comparison

问题 Is there a way to compare two geometric shapes (or any two more generic data structures), without using the brute force when a tolerance is involved? The brute force (that is comparing each value of each object against each value of the other object) works but it's slow, and I can't use it. I tried sorting the data and comparing two sorted collections. It's fast, but it only works with zero tolerance. As soon as I add the tolerance I get lost. The problem is that two values can be identical

问题 Is there a way to compare two geometric shapes (or any two more generic data structures), without using the brute force when a tolerance is involved? The brute force (that is comparing each value of each object against each value of the other object) works but it's slow, and I can't use it. I tried sorting the data and comparing two sorted collections. It's fast, but it only works with zero tolerance. As soon as I add the tolerance I get lost. The problem is that two values can be identical

问题 Is there a difference between !== and != in PHP? 回答1: The != operator compares value, while the !== operator compares type as well. That means this: var_dump(5!="5"); // bool(false) var_dump(5!=="5"); // bool(true), because "5" and 5 are of different types 回答2: != is the inverse of the == operator, which checks equality across types !== is the inverse of the === operator, which checks equality only for things of the same type. 回答3: != is for "not equal", while !== is for "not identical". For

问题 I have this simple python program. I ran it and it prints yes , when in fact I expect it to not print anything because 14 is not greater than 14 . I saw this related question, but it isn't very helpful. #! /usr/bin/python import sys hours = "14" if (hours > 14): print "yes" What am I doing wrong? 回答1: Convert the string to an integer with int : hours = int("14") if (hours > 14): print "yes" In CPython2, when comparing two non-numerical objects of different types, the comparison is performed

问题 I have trouble comparing 2 double in Excel VBA suppose that I have the following code Dim a as double Dim b as double a = 0.15 b = 0.01 After a few manipulations on b, b is now equal to 0.6 however the imprecision related to the double data type gives me headache because if a = b then //this will never trigger end if Do you know how I can remove the trailing imprecision on the double type? 回答1: You can't compare floating point values for equality. See this article on "Comparing floating point

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . I spent quite some time researching for a library that allows me to compare images to one another in Java. I didn't really find anything useful, maybe my GoogleSearch-skill isn't high enough so I thought I'd ask you guys if you could point me into a direction of where I could find something like this. Basically

问题 This question already has answers here : What are the differences between Generics in C# and Java… and Templates in C++? [closed] (13 answers) Closed 6 years ago . I have heard that the Java implementation of Generics is not as good as the C# implementation. In that the syntax looks similar, what is it that is substandard about the Java implementation, or is it a religious point of view? 回答1: streloksi's link does a great job of breaking down the differences. The quick and dirty summary

问题 This question already has answers here : What are the differences between Generics in C# and Java… and Templates in C++? [closed] (13 answers) Closed 6 years ago . I have heard that the Java implementation of Generics is not as good as the C# implementation. In that the syntax looks similar, what is it that is substandard about the Java implementation, or is it a religious point of view? 回答1: streloksi's link does a great job of breaking down the differences. The quick and dirty summary

问题 I have a time in hh:mm and it has to be entered by the user in that format. However, I want to compare the time (eg. 11:22) is it between 10am to 6pm? But how do I compare it? 回答1: Java doesn't (yet) have a good built-in Time class (it has one for JDBC queries, but that's not what you want). One option would be use the JodaTime APIs and its LocalTime class. Sticking with just the built-in Java APIs, you are stuck with java.util.Date. You can use a SimpleDateFormat to parse the time, then the

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