math

问题 Okay so I get that some numbers can't be represented properly in binary just like 1/3 can't be fully represented in decimal. So how come when I console.log(0.3) it returns 0.3 but when I console.log(0.1 + 0.2) it returns the 0.30000000000000004 How come it is accounting for the error (if it even is) when simply outputting 0.3 but doesn't when the addition occurs? 回答1: Suppose we approximate 1/3 and 2/3 in decimal. 1/3 = 0.333 2/3 = 0.667 and we add 1/3+1/3: 1/3+1/3 = 0.333 + 0.333 = 0.666 We

问题 Okay so I get that some numbers can't be represented properly in binary just like 1/3 can't be fully represented in decimal. So how come when I console.log(0.3) it returns 0.3 but when I console.log(0.1 + 0.2) it returns the 0.30000000000000004 How come it is accounting for the error (if it even is) when simply outputting 0.3 but doesn't when the addition occurs? 回答1: Suppose we approximate 1/3 and 2/3 in decimal. 1/3 = 0.333 2/3 = 0.667 and we add 1/3+1/3: 1/3+1/3 = 0.333 + 0.333 = 0.666 We

问题 I would like to iterate over all lists of length n whose elements sum to 2. How can you do this efficiently? Here is a very inefficient method for n = 10 . Ultimately I would like to do this for `n > 25'. n = 10 for L in itertools.product([-1,1], repeat = n): if (sum(L) == 2): print L #Do something with L 回答1: you only can have a solution of 2 if you have 2 more +1 than -1 so for n==24 a_solution = [-1,]*11 + [1,]*13 now you can just use itertools.permutations to get every permutation of this

问题 I would like to iterate over all lists of length n whose elements sum to 2. How can you do this efficiently? Here is a very inefficient method for n = 10 . Ultimately I would like to do this for `n > 25'. n = 10 for L in itertools.product([-1,1], repeat = n): if (sum(L) == 2): print L #Do something with L 回答1: you only can have a solution of 2 if you have 2 more +1 than -1 so for n==24 a_solution = [-1,]*11 + [1,]*13 now you can just use itertools.permutations to get every permutation of this

问题 I'm sure this has been answered before, but I can't find the thread for the life of me! I am trying to use r to produce a list of all the distances between pairs of xy coordinates in a dataframe. The data is stored something like this: ID = c('1','2','3','4','5','6','7') x = c(1,2,4,5,1,3,1) y = c(3,5,6,3,1,5,1) df= data.frame(ID,x,y) At the moment I can calculate the distance between two points using: length = sqrt((x1 - x2)^2+(y1 - y2)^2). However, I am uncertain as to where to go next.

问题 I'm sure this has been answered before, but I can't find the thread for the life of me! I am trying to use r to produce a list of all the distances between pairs of xy coordinates in a dataframe. The data is stored something like this: ID = c('1','2','3','4','5','6','7') x = c(1,2,4,5,1,3,1) y = c(3,5,6,3,1,5,1) df= data.frame(ID,x,y) At the moment I can calculate the distance between two points using: length = sqrt((x1 - x2)^2+(y1 - y2)^2). However, I am uncertain as to where to go next.

问题 I'm attempting to implement Vincenty's inverse problem as described on wiki HERE The problem is that lambda is simply not converging. The value stays the same if I try to iterate over the sequence of formulas, and I'm really not sure why. Perhaps I've just stared myself blind on an obvious problem. It should be noted that I'm new to Python and still learning the language, so I'm not sure if it's misuse of the language that might cause the problem, or if I do have some mistakes in some of the

问题 I want to draw a graph that will be something like this: alt text http://img25.imageshack.us/img25/9786/problemo.png You can see 3 pathes: a, b & c. How can I change position of elements (1,2,3...,9) to make long of the path as short as possible? I mean this lines should be as short as possible. Im very interested in it becouse I am drawing a graph with question, some kind of infographic like 'follow the lines to know the answer'. I know that its a bit about graph theory... so if its too hard

问题 Im needing the natural logarithm function for use in a .cpp (c++) source file. Now, of course I can do this with a quick google search and a simple library solution. But Im a bit confused... On the cplusplus dot com website under reference/cmath/log/ they have an example of how to use the log function, as follows /* log example */ #include <stdio.h> /* printf */ #include <math.h> /* log */ int main () { double param, result; param = 5.5; result = log (param); printf ("log(%f) = %f\n", param,

问题 I needed to convert a fractional part of a number into integer without a comma, for example I have 3.35 I want to get just 35 part without zero or a comma, Because I used the modf() function to extract the the fractional part but it gives me a 0.35 if there is any way to do that or to filter the '0.' part I will be very grateful if you show me how with the smaller code possible, 回答1: A bit more efficient than converting to a string and back again: int fractional_part_as_int(double number, int