how to find in a parent string list, the indexes corresponding to a child string list

>>> sorter = np.argsort(Shop_Products) >>>
   sorter[np.searchsorted(Shop_Products, Shop_Query, sorter = sorter)]
array([4, 1]) >>>
   Shop_Inventory[sorter[np.searchsorted(Shop_Products, Shop_Query, sorter = sorter)]]
array([8, 6])

>>> np.in1d(Shop_Products, ['Cheese', 'Bread']).nonzero()
   (array([1, 4]), ) >>>
   np.in1d(Shop_Products, ['Bread', 'Cheese']).nonzero()
   (array([1, 4]), )

>>> np.in1d(Shop_Products, Shop_Query).nonzero()
   (array([1, 4]), )

>>> i = np.in1d(Shop_Products, Shop_Query).nonzero() >>>
   Shop_Inventory[i]
array([6, 8])

>>> sorter = np.argsort(Shop_Products) >>>
   sorter[np.searchsorted(Shop_Products, Shop_Query, sorter = sorter)]
array([4, 1]) >>>
   Shop_Inventory[sorter[np.searchsorted(Shop_Products, Shop_Query, sorter = sorter)]]
array([8, 6])

use strict;
use warnings 'all';

my % side = (i => 24.56, j => 3.56);

my($sside, $bside) = $side {
   i
} < $side {
   j
} ? qw / i j / : qw / j i / ;

$side {
   $sside
} *= 20;
$side {
   $bside
} *= 21;

/**
 * @brief: Shows the subpages of the current page, or
 *         the adjacent sibling pages.
 **/
function show_subpages(){

    global $post;
    $subpages = wp_list_pages( array(
        'echo'=>0,
        'title_li'=>'',
        'depth'=>2,
        'child_of'=> ( $post->post_parent == 0 ? $post->ID : $post->post_parent)
    ));
    if ( !empty($subpages) ) {

        if ( $post->post_parent != 0 ) {
            echo '<p class="parent-link"><em>'. __('Back to') .' <a href="'. get_permalink($post->post_parent) .'">'. get_the_title($post->post_parent) .'</em></a><p>';
        }
        echo '<ul>';
        echo $subpages;
        echo '</ul>';
    } else {
        echo 'no subpages';
    }
}

d = REXML::Document.new('<root></root>')
root = d.root
e0 = REXML::Element.new('foo')
e1 = REXML::Element.new('bar')
root.add(e0) # =>
<foo />
root.add(e1) # =>
<bar />
root.add(e0) # =>
<foo />
root.add(e1) # =>
<bar />
root.index(e0) # => 0
root.index(e1) # => 1

def find_str(s, char):
   index = 0

if char in s:
   c = char[0]
for ch in s:
   if ch == c:
   if s[index: index + len(char)] == char:
   return index

index += 1

return -1

print(find_str("Happy birthday", "py"))
print(find_str("Happy birthday", "rth"))
print(find_str("Happy birthday", "rh"))

library(zoo)
na.locf(replace(string, string == "", NA))
#[1]
"1"
"2"
"3"
"3"
"5"
"6"
"6"

i1 < -which(string == "")
string[i1] < -string[i1 - 1]

1) Input: txt[] = "BACDGABCDA"
pat[] = "ABCD"
Output: Found at Index 0
Found at Index 5
Found at Index 6
2) Input: txt[] = "AAABABAA"
pat[] = "AABA"
Output: Found at Index 0
Found at Index 1
Found at Index 4

Brute Force:
   Consider the Input txt[] = "BACDGABCDA"
pat[] = "ABCD".
Occurrences of the pat[] and its permutations are found at indexes 0, 5, 6.
The permutations are BACD, ABCD, BCDA.
Let 's sort the pat[] and the permutations of pat[] in txt[].
pat[] after sorting becomes: ABCD
permutations of pat[] in txt[] after sorting becomes: ABCD, ABCD, ABCD.
So we can say that the sorted version of pat[] and sorted version of its
permutations yield the same result.

Found at Index 0
Found at Index 5
Found at Index 6

Found at Index 0
Found at Index 5
Found at Index 6

// Note that parg is an object reference to a <p> element

// First check that the element has child nodes
if (parg.hasChildNodes()) {
  let children = parg.childNodes;

  for (const node of children) {
    // Do something with each child as children[i]
    // NOTE: List is live! Adding or removing children will change the list's `length`
  }
}

// This is one way to remove all children from a node
// box is an object reference to an element
while (box.firstChild) {
   //The list is LIVE so it will re-index each call
   box.removeChild(box.firstChild);
}

user_preference = Table('user_preference', metadata_obj,
   Column('pref_id', Integer, primary_key = True),
   Column('user_id', Integer, ForeignKey("user.user_id"), nullable = False),
   Column('pref_name', String(40), nullable = False),
   Column('pref_value', String(100))
)

ForeignKey(user.c.user_id)

invoice = Table('invoice', metadata_obj,
   Column('invoice_id', Integer, primary_key = True),
   Column('ref_num', Integer, primary_key = True),
   Column('description', String(60), nullable = False)
)

invoice_item = Table('invoice_item', metadata_obj,
   Column('item_id', Integer, primary_key = True),
   Column('item_name', String(60), nullable = False),
   Column('invoice_id', Integer, nullable = False),
   Column('ref_num', Integer, nullable = False),
   ForeignKeyConstraint(['invoice_id', 'ref_num'], ['invoice.invoice_id', 'invoice.ref_num'])
)

CREATE TABLE addresses(
   id INTEGER NOT NULL,
   user_id INTEGER,
   email_address VARCHAR NOT NULL,
   PRIMARY KEY(id),
   CONSTRAINT user_id_fk FOREIGN KEY(user_id) REFERENCES users(id)
)

node = Table(
   'node', metadata_obj,
   Column('node_id', Integer, primary_key = True),
   Column(
      'primary_element', Integer,
      ForeignKey('element.element_id')
   )
)

element = Table(
   'element', metadata_obj,
   Column('element_id', Integer, primary_key = True),
   Column('parent_node_id', Integer),
   ForeignKeyConstraint(
      ['parent_node_id'], ['node.node_id'],
      name = 'fk_element_parent_node_id'
   )
)

File input = new File("/tmp/input.html");
Document doc = Jsoup.parse(input, "UTF-8", "http://example.com/");

Elements links = doc.select("a[href]"); // a with href
Elements pngs = doc.select("img[src$=.png]");
// img with src ending .png

Element masthead = doc.select("div.masthead").first();
// div with class=masthead

Elements resultLinks = doc.select("h3.r > a"); // direct a after h3

<!doctype html>
<html lang="en">

<head>
   <meta charset="utf-8">
   <title>eq demo</title>
   <script src="https://code.jquery.com/jquery-3.5.0.js"></script>
</head>

<body>
   <table border="1">
      <tr>
         <td>TD #0</td>
         <td>TD #1</td>
         <td>TD #2</td>
      </tr>
      <tr>
         <td>TD #3</td>
         <td>TD #4</td>
         <td>TD #5</td>
      </tr>
      <tr>
         <td>TD #6</td>
         <td>TD #7</td>
         <td>TD #8</td>
      </tr>
   </table>
   <script>
      $("td:eq( 2 )").css("color", "red");
   </script>
</body>

</html>

<!doctype html><html lang="en"><head>  <meta charset="utf-8">  <title>eq demo</title>  <script src="https://code.jquery.com/jquery-3.5.0.js"></script></head><body> <ul class="nav">  <li>List 1, item 1</li>  <li>List 1, item 2</li>  <li>List 1, item 3</li></ul><ul class="nav">  <li>List 2, item 1</li>  <li>List 2, item 2</li>  <li>List 2, item 3</li></ul> <script>// Applies yellow background color to a single <li>$( "ul.nav li:eq(1)" ).css( "backgroundColor", "#ff0" ); // Applies italics to text of the second <li> within each <ul class="nav">$( "ul.nav" ).each(function( index ) {  $( this ).find( "li:eq(1)" ).css( "fontStyle", "italic" );}); // Applies red text color to descendants of <ul class="nav">// for each <li> that is the second child of its parent$( "ul.nav li:nth-child(2)" ).css( "color", "red" );</script> </body></html>

<!doctype html><html lang="en"><head>  <meta charset="utf-8">  <title>eq demo</title>  <style>  .foo {    color: blue;    background-color: yellow;  }  </style>  <script src="https://code.jquery.com/jquery-3.5.0.js"></script></head><body> <ul class="nav">  <li>List 1, item 1</li>  <li>List 1, item 2</li>  <li>List 1, item 3</li></ul><ul class="nav">  <li>List 2, item 1</li>  <li>List 2, item 2</li>  <li>List 2, item 3</li></ul> <script>$( "li:eq(-2)" ).addClass( "foo" )</script> </body></html>