Echo top 5 appearing words on page
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
javascript php wordpress count word
add a comment |
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
javascript php wordpress count word
would you like to count the words on the backend (PHP/Wordpress) or frontend (JavaScript) ???
– marvinIsSacul
Nov 16 '18 at 7:19
add a comment |
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
javascript php wordpress count word
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
javascript php wordpress count word
javascript php wordpress count word
asked Nov 15 '18 at 19:33
JGRJGR
246
246
would you like to count the words on the backend (PHP/Wordpress) or frontend (JavaScript) ???
– marvinIsSacul
Nov 16 '18 at 7:19
add a comment |
would you like to count the words on the backend (PHP/Wordpress) or frontend (JavaScript) ???
– marvinIsSacul
Nov 16 '18 at 7:19
would you like to count the words on the backend (PHP/Wordpress) or frontend (JavaScript) ???
– marvinIsSacul
Nov 16 '18 at 7:19
would you like to count the words on the backend (PHP/Wordpress) or frontend (JavaScript) ???
– marvinIsSacul
Nov 16 '18 at 7:19
add a comment |
1 Answer
1
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votes
I'm not sure how useful it would be. You end up with meaningless words.
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Filtering Added
We can filter the array before counting words, so you can add whatever you like to the filters
array to get rid of those words. I probably should have thought of that to begin with...
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
add a comment |
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I'm not sure how useful it would be. You end up with meaningless words.
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Filtering Added
We can filter the array before counting words, so you can add whatever you like to the filters
array to get rid of those words. I probably should have thought of that to begin with...
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
add a comment |
I'm not sure how useful it would be. You end up with meaningless words.
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Filtering Added
We can filter the array before counting words, so you can add whatever you like to the filters
array to get rid of those words. I probably should have thought of that to begin with...
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
add a comment |
I'm not sure how useful it would be. You end up with meaningless words.
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Filtering Added
We can filter the array before counting words, so you can add whatever you like to the filters
array to get rid of those words. I probably should have thought of that to begin with...
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
I'm not sure how useful it would be. You end up with meaningless words.
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
Filtering Added
We can filter the array before counting words, so you can add whatever you like to the filters
array to get rid of those words. I probably should have thought of that to begin with...
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
const filters = ['a', 'to', 'the', 'for', 'i'];
let text = document.querySelector('#input').innerText.split(/s/);
let counts = text.map(w => w.toLowerCase()).filter(w => !filters.includes(w)).reduce((acc, cv) => {
if (cv.length > 0)
acc[cv] = acc[cv] + 1 || 1;
return acc;
}, {});
text = Object.keys(counts).sort((a, b) => counts[b] - counts[a]).slice(0, 5);
console.log(text);
<div id="input">
<p>
Is there a little function or plugin to get the top 5 most appearing words on a page? I am building a website with dynamic content and I want to check for most appearing same words.
</p>
<p>
For example: The word 'Spain' is on the page for around 12 times and the word 'Netherlands' for around 8 times. I want to have a function to check this automatically and echo this.
</p>
</div>
edited Nov 21 '18 at 18:29
answered Nov 15 '18 at 20:37
WillWill
1,79411111
1,79411111
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
add a comment |
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
Thanks man, I will check this code. Looks good. I will let you know if it works
– JGR
Nov 20 '18 at 12:52
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
One little question. The code is working but can I skip numbers? So only text being counted.
– JGR
Nov 20 '18 at 13:12
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
Yep, added filtering.
– Will
Nov 21 '18 at 18:30
add a comment |
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would you like to count the words on the backend (PHP/Wordpress) or frontend (JavaScript) ???
– marvinIsSacul
Nov 16 '18 at 7:19