy to transition to any language. In the lookup you'll see commonly confused characters are omitted. This should speed up entry. I also had the requirement to have a fixed length, but it would be easy for you to modify this.
static public class CodeGenerator
{
static Dictionary<int, char> _lookupTable = new Dictionary<int, char>();
static CodeGenerator()
{
PrepLookupTable();
}
private static void PrepLookupTable()
{
_lookupTable.Add(0,'3');
_lookupTable.Add(1,'2');
_lookupTable.Add(2,'5');
_lookupTable.Add(3,'4');
_lookupTable.Add(4,'7');
_lookupTable.Add(5,'6');
_lookupTable.Add(6,'9');
_lookupTable.Add(7,'8');
_lookupTable.Add(8,'W');
_lookupTable.Add(9,'Q');
_lookupTable.Add(10,'E');
_lookupTable.Add(11,'T');
_lookupTable.Add(12,'R');
_lookupTable.Add(13,'Y');
_lookupTable.Add(14,'U');
_lookupTable.Add(15,'A');
_lookupTable.Add(16,'P');
_lookupTable.Add(17,'D');
_lookupTable.Add(18,'S');
_lookupTable.Add(19,'G');
_lookupTable.Add(20,'F');
_lookupTable.Add(21,'J');
_lookupTable.Add(22,'H');
_lookupTable.Add(23,'K');
_lookupTable.Add(24,'L');
_lookupTable.Add(25,'Z');
_lookupTable.Add(26,'X');
_lookupTable.Add(27,'V');
_lookupTable.Add(28,'C');
_lookupTable.Add(29,'N');
_lookupTable.Add(30,'B');
}
public static bool TryPCodeDecrypt(string iPCode, out Int64 oDecryptedInt)
{
//Prep the result so we can exit without having to fiddle with it if we hit an error.
oDecryptedInt = 0;
if (iPCode.Length > 3)
{
Char[] Bits = iPCode.ToCharArray(0,iPCode.Length-2);
int CheckInt7 = 0;
int CheckInt3 = 0;
if (!int.TryParse(iPCode[iPCode.Length-1].ToString(),out CheckInt7) ||
!int.TryParse(iPCode[iPCode.Length-2].ToString(),out CheckInt3))
{
//Unsuccessful -- the last check ints are not integers.
return false;
}
//Adjust the CheckInts to the right values.
CheckInt3 -= 2;
CheckInt7 -= 2;
int COffset = iPCode.LastIndexOf('M')+1;
Int64 tempResult = 0;
int cBPos = 0;
while ((cBPos + COffset) < Bits.Length)
{
//Calculate the current position.
int cNum = 0;
foreach (int cKey in _lookupTable.Keys)
{
if (_lookupTable[cKey] == Bits[cBPos + COffset])
{
cNum = cKey;
}
}
tempResult += cNum * (Int64)Math.Pow((double)31, (double)(Bits.Length - (cBPos + COffset + 1)));
cBPos += 1;
}
if (tempResult % 7 == CheckInt7 && tempResult % 3 == CheckInt3)
{
oDecryptedInt = tempResult;
return true;
}
return false;
}
else
{
//Unsuccessful -- too short.
return false;
}
}
public static string PCodeEncrypt(int iIntToEncrypt, int iMinLength)
{
int Check7 = (iIntToEncrypt % 7) + 2;
int Check3 = (iIntToEncrypt % 3) + 2;
StringBuilder result = new StringBuilder();
result.Insert(0, Check7);
result.Insert(0, Check3);
int workingNum = iIntToEncrypt;
while (workingNum > 0)
{
result.Insert(0, _lookupTable[workingNum % 31]);
workingNum /= 31;
}
if (result.Length < iMinLength)
{
for (int i = result.Length + 1; i <= iMinLength; i++)
{
result.Insert(0, 'M');
}
}
return result.ToString();
}
}