What is the fastest and most elegant solution to building a sequence of log returns?
构建一系列日志返回的最快和最优雅的解决方案是什么?
The problem is mainly around mapping a function that takes the i'th and (i+1)'th elements as inputs for every element in the array.
问题主要在于映射一个函数,该函数将第i个和第(i + 1)个元素作为数组中每个元素的输入。
for a function and simple array I can define the log returns as follows:
对于函数和简单数组,我可以定义日志返回如下:
import numpy as np
ar = np.random.rand(10)
f_logR = lambda ri, rf: np.log(rf) - np.log(ri)
logR = np.asarray([f_logR(ar[i], rf) for i,rf in enumerate(ar[1:])])
However, I am building a list from individual numpy elements and then converting it back into a numpy array again.
但是,我正在从单个numpy元素构建一个列表,然后再将它转换回numpy数组。
I am also accessing the elements in a fairly brutish way as I have little experience with generator functions or numpy internals.
我也是以相当野蛮的方式访问这些元素,因为我对生成器函数或numpy内部有很少的经验。
1 个解决方案
#1
f_logR = lambda ri, rf: np.log(rf) - np.log(ri)
logR = np.asarray([f_logR(ar[i], rf) for i,rf in enumerate(ar[1:])])
is equivalent to
相当于
logR = np.diff(np.log(ar))
np.log takes the log of every value in ar, and np.diff takes the difference between every consecutive pair of values.
np.log获取ar中每个值的日志,np.diff取每个连续值对之间的差值。
#1
f_logR = lambda ri, rf: np.log(rf) - np.log(ri)
logR = np.asarray([f_logR(ar[i], rf) for i,rf in enumerate(ar[1:])])
is equivalent to
相当于
logR = np.diff(np.log(ar))
np.log takes the log of every value in ar, and np.diff takes the difference between every consecutive pair of values.
np.log获取ar中每个值的日志,np.diff取每个连续值对之间的差值。