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By : kjelle
Date : November 19 2020, 03:01 PM
this will help You can use apply() to go though your start-finish data.frame, check if the numbers are between start and finish values and sum up the logical vector returned from data.tables' between() function. code :
``````Numbers<-c(0.1,0.2,0.3,0.7,0.8,0.9,0.91,0.99)

sf <-
"Start   Finish
0.00    0.86
0.87    0.89
0.90    0.98
0.99    1.00",
)

apply(sf, 1, function(x) {
sum(data.table::between(Numbers, x, x))
})
``````
``````5 0 2 1
`````` ## giving lots of intervals [ai, bi], find a interval which intersect with the most number of intervals

By : Kobus van vuuren
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Suppose the intervals are given as (a1,b1), ..., (an,bn). Make a sorted array of length 2n where ties are broken by
if ai = aj, then put ai first iff bi < bj if bi = bj, then put bi first iff ai < aj if ai = bj, then put ai first ## Given a vector of intervals output any one number within the interval that has been overlapped the most times

By : user3376867
Date : March 29 2020, 07:55 AM
I hope this helps you . Here is a simple O(n log n) solution which uses events and sweep line: ## Matlab: How to turn a vector of 2928x1 into a vector of 8784x1 with specific intervals of elements?

By : Gloria
Date : March 29 2020, 07:55 AM
With these it helps I am trying to convert a vector with 2928 values into a vector with 8784 values. The first vector is a vector with info with an interval of 3hours, and I would like to have an hourly vector with those values added every 3 hours and the remaining should be filled with NaN. My first approach was to create a NaN vector with 8784 values but then I have not been able to create a 'for loop' that worked with this. , Simply index with step different to 1. In this case, step is 3.
code :
``````B_h(1:3:end) = S_3h
`````` ## Looping through a vector and finding a maximum sum of values with fixed number of indices

By : sc0
Date : March 29 2020, 07:55 AM
hop of those help? Pretend you have an infinite sequence of numbers, repeating itself. It makes the problem much easier:
code :
``````fn main() {
let d = [1, 2, 3, 4, 5, 6, 7, 8];
let mut numbers = d.iter().cycle();

let max = (0..d.len())
.map(|_| {
let sum1: i32 = numbers.by_ref().take(3).sum();
let sum2: i32 = numbers.by_ref().take(5).sum();

// Skip one so the next iteration is offset by one
numbers.next();

(sum1, sum2)
})
.map(|(sum1, sum2)| i32::abs(3 * sum1 + 5 * sum2))
.max();

println!("{:?}", max);
}
`````` ## find different intervals for a vector of values, making number of values in each interval approximately equal

By : user3190399
Date : March 29 2020, 07:55 AM
it helps some times You want the breaks at particular quantiles of your data. For example,
code :
``````v01 <- runif(1000, 0, 1)
v12 <- runif(1000, 1, 2)
v25 <- runif(1000, 2, 5)
v <- c(v01, v12, v25)
n <- 3  # Number of intervals
breaks <- quantile(v, (0:n)/n)  # These are the breakpoints.  You might
# want to round them to integers, based 