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How can I effectively curve my items inside a list?


How can I effectively curve my items inside a list?

By : Vijay
Date : January 11 2021, 03:34 PM
hop of those help? Eventually, if it ever arrives, the css shape-inside property is supposed to solve this problem. (See https://drafts.csswg.org/css-shapes-2/) In the meantime, here is a proposed solution. I am using html to create li classes and using css to style the margin-left property of each bullet. I can then use classes to simulate a gentle curve in one direction and repeat each class on the other side to simulate the other half of the curve. You can tweak the margin-left values according to your needs. You mention that margins do not work for you, but I wonder if that's because of some aspect of your code I'm not seeing, as my version works.
code :
<!DOCTYPE html>
<html>
<head>
<style>
.a {
  margin-left: 30px;
}
.b {
  margin-left: 23px;
}
.c {
  margin-left: 18px;
}
.d {
  margin-left: 16px;
}

</style>
</head>
<body>

<ul>
  <li class="a">Coffee</li>
  <li class="b">Tea</li>
  <li class="c">Coca Cola</li>
  <li class="d">Sprite</li>
  <li class="d">Something else</li>
  <li class="c">Other text</li>
  <li class="b">Espresso</li>
  <li class="a">Root beer</li>
</ul>
</body>
</html>


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Binary search tree effectively acts as linked list when items are added in an ordered way

Binary search tree effectively acts as linked list when items are added in an ordered way


By : user3806034
Date : March 29 2020, 07:55 AM
wish of those help If you don't want to use a self-balancing tree (red-black, AVL, etc.), then the "easy" solution is to just shuffle the collection before inserting it into the tree.
If you want a perfectly balanced tree, you could start with a sorted collection, then insert the middle element of that list into the tree and recursively do the same with the two remaining sorted subcollections.
I am trying to remove items from a list based upon items from another list (inside a loop). error: list index out of ran

I am trying to remove items from a list based upon items from another list (inside a loop). error: list index out of ran


By : Vivek Pandey
Date : March 29 2020, 07:55 AM
I hope this helps you . Your problem is that you change the size of the list during iteration. Which obviously is a problem since after deleting a few items your j loop variable is going to be outside the range of the new (after deletion) list length. The first time it only works because the list only contains 1 element.
Try this instead:
code :
oc = [item for item in oc if item not in remlist]
How to CPU effectively add item to a list of unique items

How to CPU effectively add item to a list of unique items


By : Ralf Dragon
Date : March 29 2020, 07:55 AM
Hope that helps Almost each language has a list that is optimized for holding only unique values. In C++ you could use a std::set instead of a list. In C# you would use a HashSet. In JavaScript you would use an object...
In your question you're doing a O(N) lookup for each element, a set or unique list will at least do a O(log(N)) which is many times faster.
How to compress/archive a temperature curve effectively?

How to compress/archive a temperature curve effectively?


By : Meghna Bhave
Date : March 29 2020, 07:55 AM
To fix this issue A simple approach would be code the temperature into a byte or two bytes, depending on the range and precision you need, and then to write the first temperature to your output, followed by the difference between temperatures for all the rest. For two-byte temperatures you can restrict the range some and write one or two bytes depending on the difference with a variable-length integer. E.g. if the high bit of the first byte is set, then the next byte contains 8 more bits of difference, allowing for 15 bits of difference. Most of the time it will be one byte, based on your description.
Then take that stream and feed it to a standard lossless compressor, e.g. zlib.
Rearranging list items based on a score to fit a function curve

Rearranging list items based on a score to fit a function curve


By : A Majhi
Date : March 29 2020, 07:55 AM
it fixes the issue I'd do sth along these lines. Sort the words by their points, take every second out, reverse that half and concat the two:
code :
>>> s = sorted(zip(map(int, points), words))
>>> new_words = [word for p, word in list(reversed(s[::2])) + s[1::2]]
# If you have lots of words you'll be better off using some 
# itertools like islice and chain, but the principle becomes evident
>>> new_words
['apple', 'car', 'older', 'values', 'exponential', 'coefficient', 'average', 'man', 'pear']
[(9999, 'apple'), (8231, 'car'), (4712, 'older'), (500, 'values'), (5, 'exponential'), (10, 'coefficient'), (3242, 'average'), (5123, 'man'), (9231, 'pear')]
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