我如何在python 2.7中使用“e”（欧拉数）和幂运算
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How can i use "e" (Euler's number) and power operation in python 2.7
提问by Thanos Smar
How can i write x.append(1-e^(-value1^2/2*value2^2))in python 2.7?
我如何x.append(1-e^(-value1^2/2*value2^2))在 python 2.7 中编写？
I don't know how to use power operator and e.
我不知道如何使用电源运算符和 e。
回答by Moinuddin Quadri
Refer mathlibrary of python. exp(x)function this library is same as e^x. Hence you may write your code as:
参考python的数学库。exp(x)此库的功能与e^x. 因此，您可以将代码编写为：
I have modified the equation by replacing 1/2as 0.5. Else for Python <2.7we have to explicitly type cast the division value to floatbecause Python round of the result of division of two int. (for example: 1/2 -> 0in python 2.7 and below)
我通过替换1/2as修改了方程0.5。否则对于 Python <2.7，我们必须明确地将除法值转换为类型，float因为 Python 是对二除法结果的舍入int。（例如：1/2 -> 0在 python 2.7 及以下版本中）
import math
x.append(1 - math.exp( -0.5 * (value1*value2)**2))
回答by karlosss
Python's power operator is **and Euler's number is math.e, so:
Python 的幂运算符是**，欧拉数是math.e，所以：
from math import e
x.append(1-e**(-value1**2/2*value2**2))
回答by Daniel
Power is **and e^is math.exp:
功率是**和e^是math.exp：
x.append(1 - math.exp(-0.5 * (value1*value2)**2))
回答by jorijnsmit
Just saying: numpyhas this too. So no need to import mathif you already did import numpy as np:
只想说：numpy这个也有。因此，math如果您已经这样做了，则无需导入import numpy as np：
>>> np.exp(1)
2.718281828459045
回答by Alexey K.
math.e or from math import e (2.718281…)
math.e 或 from math import e (2.718281…)
math.exp(x) = e**x
however:
Return e raised to the power x, where e = 2.718281… is the base of natural logarithms. This is usually more accurate than math.e ** x or pow(math.e, x). docs.python
math.exp(x) = e**x
然而：
返回 e 的 x 次幂，其中 e = 2.718281…是自然对数的底。这通常比 math.e ** x 或 pow(math.e, x) 更准确。文档.python
for power use " ** " (3**2 = 9), not " ^ "
" ^ " is a bitwise XOR operator (& and, | or), it works logicaly with bits 10^4=14 (0000 1010 ^ 0000 0100 = 0000 1110) programiz
对于电源使用“**”（3**2 = 9），而不是“^”
“^”是按位异或运算符（&和，|或），它在逻辑上与位10^4=14（0000 1010 ^ 0000 0100 = 0000 1110) 程序化
