I have a MySQL database table column that has a size value in multiple formats where users manually entered different value formats.
Using PHP I need to iterate DB table and process this field to grab a withd and height value from each column when the column value matches into the pattern we create...
Below are 90% of the values in these formats. Many are the same format but with either single or double digit to left or right side of lowercase or capital X
Usung PHP how could I match each string to strip all non numeric characters from the value on the left and right side of the X.
left = width
right side = height
1x1
1X1
1"x1"
12x12
12X12
12"x12"
12"X12"
NULL
'' ,_ empty field
I just need to get these values into a width and height variable in PHP.
If I can grab everything left of lowercase and capital X ad well as right of and strip all non numbers then I think it would work easily
There are other values as well and those one should be ignored as they will not fit the pattern. Below is example of some of those odd values I found so far...
18" channel letters
64x20 x 2
Glass Dimensions: 12"x72"
172.61 cm x 28.46 cm
230.15 cm x 42.07 cm
24x24 Interior Double Sided
These type of values should be ignored so I can manually edit these later
I've written up a function called rough_strip_all that should strip all characters in a string except for those listed. Adding this step may resolve the issue for you, but if it doesn't, you might have to look into recompiling to enable the UTF8 support for PCRE.
<?php
// Strips out all characters except for those in allowed set
function rough_strip_all( $string, $allowed_set = '0123456789x. ' )
{
// Takes the allowed set, splits it into character by character,
// then converts each character in the array to its ASCII value
$allowed_ascii = array_map( function($a) {
return ord( $a );
}, str_split( $allowed_set ) );
$return = '';
for( $i = 0, $ilen = mb_strlen( $string ); $i < $ilen; $i++ )
{
// Check if the ASCII value of current character is in the list of allowed
// ascii characters given. If it is, add it to the return string
$ascii = ord( $string{$i} );
if( in_array( $ascii, $allowed_ascii ) )
{
$return .= $string{$i};
}
}
// Returns the newly compiled string
return $return;
}
// Original string
$string = "Misc text: 35.25”x 21.00” 123 extra text 456";
// Display original string
echo "Original string: {$string}<br />";
// Strips out all characters except the following: '0123456789x. '
$string = rough_strip_all( strtolower( $string ) );
// Strip out all characters except for numbers, letter x, decimal points, and spaces
$string = preg_replace( '/([^0-9x \.])/ui', '', $string );
// Find anything that fits the number X number format (including decimal numbers)
preg_match( '/([0-9]+(\.[0-9]+)?) ?x ?([0-9]+(\.[0-9]+)?)/ui', $string, $values );
// Match found
if( !empty( $values ) )
{
// Set dimensions in easy to read variables
$dimension_a = $values[1];
$dimension_b = $values[3];
// Values returned
echo "Dimension A: {$dimension_a}<br />";
echo "Dimension B: {$dimension_b}<br />";
}
// No match found
else
{
echo "No match found.";
}
?>
This should should also work for the additional outliers you added in as it strips out all non-essential characters first, then attempts to make a match. I've also added some display logic to it so you can see the original string and what each dimension is after its been processed, or a message if there has been no match.
Even easier would be preg_match_all("/[0-9]+/", $string, $matches);
Test cases:
1x1
1\"x1
12X12
Array ( [0] => Array ( [0] => 1 [1] => 1 ) ) Array ( [0] => Array ( [0] => 1 >[1] => 1 ) ) Array ( [0] => Array ( [0] => 12 [1] => 12 ) )
I have a range of whole numbers that might or might not have some numbers missing. Is it possible to find the smallest missing number without using a loop structure? If there are no missing numbers, the function should return the maximum value of the range plus one.
This is how I solved it using a for loop:
$range = [0,1,2,3,4,6,7];
// sort just in case the range is not in order
asort($range);
$range = array_values($range);
$first = true;
for ($x = 0; $x < count($range); $x++)
{
// don't check the first element
if ( ! $first )
{
if ( $range[$x - 1] + 1 !== $range[$x])
{
echo $range[$x - 1] + 1;
break;
}
}
// if we're on the last element, there are no missing numbers
if ($x + 1 === count($range))
{
echo $range[$x] + 1;
}
$first = false;
}
Ideally, I'd like to avoid looping completely, as the range can be massive. Any suggestions?
Algo solution
There is a way to check if there is a missing number using an algorithm. It's explained here. Basically if we need to add numbers from 1 to 100. We don't need to calculate by summing them we just need to do the following: (100 * (100 + 1)) / 2. So how is this going to solve our issue ?
We're going to get the first element of the array and the last one. We calculate the sum with this algo. We then use array_sum() to calculate the actual sum. If the results are the same, then there is no missing number. We could then "backtrack" the missing number by substracting the actual sum from the calculated one. This of course only works if there is only one number missing and will fail if there are several missing. So let's put this in code:
$range = range(0,7); // Creating an array
echo check($range) . "\r\n"; // check
unset($range[3]); // unset offset 3
echo check($range); // check
function check($array){
if($array[0] == 0){
unset($array[0]); // get ride of the zero
}
sort($array); // sorting
$first = reset($array); // get the first value
$last = end($array); // get the last value
$sum = ($last * ($first + $last)) / 2; // the algo
$actual_sum = array_sum($array); // the actual sum
if($sum == $actual_sum){
return $last + 1; // no missing number
}else{
return $sum - $actual_sum; // missing number
}
}
Output
8
3
Online demo
If there are several numbers missing, then just use array_map() or something similar to do an internal loop.
Regex solution
Let's take this to a new level and use regex ! I know it's nonsense, and it shouldn't be used in real world application. The goal is to show the true power of regex :)
So first let's make a string out of our range in the following format: I,II,III,IIII for range 1,3.
$range = range(0,7);
if($range[0] === 0){ // get ride of 0
unset($range[0]);
}
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
echo $str;
The output should be something like: I,II,III,IIII,IIIII,IIIIII,IIIIIII.
I've come up with the following regex: ^(?=(I+))(^\1|,\2I|\2I)+$. So what does this mean ?
^ # match begin of string
(?= # positive lookahead, we use this to not "eat" the match
(I+) # match I one or more times and put it in group 1
) # end of lookahead
( # start matching group 2
^\1 # match begin of string followed by what's matched in group 1
| # or
,\2I # match a comma, with what's matched in group 2 (recursive !) and an I
| # or
\2I # match what's matched in group 2 and an I
)+ # repeat one or more times
$ # match end of line
Let's see what's actually happening ....
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^
(I+) do not eat but match I and put it in group 1
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^
^\1 match what was matched in group 1, which means I gets matched
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^ ,\2I match what was matched in group 1 (one I in thise case) and add an I to it
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^^ \2I match what was matched previously in group 2 (,II in this case) and add an I to it
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^^^ \2I match what was matched previously in group 2 (,III in this case) and add an I to it
We're moving forward since there is a + sign which means match one or more times,
this is actually a recursive regex.
We put the $ to make sure it's the end of string
If the number of I's don't correspond, then the regex will fail.
See it working and failing. And Let's put it in PHP code:
$range = range(0,7);
if($range[0] === 0){
unset($range[0]);
}
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
if(preg_match('#^(?=(I*))(^\1|,\2I|\2I)+$#', $str)){
echo 'works !';
}else{
echo 'fails !';
}
Now let's take in account to return the number that's missing, we will remove the $ end character to make our regex not fail, and we use group 2 to return the missed number:
$range = range(0,7);
if($range[0] === 0){
unset($range[0]);
}
unset($range[2]); // remove 2
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
preg_match('#^(?=(I*))(^\1|,\2I|\2I)+#', $str, $m); // REGEEEEEX !!!
$n = strlen($m[2]); //get the length ie the number
$sum = array_sum($range); // array sum
if($n == $sum){
echo $n + 1; // no missing number
}else{
echo $n - 1; // missing number
}
Online demo
EDIT: NOTE
This question is about performance. Functions like array_diff and array_filter are not magically fast. They can add a huge time penalty. Replacing a loop in your code with a call to array_diff will not magically make things fast, and will probably make things slower. You need to understand how these functions work if you intend to use them to speed up your code.
This answer uses the assumption that no items are duplicated and no invalid elements exist to allow us to use the position of the element to infer its expected value.
This answer is theoretically the fastest possible solution if you start with a sorted list. The solution posted by Jack is theoretically the fastest if sorting is required.
In the series [0,1,2,3,4,...], the n'th element has the value n if no elements before it are missing. So we can spot-check at any point to see if our missing element is before or after the element in question.
So you start by cutting the list in half and checking to see if the item at position x = x
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
Yup, list[4] == 4. So move halfway from your current point the end of the list.
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
Uh-oh, list[6] == 7. So somewhere between our last checkpoint and the current one, one element was missing. Divide the difference in half and check that element:
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
In this case, list[5] == 5
So we're good there. So we take half the distance between our current check and the last one that was abnormal. And oh.. it looks like cell n+1 is one we already checked. We know that list[6]==7 and list[5]==5, so the element number 6 is the one that's missing.
Since each step divides the number of elements to consider in half, you know that your worst-case performance is going to check no more than log2 of the total list size. That is, this is an O(log(n)) solution.
If this whole arrangement looks familiar, It's because you learned it back in your second year of college in a Computer Science class. It's a minor variation on the binary search algorithm--one of the most widely used index schemes in the industry. Indeed this question appears to be a perfectly-contrived application for this searching technique.
You can of course repeat the operation to find additional missing elements, but since you've already tested the values at key elements in the list, you can avoid re-checking most of the list and go straight to the interesting ones left to test.
Also note that this solution assumes a sorted list. If the list isn't sorted then obviously you sort it first. Except, binary searching has some notable properties in common with quicksort. It's quite possible that you can combine the process of sorting with the process of finding the missing element and do both in a single operation, saving yourself some time.
Finally, to sum up the list, that's just a stupid math trick thrown in for good measure. The sum of a list of numbers from 1 to N is just N*(N+1)/2. And if you've already determined that any elements are missing, then obvously just subtract the missing ones.
Technically, you can't really do without the loop (unless you only want to know if there's a missing number). However, you can accomplish this without first sorting the array.
The following algorithm uses O(n) time with O(n) space:
$range = [0, 1, 2, 3, 4, 6, 7];
$N = count($range);
$temp = str_repeat('0', $N); // assume all values are out of place
foreach ($range as $value) {
if ($value < $N) {
$temp[$value] = 1; // value is in the right place
}
}
// count number of leading ones
echo strspn($temp, '1'), PHP_EOL;
It builds an ordered identity map of N entries, marking each value against its position as "1"; in the end all entries must be "1", and the first "0" entry is the smallest value that's missing.
Btw, I'm using a temporary string instead of an array to reduce physical memory requirements.
I honestly don't get why you wouldn't want to use a loop. There's nothing wrong with loops. They're fast, and you simply can't do without them. However, in your case, there is a way to avoid having to write your own loops, using PHP core functions. They do loop over the array, though, but you simply can't avoid that.
Anyway, I gather what you're after, can easily be written in 3 lines:
function highestPlus(array $in)
{
$compare = range(min($in), max($in));
$diff = array_diff($compare, $in);
return empty($diff) ? max($in) +1 : $diff[0];
}
Tested with:
echo highestPlus(range(0,11));//echoes 12
$arr = array(9,3,4,1,2,5);
echo highestPlus($arr);//echoes 6
And now, to shamelessly steal Pé de Leão's answer (but "augment" it to do exactly what you want):
function highestPlus(array $range)
{//an unreadable one-liner... horrid, so don't, but know that you can...
return min(array_diff(range(0, max($range)+1), $range)) ?: max($range) +1;
}
How it works:
$compare = range(min($in), max($in));//range(lowest value in array, highest value in array)
$diff = array_diff($compare, $in);//get all values present in $compare, that aren't in $in
return empty($diff) ? max($in) +1 : $diff[0];
//-------------------------------------------------
// read as:
if (empty($diff))
{//every number in min-max range was found in $in, return highest value +1
return max($in) + 1;
}
//there were numbers in min-max range, not present in $in, return first missing number:
return $diff[0];
That's it, really.
Of course, if the supplied array might contain null or falsy values, or even strings, and duplicate values, it might be useful to "clean" the input a bit:
function highestPlus(array $in)
{
$clean = array_filter(
$in,
'is_numeric'//or even is_int
);
$compare = range(min($clean), max($clean));
$diff = array_diff($compare, $clean);//duplicates aren't an issue here
return empty($diff) ? max($clean) + 1; $diff[0];
}
Useful links:
The array_diff man page
The max and min functions
Good Ol' range, of course...
The array_filter function
The array_map function might be worth a look
Just as array_sum might be
$range = array(0,1,2,3,4,6,7);
// sort just in case the range is not in order
asort($range);
$range = array_values($range);
$indexes = array_keys($range);
$diff = array_diff($indexes,$range);
echo $diff[0]; // >> will print: 5
// if $diff is an empty array - you can print
// the "maximum value of the range plus one": $range[count($range)-1]+1
echo min(array_diff(range(0, max($range)+1), $range));
Simple
$array1 = array(0,1,2,3,4,5,6,7);// array with actual number series
$array2 = array(0,1,2,4,6,7); // array with your custom number series
$missing = array_diff($array1,$array2);
sort($missing);
echo $missing[0];
$range = array(0,1,2,3,4,6,7);
$max=max($range);
$expected_total=($max*($max+1))/2; // sum if no number was missing.
$actual_total=array_sum($range); // sum of the input array.
if($expected_total==$actual_total){
echo $max+1; // no difference so no missing number, then echo 1+ missing number.
}else{
echo $expected_total-$actual_total; // the difference will be the missing number.
}
you can use array_diff() like this
<?php
$range = array("0","1","2","3","4","6","7","9");
asort($range);
$len=count($range);
if($range[$len-1]==$len-1){
$r=$range[$len-1];
}
else{
$ref= range(0,$len-1);
$result = array_diff($ref,$range);
$r=implode($result);
}
echo $r;
?>
function missing( $v ) {
static $p = -1;
$d = $v - $p - 1;
$p = $v;
return $d?1:0;
}
$result = array_search( 1, array_map( "missing", $ARRAY_TO_TEST ) );
*I try to count the unique appearances of a substring inside a list of words *
So check the list of words and detect if in any words there are substrings based on min characters that occur multiple times and count them. I don't know any substrings.
This is a working solution where you know the substring but what if you do not know ?
Theres a Minimum Character count where words are based on.
Will find all the words where "Book" is a substring of the word. With below php function.
Wanted outcome instad:
book count (5)
stor count (2)
Given a string of length 100
book bookstore bookworm booking book cooking boring bookingservice.... ok
0123456789... ... 100
your algorithm could be:
Investigate substrings from different starting points and substring lengths.
You take all substrings starting from 0 with a length from 1-100, so: 0-1, 0-2, 0-3,... and see if any of those substrings accurs more than once in the overall string.
Progress through the string by starting at increasing positions, searching all substrings starting from 1, i.e. 1-2, 1-3, 1-4,... and so on until you reach 99-100.
Keep a table of all substrings and their number of occurances and you can sort them.
You can optimize by specifying a minimum and maximum length, which reduces your number of searches and hit accuracy quite dramatically. Additionally, once you find a substring save them in a array of searched substrings. If you encounter the substring again, skip it. (i.e. hits for book that you already counted you should not count again when you hit the next booksubstring). Furthermore you will never have to search strings that are longer than half of the total string.
For the example string you might run additional test for the uniquness of a string.
You'd have
o x ..
oo x 7
bo x 7
ok x 6
book x 5
booking x 2
bookingservice x 1
with disregarding stings shorter than 3 (and longer than half of total textstring), you'd get
book x 5
booking x 2
bookingservice x 1
which is already quite a plausible result.
[edit] This would obviously look through all of the string, not just natural words.
[edit] Normally I don't like writing code for OPs, but in this case I got a bit interested myself:
$string = "book bookshelf booking foobar bar booking ";
$string .= "selfservice bookingservice cooking";
function search($string, $min = 4, $max = 16, $threshhold = 2) {
echo "<pre><br/>";
echo "searching <em>'$string'</em> for string occurances ";
echo "of length $min - $max: <br/>";
$hits = array();
$foundStrings = array();
// no string longer than half of the total string will be found twice
if ($max > strlen($string) / 2) {
$max = strlen($string);
}
// examin substrings:
// start from 0, 1, 2...
for ($start = 0; $start < $max; $start++) {
// and string length 1, 2, 3, ... $max
for ($length = $min; $length < strlen($string); $length++) {
// get the substring in question,
// but search for natural words (trim)
$substring = trim(substr($string, $start, $length));
// if substring was not counted yet,
// add the found count to the hits
if (!in_array($substring, $foundStrings)) {
preg_match_all("/$substring/i", $string, $matches);
$hits[$substring] = count($matches[0]);
}
}
}
// sort the hits array desc by number of hits
arsort($hits);
// remove substring hits with hits less that threshhold
foreach ($hits as $substring => $count) {
if ($count < $threshhold) {
unset($hits[$substring]);
}
}
print_r($hits);
}
search($string);
?>
The comments and variable names should make the code explain itself. $string would come for a read file in your case. This exmaple would output:
searching 'book bookshelf booking foobar bar booking selfservice
bookingservice cooking' for string occurances of length 4 - 16:
Array
(
[ook] => 6
[book] => 5
[boo] => 5
[bookin] => 3
[booking] => 3
[booki] => 3
[elf] => 2
)
Let me know how you implement it :)
This is my first approximation: unfinished, untested, has at least 1 bug, and is written in eiffel. Well I am not going to do all the work for you.
deferred class
SUBSTRING_COUNT
feature
threshold : INTEGER_32 =5
biggest_starting_substring_length(a,b:STRING):INTEGER_32
deferred
end
biggest_starting_substring(a,b:STRING):STRING
do
Result := a.substring(0,biggest_starting_substring_length(a,b))
end
make_list_of_substrings(a,b:STRING)
local
index:INTEGER_32
this_one: STRING
do
from
a_index := b_index + 1
invariant
a_index >=0 and a_index <= a.count
until
a_index >= a.count
loop
this_one := biggest_starting_substring(a.substring (a_index, a.count-1),b)
if this_one.count > threshold then
list.extend (this_one)
end
variant
a.count - a_index
end
end -- biggest_substring
list : ARRAYED_LIST[STRING]
end
I have a list of phrases and I want to know which two words occurred the most often in all of my phrases.
I tried playing with regex and other codes and I just cannot find the right way to do this.
Can anyone help?
eg:
I am purchasing a wallet
a wallet for 20$
purchasing a bag
I'd know that
a wallet occurred 2 times
purchasing a occurred 2 times
<?
$string = "I am purchasing a wallet a wallet for 20$ purchasing a bag";
//split string into words
$words = explode(' ', $string);
//make chunks block ie [0,1][2,3]...
$chunks = array_chunk($words, 2);
//remove first array element
unset($words[0]);
//make chunks block ie [0,1][2,3]...
//but since first element is removed , the real block will be [1,2][3,4]...
$alternateChunks = array_chunk($words, 2);
//merge both chunks
$totalChunks = array_merge($chunks,$alternateChunks);
$finalChunks = array();
foreach($totalChunks as $t)
{
//change the inside chunk to pharse using +
//+ can be replaced to space, if neeced
//to keep associative working + is used instead of white space
$finalChunks[] = implode('+', $t);
}
//count the words inside array
$result = array_count_values($finalChunks);
echo "<pre>";
print_r($result);
I hesitate to suggest this, as it's an extremely brute force way to go about it:
Take your string of words, explode it using the explode(" ", $string); command, then run it through a for loop checking every two word combination against every two words in the string.
$string = "I am purchasing a wallet a wallet for 20$ purchasing a bag";
$words = explode(" ", $string);
for ($t=0; $t<count($string); $t++)
{
for ($i=0; $i<count($string); $i++)
{
if (($words[$t] . words[$t+1]) == ($words[$i] . $word[$i+1])) {$count[$words[$i].$words[$i+1]]++}
}
}
So the nested for loop steps in, grabs the first two words, compares them to each other set of two consecutive words, then grabs the next two words and does it again. Every answer will have an answer of at least 1 (it will always match itself) but sorting the resulting array by size will give you the most repeated values.
Note that this will run (n-1)*(n-1) iterations, which could get unwieldy FAST.
Place them all into an array, and access them by the current word index and next word index.
I think this should do the trick. It will grab pairs of words, unless you are at the end of the string, where you'll get only one word.
$str = "I purchased a wallet because I wanted a wallet a wallet a wallet";
$words = explode(" ", $str);
$array_results = array();
for ($i = 0; $i<count($words); $i++) {
if ($i < count($words)-1) {
$pair = $words[$i] . " " . $words[$i+1]; echo $pair . "\n";
// Have to check if the key is in use yet to avoid a notice
$array_results[$pair] = isset($array_results[$pair]) ? $array_results[$pair] + 1 : 1;
}
// At the end of the array, just use a single word
else $array_results[$words[$i]] = isset($array_results[$words[$i]]) ? $array_results[$words[$i]] + 1 : 1;
}
// Sort the results
// use arsort() instead to get the highest first
asort($array_results);
// Prints:
Array
(
[I wanted] => 1
[wanted a] => 1
[wallet] => 1
[because I] => 1
[wallet because] => 1
[I purchased] => 1
[purchased a] => 1
[wallet a] => 2
[a wallet] => 4
)
Update changed ++ to +1 above since it wasn't working when tested...
Try to put it with explode into an array and count the values with array_count_values.
<?php
$text = "whatever";
$text_array = explode( ' ', $text);
$double_words = array();
for($c = 1; $c < count($text_array); $c++)
{
$double_words[] = $text_array[$c -1] . ' ' . $text_array[$c];
}
$result = array_count_values($double_words);
?>
I updated it now to two word version. Does this work for you?
array(9) {
["I am"]=> int(1)
["am purchasing"]=> int(1)
["purchasing a"]=> int(2)
["a wallet"]=> int(2)
["wallet a"]=> int(1)
["wallet for"]=> int(1)
["for 20$"]=> int(1)
["20$ purchasing"]=> int(1)
["a bag"]=> int(1)
}
Since you used the excel tag, I thought I'd give it a shot, and it's actually really easy.
Split string using space as delimiter. Data > Text to Columns... > Delimited > Delimiter: Space. Each word is now in its own cell.
Transpose the result (not strictly required but much easier to visualize). Copy, Edit > Paste Special... > Transpose.
Make cells containing consecutive word pairs. So if your words are in cells B5:B15, cell C5 should be =B5&" "&B6 (and drag down).
Count occurence of each word pair: In cell D5, =COUNTIF($C$5:$C$15,"="&C5), drag down.
Highlight the winner(s). Select C5:D15, Format > Conditional Formatting... > Formula Is =$D5=MAX($D$5:$D$15) and choose e.g. a yellow background.
Note that there is some inefficiency in step 4 because the count of each word pair will be calculated multiple times if that word pair occurs multiple times. If this is a concern, then you can first make a list of unique word pairs using Data > Filter > Advanced Filter... > Unique records only.
An automated VBA solution could easily be crafted by recording a macro of the above followed by some minor editing.
One way to go about it is to use SPLIT or a regex to split the sentences into words and store each into an array. Then take the array and create a dictionary object. When you add a term to the dictionary, if it's already there, add 1 to the .value to tally the count.
Here is some example code (far from perfect as it's just to show the overlying concept) that will take all the string in column A and generate a word frequency list in columns B and C. It's not exactly what you want, but should give you some ideas on how you can go about doing it I hope:
Sub FrequencyList()
Dim vArray As Variant
Dim myDict As Variant
Set myDict = CreateObject("Scripting.Dictionary")
Dim i As Long
Dim cell As range
With myDict
For Each cell In range("A1", cells(Rows.count, "A").End(xlUp))
vArray = Split(cell.Value, " ")
For i = LBound(vArray) To UBound(vArray)
If Not .exists(vArray(i)) Then
.Add vArray(i), 1
Else
.Item(vArray(i)) = .Item(vArray(i)) + 1
End If
Next
Next
range("B1").Resize(.count).Value = Application.Transpose(.keys)
range("C1").Resize(.count).Value = Application.Transpose(.items)
End With
End Sub