RuntimeWarning: divide by zero encountered in log是一个常见的Python警告,通常在使用NumPy库进行对数运算时出现。这个警告表示在对数运算中遇到了除以零的情况。 相关优势 NumPy是一个强大的科学计算库,提供了高效的数组操作和数学函数。对数运算在数据分析、信号处理、机器学习等领域有广泛应用。
log10(df.x, where=x>0) # no warning, usually incorrect result np.log10(df.x, out=np.zeros_like(x), where=x>0) # warning, correct result Issue Description Warning: Python\Python310\lib\site-packages\pandas\core\arraylike.py:397: RuntimeWarning: divide by zero encountered in log10 ...
在使用NumPy计算数组中元素的对数时,当数组中包含零元素时,会出现“divide by zero encountered in log”或“divide by zero encountered in log10”这样的警告。这是因为对数函数在零值处无定义,对数为负无穷(-inf)。 这些警告是由零值引起的,它们在对数运算中会导致无穷大的结果。这在NumPy中是一个正常的行为,...
import numpy as np x=np.log([2, np.e, np.e**3, 0]) x 1. 2. 3. 输出 __main__:1: RuntimeWarning: divide by zero encountered in log array([0.69314718, 1. , 3. , -inf]) 1. 2. 在上面的代码中 首先, 我们导入了别名为np的numpy。 我们已经声明了变量” x”并分配了np.log(...
d:\anaconda3\envs\sar\lib\site-packages\ipykernel_launcher.py:3: RuntimeWarning: divide by zero encountered in log10 This is separate from the ipykernel package so we can avoid doing imports until ## 统计#均值a=np.random.randint(0,10,(2,3))print(a)print(a.sum(),a.sum(0),a....
import numpy as np np.seterr(divide = 'ignore') seberg commentedon Sep 29, 2023 seberg seberg closed this ascompletedon Sep 29, 2023 Sign up for freeto join this conversation on GitHub.Already have an account?Sign in to comment
logaddexp2(x1, x2[, out]) Logarithm of the sum of exponentiations of the inputs in base-2. 算术运算 FunctionDescribe add(x1, x2[, out])加法 reciprocal(x[, out]) 倒数 negative(x[, out]) 负数 multiply(x1, x2[, out]) 乘法 divide(x1, x2[, out]) 除法 power(x1, x2[, out...
Natural logarithm (base e), log base 10, log base 2, and log(1 + x), respectively sign Compute the sign of each element: 1 (positive), 0 (zero), or –1 (negative) ceil 计算大于等于的最大整数值 Compute the ceiling of each element (i.e., the smallest integer greater than or equ...
(3) # 自然底数e的多少次幂 np.log(20.085536) # 自然底数e对数求解 np.sin(np.pi/2) # 90度sin正弦值 np.cos(0) # 余弦值 np.tan(np.pi/6) # 正切,30度正切值 # 给两个数组,从中选取大的,或者选取小的 np.maximum(a,c) # 从a和c中选取最大的值 np.minimum(a,c) # 选取最小的值 ...
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