range

问题 I'm refactoring a function that, given a series of endpoints that implicitly define intervals, checks if a number is included in the interval, and then return a corresponding (not related in any computable way). The code that is now handling the work is: if p <= 100: return 0 elif p > 100 and p <= 300: return 1 elif p > 300 and p <= 500: return 2 elif p > 500 and p <= 800: return 3 elif p > 800 and p <= 1000: return 4 elif p > 1000: return 5 Which is IMO quite horrible, and lacks in that both

问题 Variations of this question can be found all over the internet but not an answer. I want a seekbar with two-thumb range selection. I'm willing to program this myself but I lack experience with Android. Could someone give me some pointers on where to start. I mean, I know I will have to extend something (ProgressBar probably), but how should I go about to do that? Do I really have to recreate all the functionality of a standard seekbar, or is there an easier way? Complete solutions are also

问题 Variations of this question can be found all over the internet but not an answer. I want a seekbar with two-thumb range selection. I'm willing to program this myself but I lack experience with Android. Could someone give me some pointers on where to start. I mean, I know I will have to extend something (ProgressBar probably), but how should I go about to do that? Do I really have to recreate all the functionality of a standard seekbar, or is there an easier way? Complete solutions are also

问题 Is there an unbounded version of range (or xrange for Python 2), or is it necessary to define it manually? For example squares = (x*x for x in range(n)) can only give me a generator for the squares up to (n-1)**2 , and I can't see any obvious way to call range(infinity) so that it just keeps on truckin'. 回答1: You're describing the basic use of itertools.count: import itertools squares = (x*x for x in itertools.count()) 来源： https://stackoverflow.com/questions/7186336/unbounded-range

问题 I am getting an error when running a python program: Traceback (most recent call last): File "C:\Program Files (x86)\Wing IDE 101 4.1\src\debug\tserver\_sandbox.py", line 110, in <module> File "C:\Program Files (x86)\Wing IDE 101 4.1\src\debug\tserver\_sandbox.py", line 27, in __init__ File "C:\Program Files (x86)\Wing IDE 101 4.1\src\debug\tserver\class\inventory.py", line 17, in __init__ builtins.NameError: global name 'xrange' is not defined The game is from here. What causes this error?

问题 How can I revert a range of commits in git? From looking at the gitrevisions documentation, I cannot see how to specify the range I need. For example: A -> B -> C -> D -> E -> HEAD I want to do the equivalent of: git revert B-D where the result would be: A -> B -> C -> D -> E -> F -> HEAD where F contains the reverse of B-D inclusive. 回答1: What version of Git are you using? Reverting multiple commits in only supported in Git1.7.2+: see "Rollback to an old commit using revert multiple times."

问题 xrange function doesn't work for large integers: >>> N = 10**100 >>> xrange(N) Traceback (most recent call last): ... OverflowError: long int too large to convert to int >>> xrange(N, N+10) Traceback (most recent call last): ... OverflowError: long int too large to convert to int Python 3.x: >>> N = 10**100 >>> r = range(N) >>> r = range(N, N+10) >>> len(r) 10 Is there a backport of py3k builtin range() function for Python 2.x? Edit I'm looking for a complete implementation of "lazy" range()

问题 Is it possible to use the range operator ... and ..< with if statement. Maye something like this: let statusCode = 204 if statusCode in 200 ..< 299 { NSLog("Success") } 回答1: You can use the "pattern-match" operator ~= : if 200 ... 299 ~= statusCode { print("success") } Or a switch-statement with an expression pattern (which uses the pattern-match operator internally): switch statusCode { case 200 ... 299: print("success") default: print("failure") } Note that ..< denotes a range that omits

问题 Is there a range() equivalent for floats in Python? >>> range(0.5,5,1.5) [0, 1, 2, 3, 4] >>> range(0.5,5,0.5) Traceback (most recent call last): File "<pyshell#10>", line 1, in <module> range(0.5,5,0.5) ValueError: range() step argument must not be zero 回答1: I don't know a built-in function, but writing one like this shouldn't be too complicated. def frange(x, y, jump): while x < y: yield x x += jump As the comments mention, this could produce unpredictable results like: >>> list(frange(0,

问题 Given two inclusive integer ranges [x1:x2] and [y1:y2], where x1 ≤ x2 and y1 ≤ y2, what is the most efficient way to test whether there is any overlap of the two ranges? A simple implementation is as follows: bool testOverlap(int x1, int x2, int y1, int y2) { return (x1 >= y1 && x1 <= y2) || (x2 >= y1 && x2 <= y2) || (y1 >= x1 && y1 <= x2) || (y2 >= x1 && y2 <= x2); } But I expect there are more efficient ways to compute this. What method would be the most efficient in terms of fewest