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Compare Function result to multiple values


Compare Function result to multiple values

By : loal
Date : November 22 2020, 09:00 AM
Hope that helps I'm redesigning some of my For-loops to run faster. Especially the ones that call functions to compare values with. I'm wondering if my basic Delphi knowledge is missing something really useful. , You need to extract this into a function that you can re-use:
code :
function PropValueIsTrue(const PropValue: string): Boolean;
begin
  Result := (PropValue='Yes') or (PropValue='Y') or (StrtoIntDef(PropValue,0)>0);
end;
if PropValueIsTrue(fGetPropValue(i,'Condition')) and
   PropValueIsTrue(fGetPropValue(i,'Enabled')) and
   PropValueIsTrue(fGetPropValue(i,'Visible')) and
   PropValueIsTrue(fGetPropValue(i,'AllowAccess')) then


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How to use multiple result values from a function in R?

How to use multiple result values from a function in R?


By : W. Nelsen
Date : March 29 2020, 07:55 AM
it should still fix some issue I have the following code in R: , so many options
code :
# with
with(evaluate(a), b + c + d)
[1] 18

# unlist the unnamed output object
sum(unlist(evaluate(a)))
[1] 18

# subset a named output object
result <- evaluate(a) 
result$b + result$c + result$d
[1] 18

# subset an unnamed output object
evaluate(a)$b + evaluate(a)$c + evaluate(a)$d
[1] 18

# custom function with fancy arguments
f <- function(...) {
   args <- unlist(...)
   sum(args)
}

f(evaluate(a))
[1] 18
result[["b"]] + result[["c"]] + result[["d"]]
[1] 18
PHP MySQL Query Multiple Values then Compare to HTML Input then echo based on each result

PHP MySQL Query Multiple Values then Compare to HTML Input then echo based on each result


By : David Schmidt
Date : March 29 2020, 07:55 AM
Does that help I believe you are trying to conditionalize the select statement, which you can do using the CASE syntax
code :
$sql = "
    SELECT 
        CASE 
            WHEN rma_num_status = 1 THEN ''
            WHEN rma_num_status = 2 THEN 'Already in use'
            WHEN rma_num_status = 3 THEN 'No longer Valid'
            ELSE 'RMA Number was not found'
        END CASE as 'status'
    FROM " . DB_PREFIX . "reg_product 
    WHERE rma_num = '". $rma_num ."'
    AND regpro_sn = '". $regpro_sn ."'
";
How to compare multiple values in a function that takes only 2 parameters?

How to compare multiple values in a function that takes only 2 parameters?


By : thol
Date : March 29 2020, 07:55 AM
Any of those help To avoid excessive usage of if, especially for larger array where combination will explode, you can rely on loops. Below is the solution I proposed after doing a bit of cleanup to your code:
code :
// splits the hex value to its components
function splitHex(hex) {
    return {
        r: parseInt(hex.substring(0, 2), 16),
        g: parseInt(hex.substring(2, 4), 16),
        b: parseInt(hex.substring(4, 6), 16)
    };
}

// return number between 0 (opposit colors) and 1 (same colors)
function hexColorCalc(hex1, hex2) {
    // get red/green/blue int values of hex1
    const hex1Split = splitHex(hex1);
    // get red/green/blue int values of hex2
    const hex2Split = splitHex(hex2);
    // calculate normalized differences between reds, greens and blues
    const r = 1 - Math.abs(hex1Split.r - hex2Split.r) / 255;
    const g = 1 - Math.abs(hex1Split.g - hex2Split.g) / 255;
    const b = 1 - Math.abs(hex1Split.b - hex2Split.b) / 255;

    return (r + g + b) / 3;
}

// clears array from similar colors with threshold
function hexColorDelta(array, threshold) {
    // loop through array twice as a table to get all combinations
    for (let i = 0; i < array.length; i++) {
        for (let j = 0; j < array.length; j++) {
            // only remove the color in the second loop if similar
            if (i !== j && hexColorCalc(array[i][0], array[j][0]) > threshold) {
                array.splice(j, 1);
                j--;
            }   
        }
    }
    return array;
}

// test
const similar_1_2 = [["da542f", 3117], ["da5630", 60], ["333333", 6421]];
const similar_1_3 = [["da542f", 3117], ["333333", 6421], ["da5630", 60]];
const similar_2_3 = [["333333", 6421], ["da542f", 3117], ["da5630", 60]];
const similar_all = [["da5630", 60], ["da542f", 3117], ["da5630", 60]];
console.log(JSON.stringify(hexColorDelta(similar_1_2, 0.9)));
console.log(JSON.stringify(hexColorDelta(similar_1_3, 0.9)));
console.log(JSON.stringify(hexColorDelta(similar_2_3, 0.9)));
console.log(JSON.stringify(hexColorDelta(similar_all, 0.9)));
compare multiple values inside aggregation and display result in array in mongodb

compare multiple values inside aggregation and display result in array in mongodb


By : Aftab Alam
Date : March 29 2020, 07:55 AM
Any of those help You can try below aggregation query.
First $group to get distinct order number and status combination and pick the latest time stamp.
code :
db.order_summary.aggregate([
  {"$project":{
    "schedule_datetime":1,
    "order_number ":1,
    "order_status":{"$ifNull":["$order_status",""]
    }
  }},
  {"$sort":{"schedule_datetime":-1}},
  {"$group":{
    "_id":{
      "order_number":"$order_number",
      "order_status":"$order_status"
    },
    "schedule_datetime":{"$first":"$schedule_datetime"}
  }},
  {"$group":{
    "_id":{
       "year":{"$substr":["$schedule_datetime",0,4]},
       "order_number":"$_id.order_number"
      },
     "order_statuses":{"$push":"$_id.order_status"}
   }},
  {"$redact": {"$cond":[{"$in": ["SUCCESS","$order_statuses"]},"$$KEEP","$$PRUNE"]}},
  {"$sort":{"_id.year":-1}},
  {"$project":{"_id":0, "year":"$_id.year","order_number":"$_id.order_number", "order_statuses":1 }},
])
How do I compare the result of a function to values in a list box to determine which form to open on start up?

How do I compare the result of a function to values in a list box to determine which form to open on start up?


By : George Reshetnikov
Date : March 29 2020, 07:55 AM
I hope this helps you . I'm trying to employ a basic level of security to my Access db. , Try this:
code :
Set c = Forms!Users_Form!lst_Users

For i = 0 To c.ListCount - 1
    If c.Column(0, i) = ThisUserName() Then
        y = True
    End If
Next i
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