使用 `np.diff` 但假设输入以额外的零开头 [英] Use `np.diff` but assume the input starts with an extra zero

问题描述

给定一系列事件时间 v，我可以使用 np.diff(v) 创建它们的间隔持续时间.有没有办法让 np.diff 假设系列以隐式 0. 开头，以便它生成一个与 v<具有相同长度的数组/代码>?

手动解决方法是:

def diff_from_zero(v):返回 np.diff(np.hstack(([0.], v)))

有没有办法使用 diff 或其他函数来获得相同的结果?

解决方案

举个例子:

t = np.array([1.1, 2.0, 4.5, 4.9, 5.2])

我们要计算 t 中的连续差异，包括从 0. 到 t 中第一个元素的差异.

问题给出了实现此目的的方式:

<预><代码>>>>np.diff(np.hstack((0, t)))

也可能是这样:

<预><代码>>>>np.hstack((t[0], np.diff(t)))

但是名字模糊的函数ediff1d 可以在一个函数调用中完成:

<预><代码>>>>np.ediff1d(t, to_begin=t[0])数组([ 1.1, 0.9, 2.5, 0.4, 0.3])

将 t[0] 添加到结果中与计算 t[0] - 0 的差值相同.，当然.(假设 t 为非空).

<小时>

时间(不是问题的动机，但我很好奇)

将 numpy 导入为 npt = np.random.randn(10000)%timeit np.diff(np.concatenate(([0], t)))10000 个循环，最好的 3 个:每个循环 23.1 µs%timeit np.diff(np.hstack((0, t)))10000 个循环，最好的 3 个:每个循环 31.2 µs%timeit np.ediff1d(t, to_begin=t[0])10000 个循环，最好的 3 个:每个循环 92 µs

Given a series of event times v, I can create their interval durations using np.diff(v). Is there a way to have np.diff assume the series starts with an implicit 0., so that it produces an array that has the same length as v?

A manual workaround is:

def diff_from_zero(v):
    return np.diff(np.hstack(([0.], v)))

Is there a way to use diff or another function to have the same result?

解决方案

Given for example:

t = np.array([1.1, 2.0, 4.5, 4.9, 5.2])

We want to compute the consecutive differences in t, including the diff from 0. to the first element in t.

The question gave this way of accomplishing this:

>>> np.diff(np.hstack((0, t)))

And it could be this too:

>>> np.hstack((t[0], np.diff(t)))

But the obscurely-named function ediff1d can do it in one function call:

>>> np.ediff1d(t, to_begin=t[0])
array([ 1.1,  0.9,  2.5,  0.4,  0.3])

Prepending t[0] to the result is the same as computing the difference t[0] - 0., of course. (Assuming t is nonempty).

---

Timings (not the motivation of the question, but I was curious)

import numpy as np
t = np.random.randn(10000)
%timeit np.diff(np.concatenate(([0], t)))
10000 loops, best of 3: 23.1 µs per loop
%timeit np.diff(np.hstack((0, t)))
10000 loops, best of 3: 31.2 µs per loop
%timeit np.ediff1d(t, to_begin=t[0])
10000 loops, best of 3: 92 µs per loop

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