Hi guys I am really stuck in this one situation :S I have a local .txtfile with a random sentence and my program is meant to :
I am finding it difficult to execute the third question. My code is ..
JavaScript
lengths.forEach((leng) => {
counter[leng] = counter[leng] || 0;
counter[leng]++;
});
$("#display_File_most").text(counter);
}
}
r.readAsText(f);
}
});
</script>
I have used this question for help but no luck - Using Javascript to find most common words in string?
I believe I have to store the sentence in an array and loop through it, uncertain if that is the correct step or if there is quicker way of finding the solution so I ask you guys.
Thanks for your time & I hope my question made sense :)
If you think of your solution as separated well done tasks, it would be really simple to find it. Here you have them together:
Convert the words into an array. Your guts were right about this :)
var source = "Hello world & good morning. The date is 18/09/2018";
var words = source.split(' ');
The next step is to find out the length of each word
var lengths = words.map(function(word) {
return word.length;
});
Finally the most complicated part is to get the number of occurrences for each length. One idea is to use an object to use key/value where key is the length and value is its count (source: https://stackoverflow.com/a/10541220/1505348)
Now you will see under the counter object have each word length with its repetition number on the source string.
var source = "Hello world & good morning. The date is 18/09/2018";
var words = source.split(' ');
var lengths = words.map(function(word) {
return word.length;
});
var counter = {};
lengths.forEach((leng) => {
counter[leng] = counter[leng] || 0;
counter[leng]++;
});
console.log(counter);
3.Produce a list of number of words of each length in sentence (not done).
Based on the question would this not be the solution?
var words = str.split(" ");
var count = {};
for (var i = 0; i<words.length; i++){
count[words[i].length] = (count [words[i].length] || 0) + 1
}
Related
I've spend some time working on the problem and got this close
fun lengthOfLongestSubstring(s: String): Int {
var set = HashSet<Char>()
var initalChar = 0
var count = 0
s.forEach {r ->
while(!set.add(s[r]))
set.remove(s[r])
initalChar++
set.add(s[r])
count = maxOf(count, r - initialChar + 1)
}
return count
}
I understand that a HashSet is needed to answer the question since it doesn't allow for repeating characters but I keep getting a type mismatch error. I'm not above being wrong. Any assistance will be appreciated.
Your misunderstanding is that r represents a character in the string, not an index of the string, so saying s[r] doesn't make sense. You just mean r.
But you are also using r on its own, so you should be using forEachIndexed, which lets you access both the element of the sequence and the index of that element:
s.forEach { i, r ->
while(!set.add(r))
set.remove(r)
initialChar++
set.add(r)
count = maxOf(count, i - initialChar + 1)
}
Though there are still some parts of your code that doesn't quite make sense.
while(!set.add(r)) set.remove(r) is functionally the same as set.add(r). If add returns false, that means the element is already in the set, you remove it and the next iteration of the loop adds the element back into the set. If add returns true, that means the set didn't have the element and it was successfully added, so in any case, the result is you add r to the set.
And then you do set.add(r) again two lines later for some reason?
Anyway, here is a brute-force solution that you can use as a starting point to optimise:
fun lengthOfLongestSubstring(s: String): Int {
val set = mutableSetOf<Char>()
var currentMax = 0
// for each substring starting at index i...
for (i in s.indices) {
// update the current max from the previous iterations...
currentMax = maxOf(currentMax, set.size)
// clear the set to record a new substring
set.clear()
// loop through the characters in this substring
for (j in i..s.lastIndex) {
if (!set.add(s[j])) { // if the letter already exists
break // go to the next iteration of the outer for loop
}
}
}
return maxOf(currentMax, set.size)
}
I absolutely disagree that this question is a duplicate! I am asking for an efficiency way to replace hundreds of words at once. This is an algorithm question! All the provided links are about to replace one word. Should I repeat that expensive operation hundreds of times? I'm sure that there are better ways as a suffix tree where I sort out html while building that tree. I removed that regex tag since for no good reason you are focusing on that part.
I want to translate a given set of words (more then 100) and translate them. My first idea was to use a simple regular expression that works better then expected. As sample:
const input = "I like strawberry cheese cake with apple juice"
const en2de = {
apple: "Apfel",
strawberry: "Erdbeere",
banana: "Banane",
/* ... */}
input.replace(new RegExp(Object.keys(en2de).join("|"), "gi"), match => en2de[match.toLowerCase()])
This works fine on the first view. However it become strange if you words which contains each other like "pineapple" that would return "pineApfel" which is totally nonsense. So I was thinking about checking word boundaries and similar things. While playing around I created this test case:
Apple is a company
That created the output:
Apfel is a company.
The translation is wrong, which is somehow tolerable, but the link is broken. That must not happen.
So I was thinking about extend the regex to check if there is a quote before. I know well that html parsing with regex is a bad idea, but I thought that this should work anyway. In the end I gave up and was looking for solutions of other devs and found on Stack Overflow a couple of questions, all without answers, so it seems to be a hard problem (or my search skills are bad).
So I went two steps back and was thinking to implement that myself with a parser or something like that. However since I have multiple inputs and I need to ignore the case I was thinking what the best way is.
Right now I think to build a dictionary with pointers to the position of the words. I would store the dict in lower case so that should be fast, I could also skip all words with the wrong prefix etc to get my matches. In the end I would replace the words from the end to the beginning to avoid breaking the indices. But is that way efficiency? Is there a better way to achieve that?
While my sample is in JavaScript the solution must not be in JS as long the solution doesn't include dozens of dependencies which cannot be translated easy to JS.
TL;DR:
I want to replace multiple words by other words in a case insensitive way without breaking html.
You may try a treeWalker and replace the text inplace.
To find words you may tokenize your text, lower case your words and map them.
const mapText = (dic, s) => {
return s.replace(/[a-zA-Z-_]+/g, w => {
return dic[w.toLowerCase()] || w
})
}
const dic = {
'grodzi': 'grodzila',
'laaaa': 'forever',
}
const treeWalker = document.createTreeWalker(
document.body,
NodeFilter.SHOW_TEXT
)
// skip body node
let currentNode = treeWalker.nextNode()
while(currentNode) {
const newS = mapText(dic, currentNode.data)
currentNode.data = newS
currentNode = treeWalker.nextNode()
}
p {background:#eeeeee;}
<p>
grodzi
LAAAA
</p>
The link stay untouched.
However mapping each word in an other language is bound to fail (be it missing representation of some word, humour/irony, or simply grammar construct). For this matter (which is a hard problem on its own) you may rely on some tools to translate data for you (neural networks, api(s), ...)
Here is my current work in progress solution of a suffix tree (or at least how I interpreted it). I'm building a dictionary with all words, which are not inside of a tag, with their position. After sorting the dict I replace them all. This works for me without handling html at all.
function suffixTree(input) {
const dict = new Map()
let start = 0
let insideTag = false
// define word borders
const borders = ' .,<"\'(){}\r\n\t'.split('')
// build dictionary
for (let end = 0; end <= input.length; end++) {
const c = input[end]
if (c === '<') insideTag = true
if (c === '>') {
insideTag = false
start = end + 1
continue
}
if (insideTag && c !== '<') continue
if (borders.indexOf(c) >= 0) {
if(start !== end) {
const word = input.substring(start, end).toLowerCase()
const entry = dict.get(word) || []
// save the word and its position in an array so when the word is
// used multiple times that we can use this list
entry.push([start, end])
dict.set(word, entry)
}
start = end + 1
}
}
// last word handling
const word = input.substring(start).toLowerCase()
const entry = dict.get(word) || []
entry.push([start, input.length])
dict.set(word, entry)
// create a list of replace operations, we would break the
// indices if we do that directly
const words = Object.keys(en2de)
const replacements = []
words.forEach(word => {
(dict.get(word) || []).forEach(match => {
// [0] is start, [1] is end, [2] is the replacement
replacements.push([match[0], match[1], en2de[word]])
})
})
// converting the input to a char array and replacing the found ranges.
// beginning from the end and replace the ranges with the replacement
let output = [...input]
replacements.sort((a, b) => b[0] - a[0])
replacements.forEach(match => {
output.splice(match[0], match[1] - match[0], match[2])
})
return output.join('')
}
Feel free to leave a comment how this can be improved.
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.
List<String> checkLength(List<String> input) {
if (input.length > 6) {
var tempOutput = input;
while (tempOutput.length > 6) {
var difference = (tempOutput.length/6).round() + 1;
for (int i = 0; i < tempOutput.length - 1; i + difference) {
tempOutput.removeAt(i); //Removing the value from the list
}
}
return tempOutput; //Return Updated list
} else {
return input;
}
}
I am trying to delete something out of a temporary list. Why does it not work? I do not see how it is fixed, in other problems I have solved, I used a similar approach and it worked (Even identical nearly)
Please note I am kind of new to Dart, so please forgive me this sort of question, but I couldn't figure out the solution.
Find the Code available in the Dart Link
Code in Dart
You can ensure that tempOutput is not a fixed-length list by initializing it as
var tempOutput = new List<String>.from(input);
thereby declaring tempOutput to be a mutable copy of input.
FYI it also looks like you have another bug in your program since you are doing i + difference in your for-loop update step but I think you want i += difference.
Can you please try this code and let me know is that works?
List<String> checkLength(List<String> input) {
if (input.length > 6) {
var tempOutput = input;
while (tempOutput.length > 6) {
var difference = (tempOutput.length/6).round() + 1;
for (int i = 0; i < tempOutput.length - 1; i = i + difference) {
tempOutput.removeAt(i); //Removing the value from the list
}
}
return tempOutput.toList(); //Return Updated list
} else {
return input.toList();
}
}
Note: You used "i + difference" which is same value say for example in first iteration you i=1 and difference = 1, then "tempOutput.removeAt(i)" will remove the value at "1" position, again in the second iteration you are trying to remove the same position, so the error clearly states "Cannot remove from the Fixed length"
Here the i value has to be incremented or decremented for each iteration process, in your for loop that is missing.
The answer of #harry-terkelsen was very helpful for solving the fixed-length problem.
For those who were asking about my algorithm:
The difference is for skipping the amount of characters when wanting to remove some. Also, I had to change the for-loop, as it did not quite do what I wanted it to.
The fix is here! https://github.com/luki/wordtocolor/blob/master/web/algorithms.dart
Thank you for understanding me!
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]);
}