floating-accuracy

问题 I encountered negative zero in output from python; it\'s created for example as follows: k = 0.0 print(-k) The output will be -0.0 . However, when I compare the -k to 0.0 for equality, it yields True. Is there any difference between 0.0 and -0.0 (I don\'t care that they presumably have different internal representation; I only care about their behavior in a program.) Is there any hidden traps I should be aware of? 回答1: Check out : −0 (number) in Wikipedia Basically IEEE does actually define a

问题 My code: a = \'2.3\' I wanted to display a as a floating point value. Since a is a string, I tried: float(a) The result I got was : 2.2999999999999998 I want a solution for this problem. Please, kindly help me. I was following this tutorial. 回答1: I think it reflects more on your understanding of floating point types than on Python. See my article about floating point numbers (.NET-based, but still relevant) for the reasons behind this "inaccuracy". If you need to keep the exact decimal

问题 when i make this multiplication 0.94 * 8700 the output is 8177.999999999999 but it should have been 8178 i\'m using java , but i don\'t think this error is related to a particular Programming language now my question is ... why this happened ?? and what other numbers (just as an example) cause the same error? 回答1: The specific reason in your case is that the real number 0.94 cannot be represented exactly in a double precision floating point. When you type 0.94 , the actual number stored is 0

问题 In my course, I am told: Continuous values are represented approximately in memory, and therefore computing with floats involves rounding errors. These are tiny discrepancies in bit patterns; thus the test e==f is unsafe if e and f are floats. Referring to Java. Is this true? I\'ve used comparison statements with double s and float s and have never had rounding issues. Never have I read in a textbook something similar. Surely the virtual machine accounts for this? 回答1: It is true. It is an

问题 This question already has an answer here: Is floating point math broken? 31 answers i am kind of confused why python add some additional decimal number in this case, please help to explain >>> mylist = [\"list item 1\", 2, 3.14] >>> print mylist [\'list item 1\', 2, 3.1400000000000001] 回答1: Floating point numbers are an approximation, they cannot store decimal numbers exactly. Because they try to represent a very large range of numbers in only 64 bits, they must approximate to some extent. It

问题 Possible Duplicate: Why are these numbers not equal? The below expression, which evaluates to 0.1, is considered larger than 0.1. > round(1740/600,0) - 1740/600 [1] 0.1 > (round(1740/600,0) - 1740/600) <= 0.1 [1] FALSE //???!!??? > (round(1740/600,0) - 1740/600) <= 0.1000000000000000000000000000000000000001 [1] TRUE Thinking that the issue might be due to rounding I tried this with the same result: > 3 - 2.9 [1] 0.1 > (3 - 2.9) <=0.1 [1] FALSE So, what gives and how do I fix it without

问题 Comparing two floating point number by something like a_float == b_float is looking for trouble since a_float / 3.0 * 3.0 might not be equal to a_float due to round off error. What one normally does is something like fabs(a_float - b_float) < tol . How does one calculate tol ? Ideally tolerance should be just larger than the value of one or two of the least significant figures. So if the single precision floating point number is use tol = 10E-6 should be about right. However this does not

问题 In Java the floating point arithmetic is not represented precisely. For example this java code: float a = 1.2; float b= 3.0; float c = a * b; if(c == 3.6){ System.out.println(\"c is 3.6\"); } else { System.out.println(\"c is not 3.6\"); } Prints \"c is not 3.6\". I\'m not interested in precision beyond 3 decimals (#.###). How can I deal with this problem to multiply floats and compare them reliably? 回答1: It's a general rule that floating point number should never be compared like (a==b) , but

问题 I\'ve heard of \"error\" when using floating point variables. Now I\'m trying to solve this puzzle and I think I\'m getting some rounding/floating point error. So I\'m finally going to figure out the basics of floating point error. What is a simple example of floating point/rounding error (preferably in C++) ? Edit: For example say I have an event that has probability p of succeeding. I do this event 10 times (p does not change and all trials are independent). What is the probability of

问题 In the C++ standard libraries I found only a floating point log method. Now I use log to find the level of an index in a binary tree ( floor(2log(index)) ). Code (C++): int targetlevel = int(log(index)/log(2)); I am afraid that for some of the edge elements (the elements with value 2^n) log will return n-1.999999999999 instead of n.0. Is this fear correct? How can I modify my statement so that it always will return a correct answer? 回答1: You can use this method instead: int targetlevel = 0;