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Use two lists instead of .replace()


Use two lists instead of .replace()

By : Syed Ammad Hassan
Date : November 21 2020, 04:01 AM
this will help If you've got a string containing for example "how are you?", I would do stringname.replace("how", "How") to have the first word written in Capital H. , If you want to capitalize the string you can use .capitalize()
code :
self.weather= self.weather.capitalize()
dictionary={'cloudy':'Cloudy','foggy':'Foggy','sunny':'Sunny','rain':'Rain'}
if self.weather in dictionary:
    self.weather=dictionary[self.weather]


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How do I parse content to replace fake lists with real lists using PHP?

How do I parse content to replace fake lists with real lists using PHP?


By : user3488972
Date : March 29 2020, 07:55 AM
will help you UPDATE: I've run in to this problem before where there's a bunch of data with 'fake' lists using indenting and different chars as the bullet so I just made this little function.
code :
function make_real_list($regex, $content, $type="unordered"){

    preg_match_all($regex, $content, $matches);

    $matches    = $matches[0];
    $count  = sizeof($matches);

    if($type=="unordered"):
        $outer_start    = "<ul>";
        $outer_end      = "</ul>";

    else:
        $outer_start    = "<ol>";
        $outer_end      = "</ol>";

    endif;

    $i = 1;
    foreach($matches as $match):

        if($i==1):
            $replace    = preg_replace($regex, '<li>$1</li>', $match, 1);
            $match  = preg_quote($match, "/");
            $content    = preg_replace("/$match/", ($outer_start?$outer_start:'').$replace, $content);

        elseif($i==$count):
            $replace    = preg_replace($regex, '<li>$1</li>', $match, 1);
            $match  = preg_quote($match, "/");
            $content    = preg_replace("/$match/", $replace.($outer_end?$outer_end:''), $content);

        else:
            $content    = preg_replace($regex, '<li>$1</li>', $content, 1);

        endif;
        $i++;

    endforeach;

    return $content;

}

$content = "<p>STUFF BEFORE</p>
<p>&ntilde; FIRST LIST ITEM</p>
<p>&ntilde; MIDDLE LIST ITEM</p>
<p>&ntilde; LAST LIST ITEM</p>
<p>STUFF AFTER</p>";

echo make_real_list("/\<p\>&ntilde; (.*?)\<\/p\>/", $content);

//OUTPUT
<p>STUFF BEFORE</p> 
<ul>
    <li>FIRST LIST ITEM</li> 
    <li>MIDDLE LIST ITEM</li> 
    <li>LAST LIST ITEM</li>
</ul> 
<p>STUFF AFTER</p>
How to replace the lists with mvc4 actionlink lists

How to replace the lists with mvc4 actionlink lists


By : Robert Lu
Date : March 29 2020, 07:55 AM
NetLogo : How to do multiple operations on lists (find, get , replace, remove , search elements within lists , ....)

NetLogo : How to do multiple operations on lists (find, get , replace, remove , search elements within lists , ....)


By : Joel
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further map, filter, reduce, foreach, n-values, and sort-by provide customizable operations on lists, using tasks. See http://ccl.northwestern.edu/netlogo/docs/programming.html#tasks.
Here's your example using map:
code :
observer> show member? 3 map first [[1 2] [2 2] [2 3]]
observer: false
observer> show member? 3 map last [[1 2] [2 2] [2 3]]
observer: true
Replace empty columns with NA's in nested lists within lists

Replace empty columns with NA's in nested lists within lists


By : Luis A. Martinez
Date : March 29 2020, 07:55 AM
wish helps you I have produced 10 nested lists within lists using two functions (below) - produced from dummy data called normalised_scores. Each nested list contains two empty columns called Predicted and Actual (code below). I would like to replace these empty columns with NA's. I tried to achieve this with a function (Shuffle100) but got error messages (below). After looking online for a long time, I cannot find an efficient method apart from using long-winded code (below). If anyone can help, then thank you. , We can use transform
code :
lapply(my_list, transform, Predicted=NA, Actual=NA)
Replace numeric(0) with NAs in all lists in a column of lists in r

Replace numeric(0) with NAs in all lists in a column of lists in r


By : user1612639
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I have a column in a dataframe that stores lists. Here is an example below: , Modify your lapply statement like so
code :
lapply(example$col, function(x){ y <- x[x != 9]; if(length(y) < 1) { y <- NA }; y })
remove_num <- function(vec, num) {
   x <- vec[vec != num]
   if (length(x) < 1) { x <- NA }
   x
}

lapply(example$col, remove_num, 9)
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