我如何计算一个数组y （比方说）相对于另一个数组 x （比方说）的导数 - 来自某个实验的两个数组？
例如
y = [1,2,3,4,4,5,6]和x = [.1,.2,.5,.6,.7,.8,.9];
我想得到dy/dx！
问问题
94019 次
4 回答
43
Use numpy.diff
If dx is constant
from numpy import diff
dx = 0.1
y = [1, 2, 3, 4, 4, 5, 6]
dy = diff(y)/dx
print dy
array([ 10., 10., 10., 0., 10., 10.])
dx is not constant (your example)
from numpy import diff
x = [.1, .2, .5, .6, .7, .8, .9]
y = [1, 2, 3, 4, 4, 5, 6]
dydx = diff(y)/diff(x)
print dydx
[10., 3.33333, 10. , 0. , 10. , 10.]
Note that this approximated "derivative" has size n-1 where n is your array/list size.
Don't know what you are trying to achieve but here are some ideas:
- If you are trying to make numerical differentiation maybe finite differences formulation might help you better.
- The solution above is like a first-order accuracy approximation for the forward schema of finite differences with a non-uniform grid/array.
于 2013-10-18T21:14:20.810 回答
8
利用numpy.gradient()
请注意，有比简单地使用更高级的方法来计算数值导数diff。我建议使用numpy.gradient，就像在这个例子中一样。
import numpy as np
from matplotlib import pyplot as plt
# we sample a sin(x) function
dx = np.pi/10
x = np.arange(0,2*np.pi,np.pi/10)
# we calculate the derivative, with np.gradient
plt.plot(x,np.gradient(np.sin(x), dx), '-*', label='approx')
# we compare it with the exact first derivative, i.e. cos(x)
plt.plot(x,np.cos(x), label='exact')
plt.legend()
于 2021-02-16T10:57:56.883 回答
5
我假设这就是你的意思：
>>> from __future__ import division
>>> x = [.1,.2,.5,.6,.7,.8,.9]
>>> y = [1,2,3,4,4,5,6]
>>> from itertools import izip
>>> def pairwise(iterable): # question 5389507
... "s -> (s0,s1), (s2,s3), (s4, s5), ..."
... a = iter(iterable)
... return izip(a, a)
...
>>> for ((a, b), (c, d)) in zip(pairwise(x), pairwise(y)):
... print (d - c) / (b - a)
...
10.0
10.0
10.0
>>>
也就是说，定义dx为 中相邻元素之间的差异x。
于 2013-05-30T17:01:48.970 回答
4
x中相邻元素之间的差异
就像@tsm 的回答一样。结果，您得到一个比原始数组短 1 个元素的数组。这当然是有道理的，因为您只能开始计算与第一个索引的差异（需要 1 个“历史元素”）。
>>> x = [1,3,4,6,7,8]
>>> dx = numpy.diff(x)
>>> dx
array([2, 1, 2, 1, 1])
>>> y = [1,2,4,2,3,1]
>>> dy = numpy.diff(y)
>>> dy
array([ 1, 2, -2, 1, -2])
现在您可以划分这 2 个结果数组以获得所需的导数。
>>> d = dy / dx
>>> d
array([ 0.5, 2. , -1. , 1. , -2. ])
如果由于某种原因，您需要相对（相对于 y 值）增长，您可以通过以下方式进行：
>>> d / y[:-1]
array([ 0.5 , 1. , -0.25 , 0.5 , -0.66666667])
解释为 50% 增长、100% 增长、-25% 增长等。
完整代码：
import numpy
x = [1,3,4,6,7,8]
y = [1,2,4,2,3,1]
dx = numpy.diff(x)
dy = numpy.diff(y)
d = dy/dx
于 2018-01-25T19:40:33.783 回答