absolute-value

问题 How do I make a program that asks for one floating point number in Python and then calculates the absolute value? I've already tried the asking, but I can't make the line where it calculates the absolute value. 回答1: You can do it with the built-in abs() function: absolute_val = abs(x) Otherwise, without the built-in function, use math: absolute_val = (x ** 2) ** 0.5 来源： https://stackoverflow.com/questions/32540121/calculating-absolute-value-in-python

问题 How do I make a program that asks for one floating point number in Python and then calculates the absolute value? I've already tried the asking, but I can't make the line where it calculates the absolute value. 回答1: You can do it with the built-in abs() function: absolute_val = abs(x) Otherwise, without the built-in function, use math: absolute_val = (x ** 2) ** 0.5 来源： https://stackoverflow.com/questions/32540121/calculating-absolute-value-in-python

问题 Is there any way to find the absolute value of a number without using the Math.abs() method in java. 回答1: If you look inside Math.abs you can probably find the best answer: Eg, for floats: /* * Returns the absolute value of a {@code float} value. * If the argument is not negative, the argument is returned. * If the argument is negative, the negation of the argument is returned. * Special cases: * <ul><li>If the argument is positive zero or negative zero, the * result is positive zero. * <li

问题 In C/C++, why should one use abs() or fabs() to find the absolute value of a variable without using the following code? int absoluteValue = value < 0 ? -value : value; Does it have something to do with fewer instructions at lower level? 回答1: The "conditional abs" you propose is not equivalent to std::abs (or fabs ) for floating point numbers, see e.g. #include <iostream> #include <cmath> int main () { double d = -0.0; double a = d < 0 ? -d : d; std::cout << d << ' ' << a << ' ' << std::abs(d)

问题 How do I get the absolute value of a number without using math.abs? This is what I have so far: function absVal(integer) { var abs = integer * integer; return abs^2; } 回答1: You can use the conditional operator and the unary negation operator: function absVal(integer) { return integer < 0 ? -integer : integer; } 回答2: You can also use >> (Sign-propagating right shift) function absVal(integer) { return (integer ^ (integer >> 31)) - (integer >> 31);; } Note: this will work only with integer 回答3:

问题 I use the following script for plotting: import matplotlib import matplotlib.pyplot as plt import numpy as np import pylab as pl import math import matplotlib as mpl from matplotlib.ticker import MultipleLocator from matplotlib.colors import LinearSegmentedColormap cdict1 = {'red': ((0.0, 1.0, 1.0), (0.4, 1.0, 1.0), (0.7, 0.0, 0.0), (1.0, 0.0, 0.0)), 'green': ((0.0, 1.0, 1.0), (0.1, 0.0, 0.0), (1.0, 0.0, 0.0)), 'blue': ((0.0, 1.0, 1.0), (0.1, 0.0, 0.0), (0.4, 0.0, 0.0), (1.0, 1.0, 1.0)) }

问题 I am aware of 3 methods, but as far as I know, only the first 2 are generally used: Mask off the sign bit using andps or andnotps . Pros: One fast instruction if the mask is already in a register, which makes it perfect for doing this many times in a loop. Cons: The mask may not be in a register or worse, not even in a cache, causing a very long memory fetch. Subtract the value from zero to negate, and then get the max of the original and negated. Pros: Fixed cost because nothing is needed to

问题 I ran some simple tests on abs() and fabs() functions and I don't understand what are the advantages of using fabs(), if it is: 1) slower 2) works only on floats 3) will throw an exception if used on a different type In [1]: %timeit abs(5) 10000000 loops, best of 3: 86.5 ns per loop In [3]: %timeit fabs(5) 10000000 loops, best of 3: 115 ns per loop In [4]: %timeit abs(-5) 10000000 loops, best of 3: 88.3 ns per loop In [5]: %timeit fabs(-5) 10000000 loops, best of 3: 114 ns per loop In [6]:

问题 In operators module we have a helper method operator.abs . But in python abs is already a function, and the only way I know to invoke the __abs__ method on an object is by function call anyway. Is there some other fancy way I don't know about to take the absolute value of a number? If not, why does operator.abs exist in the first place, and what is an example where you would have to use it instead of plain old abs ? 回答1: abs (...) abs(a) -- Same as abs(a). No, I don’t think there’s some other

问题 There's an array A containing (positive and negative) integers. Find a (contiguous) subarray whose elements' absolute sum is minimal, e.g.: A = [2, -4, 6, -3, 9] |(−4) + 6 + (−3)| = 1 <- minimal absolute sum I've started by implementing a brute-force algorithm which was O(N^2) or O(N^3) , though it produced correct results. But the task specifies: complexity: - expected worst-case time complexity is O(N*log(N)) - expected worst-case space complexity is O(N) After some searching I thought that