logarithm

问题 For homework, using C, I'm supposed to make a program that finds the log base 2 of a number greater than 0 using only the operators ! ~ & ^ | + << >> . I know that I'm supposed to shift right a number of times, but I don't know how to keep track of the number of times without having any loops or if s. I've been stuck on this question for days, so any help is appreciated. int ilog2(int x) { x = x | (x >> 1); x = x | (x >> 2); x = x | (x >> 4); x = x | (x >> 8); x = x | (x >> 16); } This is

问题 I'm trying to get a logarithmic scale for the y-axis of a morris.js line chart. http://www.oesmith.co.uk/morris.js/lines.html I already tried playing with the yLabelFormat option, but it's not what I need. Any hint is appreciated. If there is no way of doing this with morris.js, you can suggest another lightweight javascript library to make simple line charts with logarithmic scale. 回答1: You can extend Morris and modify the transY function to do the logarithmic scale. I also added the

问题 I have a normal distribution plot and a histogram plot with x axis in log scale displaying 0, 10^0, 10^1 ... I want to include minor ticks between the major ones. Actually I was able to change the major ticks format from 1, 2, 3 and so on to 10^0, 10^1, 10^2, 10^3 using the solution given to me in my previous question. I used the following code for the major ticks : major.ticks <- axTicks(1) labels <- sapply(major.ticks,function(i) as.expression(bquote(10^ .(i))) ) axis(1,at=major.ticks

问题 I have a normal distribution plot and a histogram plot with x axis in log scale displaying 0, 10^0, 10^1 ... I want to include minor ticks between the major ones. Actually I was able to change the major ticks format from 1, 2, 3 and so on to 10^0, 10^1, 10^2, 10^3 using the solution given to me in my previous question. I used the following code for the major ticks : major.ticks <- axTicks(1) labels <- sapply(major.ticks,function(i) as.expression(bquote(10^ .(i))) ) axis(1,at=major.ticks

问题 I am trying to calculate logarithms using the math module of python (math.log(x,[base]) , however when I use float(raw_input) for the x and base values, it gives me the error, ZeroDivisionError: float division by zero . x = 9.0 base = 3.0 回答1: Nonsense, it works perfectly well >>> import math >>> x=float(raw_input()) 9.0 >>> base=float(raw_input()) 3.0 >>> math.log(x, base) 2.0 Why don't you show us exactly how you reproduce the problem? wim is quite correct - a base of 1 will give that error

问题 Today, I came across quite strange problem. I needed to calculate string length of a number, so I came up with this solution // say the number is 1000 (int)(log(1000)/log(10)) + 1 This is based on mathematical formula log 10 x = log n x/log n 10 (explained here) But I found out, that in C, (int)(log(1000)/log(10)) + 1 is NOT equal to (int) log10(1000) + 1 but it should be. I even tried the same thing in Java with this code (int) (Math.log(1000) / Math.log(10)) + 1 (int) Math.log10(1000) + 1

问题 can someone please tell me How to find antilog for a number using java program? i am new to this java Math.log(10) gives the log value. now I want to take this output and verify using antilog that program is giving right value.please help me. 回答1: mathematically: e^(ln x) = x in java: Math.pow(Math.E, (Math.log(x)) == x; //equals true 回答2: You can use Logarithm class in Java to calculate log and antilog. More on this is provided here. 来源： https://stackoverflow.com/questions/18209676/how-to

问题 I'm trying to calculate the time complexity of a simple algorithm in big O notation, but one part of it is seriously boggling my mind. Here is a simplified version of the algorithm: int a=n while(a>0) { //for loop with time complexity n^3 a = a/8 } In this instance, it's integer division, so the while loop will terminate after a's value drops below 8. I'm not sure how to express this in terms of n. I'd also like to know how to tackle future calculations like this, where the number of loops

问题 I'm trying to figure out how to round up a number (greater than 0) to the nearest power of 10. Examples: roundUp(23.4) = 100 roundUp(2.34) = 10 roundUp(.234) = 1 roundUp(0.0234) = 0.1 roundUp(0.00234) = 0.01 For numbers greater than 1, I believe this works: 10^(ceil(log10(x))) But for numbers between 0 and 1, I'm not sure how to arrive at the answer. 回答1: Oops. I didn't realize the function actually DOES work for numbers between 0 & 1. It was a brain fart that I saw a negative number for

问题 Let's look at a swarmplot, made with Python 3.5 and Seaborn on some data (which is stored in a pandas dataframe df with column lables stored in another class. This does not matter for now, just look at the plot): ax = sns.swarmplot(x=self.dte.label_temperature, y=self.dte.label_current, hue=self.dte.label_voltage, data = df) Now the data is more readable if plotted in log scale on the y-axis because it goes over some decades. So let's change the scaling to logarithmic: ax.set_yscale("log") ax