How can I claulate the rank of each candidate when I have the total candidates and votes secured by each?
I've managed the percentage part, but calculating the rank has me stuck.
I'll be using MySql in the end for this, but right now I only need the formula or method to calculate ranks.
Id be glad if you could help with just the formula. Just like the formula for interest is PTR/100.
Total Candidates
5
Total Votes
75
Votes
Name Marks Percentage Rank(What I'm trying to calculate)
A 25 33.34 1/5 ->Rank 1/5 has the most votes
B 20 26.67 2/5 ->And so on
C 10 13.34 4/5
D 5 6.67 5/5
E 15 20.00 3/5
There is a previous question on SO that addresses this, using MySQL and a ranking variable. There is some lovely stuff in the answers
MySQL rank function
Related
How to calculate percent of count of UNIQUE values?
E.g. I have a dataset with people who can pick multiple symptoms (i.e each person can have 0 to 10 values).
person 1 - symptom A, B
person 2 - symptom B, C, D
person 3 - no symptoms
person 4 - symptom A
etc.
E.g. if total UNIQUE count of people is 4 and 2 of them have picked symptom A, then I'd like to see:
A = 2/4 = 50%
Currently QuickSight is able to calculate shares based on total count of people (not unique count) as one person can have multiple symptoms, so A is 2/6 = 33% (not what I need).
As much as I've tried, QuickSight doesn't enable that??
I'm in need of some kind of algorithm I can't figure out on my own sadly.
My biggest problem is that I have no good way to describe the problem... :/
I will try like this:
Imagine you have a racing game where everyone can try to be the fastest on a track or map. Every Map is worth 100 Points in total. If someone finished a map in some amount of time he gets a record in a database. If the player is the first and only player to finish this map he earns all the 100 points of this map.
Now, that's easy ;) but...
Now another player finishes the map. Let's imagine the first player finishes in 50 Seconds and the 2nd player finishes in 55 seconds, so a bit slower. I now need a calculation depending on both records in the database. Each of both players now earn a part of the 100 points. The faster player a bit more then the slower player. Let's say they finished the exact same time they both would get 50 points from 100, but as the first one is slightly faster, he now earns something around 53 of the points and the slower player just 47.
I started to calculate this like this:
Sum of both records is 105 seconds, the faster player took 50/105 in percent of this, so he earns 100-(50/105*100) points and the slower player 100-(55/105*100) points. The key to this is, that all points distributed among the players always equals to 100 in total. This works for 2 players, but it breaks at 3 and more.
For example:
Player 1 : 20 seconds
Player 2 : 20 seconds
Player 3 : 25 seconds
Calculation would be:
Player 1: 100-(20/65*100) = 69 points
Player 2: 100-(20/65*100) = 69 points
Player 3: 100-(25/65*100) = 61 points
This would no longer add up to 100 points in total.
Fair would be something around values of:
Player 1 & 2 (same time) = 35 points
Player 3 = 30 points
My problem is i can't figure out a algorithm which solves this.
And I need the same algorithm for any amount of players. Can someone help with an idea? I don't need a complete finished algorithm, maybe just an idea at which step i used the wrong idea, maybe the sum of all times is already a bad start.
Thx in advance :)
We can give each player points proportional to the reciprocal of their time.
One player with t seconds gets 100 × (1/t) / (1/t) = 100 points.
Of the two players, the one with 50 seconds gets 100 × (1/50) / (1/50 + 1/55) ≈ 52.4, and the one with 55 gets 100 × (1/55) / (1/50 + 1/55) ≈ 47.6.
Of the three players, the ones with 20 seconds get 100 × (1/20) / (1/20 + 1/20 + 1/25) ≈ 35.7, and the one with 25 seconds gets 100 × (1/25) / (1/20 + 1/20 + 1/25) ≈ 28.6.
Simple observation: Let the sum of times for all players be S. A person with lower time t would have a higher value of S-t. So you can reward points proportional to S-t for each player.
Formula:
Let the scores for N players be a,b,c...,m,n. Total sum S = a+b+c...+m+n. Then score for a given player would be
score = [S-(player's score)]/[(N-1)*S] * 100
You can easily see that using this formula, the sum of scores of all players will be always be 100.
Example 1:
S = 50 + 55 = 105, N-1 = 2-1 = 1
Player 1 : 50 seconds => score = ((105-50)/[1*105])*100 = 52.38
Player 2 : 55 seconds => score = ((105-55)/[1*105])*100 = 47.62
Similarly, for your second example,
S = 20 + 20 + 25 = 65
N - 1 = 3 - 1 = 2
For Player 1, (S-t) = 65-20 = 45
Player 1's score => (45/(2*65))*100 = 34.6
Player 2 => same as Player 1
For Player 3, (S-t) = 65-25 = 40
Player 3's score => (40/(2*65))*100 = 30.8
This method avoids any division in the intermediate states, so there will be no floating point issues for the calculations.
Say I need to place n=30 students into groups of between 2 and 6, and I collect the following preference data from each student:
Student Name: Tom
Likes to sit with: Jimi, Eric
Doesn't like to sit with: John, Paul, Ringo, George
It's implied that they're neutral about any other student in the overall class that they haven't mentioned.
How might I best run a large number of simulations of many different/random grouping arrangements, to be able to determine a score for each arrangement, through which I could then pick the "most optimal" score/arrangement?
Alternatively, are there any other methods by which I might be able to calculate a solution that satisfies all of the supplied constraints?
I'd like a generic method that can be reused on different class sizes each year, but within each simulation run, the following constants and variables apply:
Constants: Total number of students, Student preferences
Variables: Group sizes, Student Groupings, Number of different group arrangements/iterations to test
Thanks in advance for any help/advice/pointers provided.
I believe you can state this as an explicit mathematical optimization problem.
Define the binary decision variables:
x(p,g) = 1 if person p is assigned to group g
0 otherwise
I used:
I used your data set with 28 persons, and your preference matrix (with -1,+1,0 elements). For groups, I used 4 groups of 6 and 1 group of 4. A solution can look like:
---- 80 PARAMETER solution using MIQP model
group1 group2 group3 group4 group5
aimee 1
amber-la 1
amber-le 1
andrina 1
catelyn-t 1
charlie 1
charlotte 1
cory 1
daniel 1
ellie 1
ellis 1
eve 1
grace-c 1
grace-g 1
holly 1
jack 1
jade 1
james 1
kadie 1
kieran 1
kristiana 1
lily 1
luke 1
naz 1
nibah 1
niko 1
wiki 1
zeina 1
COUNT 6 6 6 6 4
Notes:
This model can be linearized, so it can be fed into a standard MIP solver
I solved this directly as a MIQP model (actually the solver reformulated the model into a MIP). The model solved in a few seconds.
Probably we need to add extra logic to make sure one person is not getting a really bad assignment. We optimize here only the total sum. This overall sum may allow an individual to get a bad deal. It is an interesting exercise to take this into account in the model. There are some interesting trade-offs.
1st approach should be, create matrix n x n where n is total number of students, indexes for row and columns are ordinals for every student, and each column representing preferences for sitting with the others students. Fills the cells with values 1=Like to sit, -1 = the Opposite, 0 = neutral. Zeroes to be filled too on main diagonal (i,i)
------Mark Maria John Peter
Mark 0 1 -1 1
Maria 0 0 -1 1
John -1 1 0 1
Peter 0
Score calculations are based on sums of these values. So ie: John likes to sit with Maria, = 1, but Maria doesn't like to sit with John -1, result is 0. Best result is when both score (sum) 2.
So on, based on Group Sizes, calculate Score of each posible combination. Bigger the score, better the arrangement. Combinations discriminate values on main diagonal. ie: John grouped with the same John is not a valid combination/group.
In a group size of 2, best score is 2
In a group size of 3, best score is 6,
In a group size of 4, best score is 12
In a group size of n, best score would be (n-1)*n
Now in ordered list of combinations / groups, you should take first the best tuples with highest scores, but avoiding duplicates of students between tuples.
In a recent research, a PSO was implemented to classify students under unknown number of groups of 4 to 6. PSO showed improved capabilities compared to GA. I think that all you need is the specific research.
The paper is: Forming automatic groups of learners using particle swarm optimization for applications of differentiated instruction
You can find the paper here: https://doi.org/10.1002/cae.22191
Perhaps the researchers could guide you through researchgate: https://www.researchgate.net/publication/338078753
Regarding the optimal sitting you need to specify an objective function with the specific data
I have some problems with defining a algorithm that will calculate a ranking number for a dentist.
Assume, we have three different dentists:
dentist number 1: Got 125 patients and out of the 125 patients the
dentist have booked a time with 75 of them. 60% of them got a time.
dentist number 2: Got 5 patients and out of the 5 patients the
dentist have booked a time with 4 of them. 80% of them got a time.
dentist number 3: Got 25 patients and out of the 14 patients the
dentist have booked a time with 14 of them. 56% got a time.
If we use the formula:
patients booked time with / totalpatients * 100
it will not be the right way to calculate the ranking, as we will get an output of the higher percentage is, the better the dentist is, but it's wrong. By doing it in that way, the dentists would have a ranking:
dentist number 2 would have a ranking of 1. (80% got a time).
dentist number 1 would have a ranking of 2 (60% got a time).
dentist number 3 would have a ranking of 3. (56% got a time).
But, it should be in this way:
dentist number 1 = ranking 1
dentist number 2 = ranking 2
dentist number 3 = ranking 3
I don't know to make a algorithm that also takes the amount of patients as a factor to the ranking-calculation.
It is quite arbitrary how you define what makes a better dentist in terms of number of patients and the percentage of those that have an appointment with them.
Let's call the number of patients P, the number of those that have an appointment A, and the function determining how "good" a dentist is f. So f would be a function of P and A: f(P, A).
One component of f could indeed be what you already calculated: A/P.
Another component would have to be P, but I would think that the effect on f(P, A) of increasing P with 1 would be much higher for a low P, than for a high P, so this component should not be a linear function. It would also be practical if this component would have a value between 0 and 1, just like the other component.
Taking all this together, I suggest this definition of f, which will give a number between 0 and 1:
f(P,A) = 1/3 * P/(10 + P) + 2/3 * A/P
For the different dentists, this results in:
1: 1/3 * 125/135 + 2/3 * 75/125 = 0.7086419753...
2: 1/3 * 5/15 + 2/3 * 4/5 = 0.6444444444...
3: 1/3 * 25/35 + 2/3 * 14/25 = 0.6114285714...
You could play a bit with the constant factors in the formula, like increasing the term 10. Or you could change the factors 1/3 and 2/3 making sure that their sum is 1.
This is just one way to do it. There are an infinity of other ways...
R - Problem: to find the optimum number of non-uniform bins to show a range of data points.
I have a bunch of data points (let us assume different prices of different mobiles). I need to categorize these mobile phones into some categories (based on the price). The bin size (in this example refers to the price range) need not be uniform (there might be lots of mobiles in the low price category and few in the long tail category).
Is there any efficient algorithm to find the optimum number of bins required and the number of data points (in this case mobile phones) which shall go into each category.
This is not a standard formula, but wanted to post as it seem to work well with data set i tested.
Find the average price of all the mobiles.
Ex: 5 mobiles with prices 10, 20, 40, 80, 200
Avg is 350/5 = 70
Subtract minimum price from average price: 70 - 10 = 60 -> name it N1
Subtract avg price from Max price: 200 - 70 = 130 -> name it N2
Find the ratio N2/N1 : 130/60: Roughly 2
This indicates that it is better to have 2 bins at the lower price range for every 1 bin at higher range.
So, for example take 2 bins below 70. Range 0 - 35(2 mobiles), 36 - 70(1 mobile)
1 bin above 70: Range 71 - 200(2 mobiles)
As you can see, number of bins and bin sizes are reasonably optimal.