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Rewriting a for loop in pure NumPy to decrease execution time


Rewriting a for loop in pure NumPy to decrease execution time

By : Joshua Budman
Date : January 02 2021, 06:48 AM


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Decrease WMI query execution time

Decrease WMI query execution time


By : Hiran Amarasinghe
Date : March 29 2020, 07:55 AM
help you fix your problem I would recommend querying only the properties you really need. So, if you only need the LicenseStatus value of the SoftwareLicensingProduct WMI class then use the following query:
code :
SelectQuery searchQuery = new 
            SelectQuery("SELECT LicenseStatus FROM SoftwareLicensingProduct");
Decrease execution time of SQL query

Decrease execution time of SQL query


By : Trushart
Date : March 29 2020, 07:55 AM
To fix this issue I've got a question in terms of processing and making a query more efficient whilst maintaining its accuracy. Before I display the query I'd like to point out some basics of it. , Your problem is this condition cant use INDEX
code :
AND TF.readyforwork = CASE  -- HERE IS THE PROBLEM
                        WHEN TF.trk_trackerstatus_lkid2 IS NULL THEN 0
                        ELSE 1
                      END
AND ( TF.readyforwork = 0 and TF.trk_trackerstatus_lkid2 IS NULL 
   OR TF.readyforwork = 1 and TF.trk_trackerstatus_lkid2 IS NOT NULL
    )
How to loop numpy ndarray by using cupy? Is will really improve execution time?

How to loop numpy ndarray by using cupy? Is will really improve execution time?


By : Alan Rubin
Date : March 29 2020, 07:55 AM
it should still fix some issue You will experience a huge boost in performance if you vectorize the option with NumPy's built-in functions:
code :
Array_B[array_A < 200] = -1000
How to increase or decrease a progress bar width within given time with pure JavaScript?

How to increase or decrease a progress bar width within given time with pure JavaScript?


By : Hoang
Date : March 29 2020, 07:55 AM
hop of those help? I would recommend using requestAnimationFrame first. Next, use a timer instead of counting how many times it is called. I made a few minor adjustments (you can call it with a different number to have a different delay).
RAF docs: https://developer.mozilla.org/en-US/docs/Web/API/window/requestAnimationFrame
code :
function move(delay) {
  var elem = document.getElementById("myBar");
  var end = Date.now() + delay;
  var frame = () => {
    var timeleft = Math.max(0, end - Date.now());
    elem.style.width = (100*timeleft)/delay + '%'; 
    elem.innerHTML = (timeleft/1000).toFixed(1)  + 's';
    if (timeleft === 0) return;
    requestAnimationFrame(frame);
  };
  requestAnimationFrame(frame);
}
#myProgress {
  width: 100%;
  background-color: #ddd;
}

#myBar {
  width: 100%;
  height: 30px;
  background-color: #4CAF50;
  text-align: center;
  line-height: 30px;
  color: white;
}
<!DOCTYPE html>
<html>
<body>
<div id="myProgress">
  <div id="myBar">7.0s</div>
</div>
<br>
<button onclick="move(7000)">Start</button> 
</body>
</html>
How to rewrite double for loop in Numpy to save execution time?

How to rewrite double for loop in Numpy to save execution time?


By : Narayana Reddy Y
Date : March 29 2020, 07:55 AM
With these it helps Somebody beat me to it in the comments, but I have some timing information for you. Note that your times may vary, but relative times should be fairly representative of the performance you might see.
First your code, which I fixed to make runnable:
code :
cards = []
for figure in range(2, 15):
    for suite in [1, 2, 3, 4]:
        card = [figure, suite]
        cards.append(card)
# 8.04 µs ± 27.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
import numpy as np
cards = np.mgrid[1:5, 2:15]
# 20.5 µs ± 320 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
import itertools as it
figures = range(2, 15)
suits = [1, 2, 3, 4]
cards = list(it.product(suits, figures))
# 2.5 µs ± 27.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
cards_it = it.product(suits, figures)  # notice no 'list'
# 479 ns ± 9.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
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