Authors | Paper | Method | Region of Interest | Application |
---|---|---|---|---|
Zheng et al. | Deep Learning-based Detection for COVID-19 from Chest CT using Weak Label | U-Net | Lung | Diagnosis |
Cao et al. | Longitudinal Assessment of COVID-19 Using a Deep Learning–based Quantitative CT Pipeline: Illustration of Two Cases | U-Net | Lung, Lesion | Quantification |
Huang et al. | Serial Quantitative Chest CT Assessment of COVID-19: Deep-Learning Approach | U-Net | Lung, Lung lobes, Lesion | Quantification |
Qi et al. | Machine learning-based CT radiomics model for predicting hospital stay in patients with pneumonia associated with SARS-CoV-2 infection: A multicenter study | U-Net | Lung lobes, Lesion | Quantification |
Gozes et al. | [Rapid ai development cycle for the c |
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/* | |
Project Euler: Problem 8: Largest product in a series | |
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832. | |
Find the `n` adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product? | |
*/ | |
class Tester { | |
constructor(func_call, test_values) { | |
this.timer_start = performance.now(); | |
this.timer_end = performance.now(); |
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/* | |
Project Euler: Problem 7: 10001st prime | |
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. | |
What is the nth prime number? | |
*/ | |
// list of numbers we wanna test | |
var test_values = [6, 10, 100, 1000, 10001]; | |
// this function execute the code and records the time to execute |
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/* | |
Project Euler: Problem 6: Sum square differencePassed | |
The sum of the squares of the first ten natural numbers is, | |
12 + 22 + ... + 102 = 385 | |
The square of the sum of the first ten natural numbers is, | |
(1 + 2 + ... + 10)2 = 552 = 3025 | |
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. |
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/* | |
Project Euler: Problem 5: Smallest multiplePassed | |
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. | |
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to n? | |
*/ | |
// list of numbers we wanna test | |
var test_values = [5, 7, 10, 13, 20]; | |
// this function execute the code and records the time to execute |
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/* | |
Project Euler: Problem 4: Largest palindrome productPassed | |
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. | |
Find the largest palindrome made from the product of two n-digit numbers. | |
*/ | |
// list of numbers we wanna test | |
var test_values = [2, 3]; | |
// this function execute the code and records the time to execute |
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/* | |
Project Euler: Problem 3: Largest prime factorPassed | |
The prime factors of 13195 are 5, 7, 13 and 29. | |
What is the largest prime factor of the given number? | |
*/ | |
// list of numbers we wanna test | |
var test_values = [2, 3, 5, 7, 13195, 600851475143]; | |
// this function execute the code and records the time to execute |
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/* | |
Project Euler: Problem 2: Even Fibonacci NumbersPassed | |
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: | |
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... | |
By considering the terms in the Fibonacci sequence that do not exceed the nth term, find the sum of the even-valued terms. | |
*/ | |
var test_values = [10, 18, 23, 43]; | |
function run_function(func, test_values) { |
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/* | |
Project Euler: Problem 1: Multiples of 3 and 5Passed | |
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. | |
Find the sum of all the multiples of 3 or 5 below the provided parameter value number. | |
*/ | |
var test_number = 1000; | |
function run_function(func) { | |
var t0 = performance.now(); |
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# KMeans | |
from sklearn.cluster import KMeans | |
kmeans = KMeans(n_clusters=int(num_days), random_state=0).fit(places_lat_long) | |
group = list(kmeans.labels_) | |
# MeanShift | |
from sklearn.cluster import MeanShift, estimate_bandwidth | |
bandwidth = estimate_bandwidth(places_lat_long, quantile=0.2) | |
clustering = MeanShift(bandwidth=bandwidth).fit(places_lat_long) | |
group = clustering.labels_ |