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C++ Need to compare one string to 200.000 words


C++ Need to compare one string to 200.000 words

By : user3850139
Date : October 20 2020, 11:12 PM
around this issue Use TRIE data structure for this. You should need some memory for constructing the data structure. But your objective will be most efficient.
code :


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Compare two string and display number of mismatching words from string

Compare two string and display number of mismatching words from string


By : Mahathir
Date : March 29 2020, 07:55 AM
it should still fix some issue This is as simple as converting both the strings to an array and then computing the difference of arrays.
code :
$arr1 = explode(' ', strtolower($str1));
$arr2 = explode(' ', strtolower($str2));
echo 'You missed ' . count(array_diff($arr1, $arr2)) . ' words from str 1';
How to compare words of one string in Java with words of another string and separate the words on match?

How to compare words of one string in Java with words of another string and separate the words on match?


By : petrt88
Date : March 29 2020, 07:55 AM
it should still fix some issue The issue you're having (and the reason the first string succeeded and the second does not) is to do with the order of words in the dict. Your current implementation checks if the words in the dict appear in the string exactly in the order they were entered into the dict - after you found the first word, put in a space and proceed to find the second word. If you did not find the next word, you do not proceed with the process.
There are many ways to rewrite the code to get what you want, but the minimal change is:
code :
public class Sample1 {
public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
int i,j,k,len;
    String[] dict= {"how","are","you","something","needs","to","be","done"};
    //StringBuilder str=new StringBuilder("howareyou");
    StringBuilder str=new StringBuilder("somethingneedstobedone");
    len=str.length();
    for(i=0,j=0;i<len;i++) //removed k from here
    {
        for(j=i+1;j<len;j++)
        {
          for (k=0;k<dict.length;k++) { //added this loop!
            if(dict[k].toString().equals(str.substring(i, j)))
            {
                str.insert(j, " ");
            }
          } //Loop closing for k - the dictionary
        }
    }
    System.out.println(str); 
    sc.close();
}
How to compare string with array of words and highlight words in string that match?

How to compare string with array of words and highlight words in string that match?


By : Gazeld
Date : March 29 2020, 07:55 AM
it should still fix some issue Convert the array to a regular expression, and use String#Replace to wrap the words with a span:
code :
const words = ['prueba', 'etiquetas'];
const product = 'Prueba Pruebaa de etiquetas aetiquetas';

// convert the array to a regular expression that looks for any word that is found in the list, regardless of case (i), over all the string (g)
const regexp = new RegExp(`\\b(${words.join('|')})\\b`, 'gi');

// replace the found words with a span that contains each word
const html = product.replace(regexp, '<span class="highlight">$&</span>');

demo.innerHTML = html;
.highlight {
  background: yellow;
}
<div id="demo"></div>
compare all the words in string against words in a list

compare all the words in string against words in a list


By : Anime Epi
Date : March 29 2020, 07:55 AM
may help you . 1.I have a list of keywords, I am keeping them in list named dict as below: , There are many ways to accomplish this. One of them is (python3):
code :
list_a = ["renovation","homeloan","personalloan"]
list_b = ["customer","needed","the","loan","for","renovation","of","his","home"]
result = {}
for index in list_a:
    result[index] = list_b.count(index)

print(result)
compare list of words, similar words deleted and added into new string

compare list of words, similar words deleted and added into new string


By : Alex
Date : March 29 2020, 07:55 AM
To fix the issue you can do I have 2 strings which I have converted into a list of words A and B. I am trying to make an algorithm that selects a word starting from the left. If that word appears as a word or part of a word in the second string then add that word to a new common string and delete the entire first occurrence of the word where it was found in the second string. Upper and lowercase letters are considered different letters. I call this algorithm Diff. , This could work
code :
common_str = ""

a =  " The quick brown fox did jump over a log"
b = " The brown rabbit quickly did outjump the fox"
alist = a.split(" ")
blist = b.split(" ")
for item in blist:
    if item in alist:
        #add word to common string
        common_str += item + " "
        #remove from second string
        i = item + " " #item plus trailing space
        a.replace(i, "")
print(common_str)
print(a)
print(b)
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