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How do I break a string into an array based on a token?


How do I break a string into an array based on a token?

By : user2950054
Date : November 17 2020, 11:52 AM
like below fixes the issue This would be a possible solution. You could write it in a single line, but I have split it into separate statements to demonstrate how it works:
code :
let tgt     = "roEuroroparo"
let token   = "ro"

// Split into an array of strings:
let t1 = tgt.componentsSeparatedByString(token)
println(t1)                   // [, Eu, , pa, ]

// Convert each string to an array:
let t2 = map(t1) { [$0] }
println(t2)                   // [[], [Eu], [], [pa], []]

// Interpose the token:
let t3 = [token].join(t2)
println(t3)                   // [, ro, Eu, ro, , ro, pa, ro, ]

// Remove empty strings:
let result = filter(t3) { countElements($0) > 0 }
println(result)               // [ro, Eu, ro, ro, pa, ro]


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use first token in a string to create separate array lists, second token will

use first token in a string to create separate array lists, second token will


By : Mayank Agarwal
Date : March 29 2020, 07:55 AM
Any of those help It looks like you would probably be better served by a map. The first element is the key, and the second element the value.
code :
ArrayList<String> lines = new ArrayList<String>();
String line;
while ((line = reader.readLine()) != null) {
    lines.add(line);
}
HashMap<String, String> votes = new HashMap<String, String>();
for(String line: lines) {
    String[] terms = line.split(">");
    votes.put(terms[0], terms[1]);
}
Break a string in two based on one of the prefixes in the array

Break a string in two based on one of the prefixes in the array


By : Pelle Alstermo
Date : March 29 2020, 07:55 AM
seems to work fine If it is true that no license string is a part of any other (you don't have both "A" and "AB" in the list), then just loop through and find the first match as follows:
code :
function parts($licenseString) {
    $foreach($license_letters AS $prefix) {
        if(strpos($licenseString, $prefix) === 0) {
            return [$prefix, substr($licenseString, strlen($prefix)];
        }
    }
}
Break an array to array of arrays based on the some of item's property

Break an array to array of arrays based on the some of item's property


By : Geo Vlas
Date : March 29 2020, 07:55 AM
Hope that helps A proposal which uses an array for the count and it uses all spaces, not only the last slot.
code :
var data = [{ id: 0, width: 2 }, { id: 1, width: 2 }, { id: 2, width: 4 }, { id: 3, width: 1 }, { id: 4, width: 3 }],
    max = 4,
    grouped = [];

data.forEach((count => a => {
    var index = count.findIndex(b => b + a.width <= max);
    if (index === -1) {
        index = grouped.push([]) - 1;
        count[index] = 0;
    }
    grouped[index].push(a);
    count[index] += a.width;
})([]));
console.log(grouped);
In Ruby, how do I break up a string given an array of indexes I want to break the string on?

In Ruby, how do I break up a string given an array of indexes I want to break the string on?


By : ZXFlux
Date : March 29 2020, 07:55 AM
wish helps you I have this expression for getting the indexes of spaces in a given string … , This should be the easiest way to achieve your goal:
code :
def split_by_indices(indices, string)
    result = []
    indices.unshift(0)
    indices.each_with_index do |val, index|
      result << string[val...(index == indices.length - 1 ? string.length : indices[index+1])]
    end
    result
end
lodash break down array based on sub-array retaining the array record

lodash break down array based on sub-array retaining the array record


By : Adwait Kumar
Date : March 29 2020, 07:55 AM
around this issue Actually you don't need lodash to achieve the result you're expecting:
code :
const books = [
  {
    'book_name': 'book1',
    'authors': [
      {
        author_id: 'value2',
        author_name: 'Name2',
      },
    ],
  },
  {
    'book_name': 'book2',
    'authors': [
      {
        author_id: 'value1',
        author_name: 'Name1',
      },
      {
        author_id: 'value2',
        author_name: 'Name2',
      }
    ],
  },
];

const transformed = books.reduce((accumulator, book) => {
  const authors = book.authors.map(author => ({
      ...book,
      authors: [author]
  }));

  return accumulator.concat(authors);
}, []);

console.log('original', books);
console.log('transformed', transformed);
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