Sort them by one of the terms (start, for example), then check for overlaps with its (right-hand) neighbor as you move through the list. class tp: def __repr__(self): return “(%d,%d)” % (self.start, self.end) def __init__(self, start, end): self.start = start self.end = end s = [tp(5, 10), tp(7, 8), tp(0, 5)] s.sort(key=lambda self: self.start) y … Read more
Edit Apr 19, 2013: Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea: static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348; static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696; // Floating-point modulo // The result (the remainder) has same sign as the divisor. // Similar to matlab’s mod(); Not similar to fmod() – Mod(-3,4)= 1 fmod(-3,4)= -3 … Read more
Probably the easiest way is: SELECT EXTRACT(epoch FROM my_interval)/3600
A simple option is to manually keep track of time using millis(). You would use two variables: one to store elapsed time one to store the wait/delay time you need In the draw() method you would check if the difference between the current time (in millis.) and the previously stored time is greater(or equal) to … Read more
This is a general purpose solution which allows rounding to arbitrary resolutions. For your specific case, you just need to provide 0.25 as the resolution but other values are possible, as shown in the test cases. def roundPartial (value, resolution): return round (value / resolution) * resolution print “Rounding to quarters” print roundPartial (10.38, 0.25) … Read more
JavaScript is not the tool for this. If you want something to run at a specific time every day, you’re almost certainly looking for something that runs locally, like python or applescript. However, let’s consider for a moment that JavaScript is your only option. There are a few ways that you could do this, but … Read more
This is a weighted variation of interval scheduling; it’s solvable in O(N log N) with dynamic programming. Let an interval be g(start, stop, score), and let them be sorted by stop. For simplicity, let’s assume for now that all stop is unique. Let best[i] be the best score we can get when we’re allowed to … Read more
Here’s a possible solution using the new “non-equi” joins in data.table (v>=1.9.8). While I doubt you’ll like the syntax, it should be very efficient soluion. Also, regarding findInterval, this function assumes continuity in your intervals, while this isn’t the case here, so I doubt there is a straightforward solution using it. library(data.table) #v1.10.0 setDT(intervals)[data.table(elements), on … Read more
you could cast the result if you want less precision: SQL> SELECT TO_DSINTERVAL(’10 10:00:00′) t_interval FROM dual; T_INTERVAL ———————————————————– +000000010 10:00:00.000000000 SQL> SELECT CAST(TO_DSINTERVAL(’10 10:00:00′) 2 AS INTERVAL DAY(2) TO SECOND(3)) t_interval 3 FROM dual; T_INTERVAL ———————————————————– +10 10:00:00.000 Edit following OP comment: From The Oracle Documentation (11gr1): Interval datatypes do not have format models. … Read more
Here is a plotrix solution: set.seed(0815) x <- 1:10 F <- runif(10,1,2) L <- runif(10,0,1) U <- runif(10,2,3) require(plotrix) plotCI(x, F, ui=U, li=L) And here is a ggplot solution: set.seed(0815) df <- data.frame(x =1:10, F =runif(10,1,2), L =runif(10,0,1), U =runif(10,2,3)) require(ggplot2) ggplot(df, aes(x = x, y = F)) + geom_point(size = 4) + geom_errorbar(aes(ymax = … Read more