By : DoodleNan
Date : November 20 2020, 03:01 PM

should help you out Say I have an app that's trying to relate a string with an int. There are many strings and I want to keep a list of the top N that have occurred. , I was looking for LeastFrequently Used code :
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Combinations of weighted elements in a set where weighted sum equal to fixed integer (in python)
By : user3728308
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Looping through all n! permutations is much too expensive. Instead, generate all 2^n subsets. code :
from itertools import chain, combinations
def weight(A):
return sum(weights[x] for x in A)
# Copied from example at http://docs.python.org/library/itertools.html
def powerset(iterable):
"powerset([1,2,3]) > () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
s = list(iterable)
return chain.from_iterable(combinations(s, r) for r in xrange(len(s) + 1))
[x for x in powerset({'A', 'B', 'C', 'D', 'E'}) if weight(x) == W]
[('A', 'D'), ('C', 'B'), ('C', 'E'), ('A', 'B', 'E'), ('B', 'E', 'D')]

Efficently simulate rolling weighted dice (or traversing a weighted graph), with frequent updates
By : arpit pathak
Date : March 29 2020, 07:55 AM
it should still fix some issue I think you can do it with log(k) complexity where k is number of faces in the dice. for one particular node let p1, p2, ..., pk be the relative probabilities. Let p1+p2,...,+pk = p. code :
====Tree Example====
220
/ \
130 90
/ \ / \
50 80 20 70
   
1 2 3 4
==== Lazy update variation ====

Weighted, Undirected Adjacency List: Maximum Weighted Edge in a Single Cycle in O(n)
By : Yann
Date : March 29 2020, 07:55 AM
will be helpful for those in need You can return in the DFS which node was found in the cycle, then go back marking every node up in the DFS tree as part of the cycle (until the found node itself). Something like this: code :
DFS(v):
mark v as visited
for edges (v, w) in E:
if w is not visited:
last_node = DFS(w)
if last_node != 1:
test (v, w) as maximum edge
if last_node != v:
return last_node
else:
return 1
else:
test (v, w) as maximum edge
return w
return 1

nw:weightedpathto, nw:turtlesonweightedpathto and multiple equally weighted paths
By : user3572741
Date : March 29 2020, 07:55 AM
Does that help Good question! You can generate either list from the other, but I think turtlepath to linkpath is easier: code :
;; Construct the turtle path, putting the current turtle on the front:
let turtlepath fput self nw:turtlesonweightpathto turtle 3 "weight"
;; Iterate through pairs of turtles, getting the link connecting them
let linkpath (map [[linkwith ?2] of ?1] butlast turtlepath butfirst turtlepath)
reduce [ lput [[otherend] of ?2] of (last ?1) ?1 ] fput (list self) nw:weightedpathto turtle 3 "weight"
toreport addlinktoturtlepath [ turtlepath nextlink ]
let lastturtle last turtlepath
report lput [[otherend] of nextlink] of lastturtle
end
;; turtleprocedure  Assumes the current turtle is the starting point of the path
toreport linkpathtoturtlepath [ linkpath ]
let startofpath (list self)
report reduce addlinktoturtlepath fput startofpath linkpath
end

Agilemethodology in Project and QueryDriven methodology in Cassandra?
By : Cordvision
Date : March 29 2020, 07:55 AM
Does that help How can we manage our project gathering both rules together? the first one accept changes easily, but, the second one, want us to know every query that will be answered in our project. New requirements, causes new queries, that will changes our DB, and it will influence the quality



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