什么是以numpy方式获取日志返回的最有效的方式 [英] What is the most efficient way to get log returns in numpy
问题描述
问题主要在于映射一个函数,该函数需要第i个和(和i + 1)'元素作为数组中每个元素的输入。
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:])] )
但是,我正在从单个numpy元素构建列表,然后将其转换为numpy数组再次。
我也以相当粗暴的方式访问元素,因为我对生成器函数或numpy内部没有经验。
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:])])
相当于
logR = np.diff(np.log(ar))
np.log
将每个值的日志记录在 ar
和 np.diff
取每个连续的值之间的差异。
What is the fastest and most elegant solution to building a sequence of log returns?
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.
for a function and simple array I can define the log returns as follows:
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.
I am also accessing the elements in a fairly brutish way as I have little experience with generator functions or numpy internals.
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
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.
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