range

When using the triple dot notation in a ruby Range object, I get this: (0...5).each{|n| p n} 0 1 2 3 4 When I use the 'last' method I get: (0...5).last => 5 I would have expected 4 Is this is a bug? Or is there something I don't understand about the the concept of a Range object? This is by design. The Ruby 2.0 documentation is more specific : Note that with no arguments last will return the object that defines the end of the range even if exclude_end? is true . (10..20).last #=> 20 (10...20).last #=> 20 (10..20).last(3) #=> [18, 19, 20] (10...20).last(3) #=> [17, 18, 19] As Stefan has

Here's a custom function that allows stepping through decimal increments: def my_range(start, stop, step): i = start while i < stop: yield i i += step It works like this: out = list(my_range(0, 1, 0.1)) print(out) [0, 0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7, 0.7999999999999999, 0.8999999999999999, 0.9999999999999999] Now, there's nothing surprising about this. It's understandable this happens because of floating point inaccuracies and that 0.1 has no exact representation in memory. So, those precision errors are understandable. Take numpy on the other hand: import numpy as np out =

From pd.date_range('2016-01', '2016-05', freq='M', ).strftime('%Y-%m') , the last month is 2016-04 , but I was expecting it to be 2016-05 . It seems to me this function is behaving like the range method, where the end parameter is not included in the returning array. Is there a way to get the end month included in the returning array, without processing the string for the end month? A way to do it without messing with figuring out month ends yourself. pd.date_range(*(pd.to_datetime(['2016-01', '2016-05']) + pd.offsets.MonthEnd()), freq='M') DatetimeIndex(['2016-01-31', '2016-02-29', '2016-03

Let's say I have a range from 0 to 10: range = 0...10 Three dots mean, that the last value (10) is excluded: range.include? 10 => false Now, is there a similar and elegant way to exclude the first value? For the above example, this would mean to include all values that are bigger ( > , not >= ) than 0 and smaller than 10. No. ((0+1)..10) I have two suggestions for you, they're not very ideal but they're the best I can think of. First you can define a new method on the Range class that does what you describe. It'd look like this: class Range def have?(x) if x == self.begin false else include?(x

I want to generate 10000 integer random numbers between 0 and 10^12. Usually, the code would look like this: x <- sample(0:1000000000000,10000,replace=T) But I get following error message: Error in 0:1000000000000 : result would be too long a vector Is there a more memory efficient method that doesn't have to put 10^12 integers in a vector just to get a sample of size 10000? If not, is there a way to increase the max size of the vector? I'm working on a 64bit OS with 12GB of free RAM. PascalVKooten The real problem lies in the fact that you cannot store the sequence of 0:10^12 into memory. By

I am using the following code to get a String substring from an NSRange : func substring(with nsrange: NSRange) -> String? { guard let range = Range.init(nsrange) else { return nil } let start = UTF16Index(range.lowerBound) let end = UTF16Index(range.upperBound) return String(utf16[start..<end]) } (via: https://mjtsai.com/blog/2016/12/19/nsregularexpression-and-swift/ ) When I compile with Swift 4 (Xcode 9b4), I get the following errors for the two lines that declare start and end : 'init' is unavailable 'init' was obsoleted in Swift 4.0 I am confused, since I am not using an init. How can I