Numpy diff的反向操作是什么？

Question

使用numpy.diff函数工作，假设有以下简单案例：

使用 numpy.diff 函数工作，假设有以下简单案例：

>>> x = np.array([1, 2, 4, 7, 0])
>>> x_diff = np.diff(x)
array([ 1,  2,  3, -7])

如何轻松将x还原为原始比例而不是差分？ 我想这与 numpy.cumsum() 有关。

- Adelson Araújo

Answer 1

将第一个元素与其余元素连接起来，然后使用cumsum-

np.r_[x[0], x_diff].cumsum()

对于串联，我们也可以使用np.hstack，像这样 -

np.hstack((x[0], x_diff)).cumsum()

可以使用np.concatenate进行连接 -

np.concatenate(([x[0]], x_diff)).cumsum()

- Divakar

Answer 2

性能基准测试

鉴于Divakar提出了几种解决方案，我在想应该选择哪种方案，在此提供性能基准测试结果。还添加了这篇答案。

结果

简而言之 - 只需使用：np.concatenate(([x[0]], x_diff)).cumsum()。

x: 问题规模, y: 每1000次运行的计算时间

代码

import timeit
import random
import numpy as np
import matplotlib.pyplot as plt

cmds = [
    'np.r_[x[0], x_diff].cumsum()',
    'np.hstack((x[0], x_diff)).cumsum()',
    'np.concatenate(([x[0]], x_diff)).cumsum()',
    'csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])',
    ]
test_range = [1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6]
# test_range = [1e0, 1e1, 1e2]

ts = np.empty((len(cmds), len(test_range)), dtype=float)
for tt, size_float in enumerate(test_range):
    size = round(size_float)
    print('array size:', size)
    x = np.random.randint(low=0, high=100, size=size)
    x_diff = np.diff(x)

    n_trials = 1000
    for cc, cmd in enumerate(cmds):

        t = timeit.Timer(cmd, globals={**globals(), **locals()})
        t = t.timeit(n_trials)
        ts[cc, tt] = t
        print('time for {:d}x \"{:}\": {:.6f}'.format(n_trials, cmd, t))


fig, ax = plt.subplots(1, 1, figsize=(15, 10))
for cc, cmd in enumerate(cmds):
    ax.plot(test_range, ts[cc, :], label=cmd)
    print(cmd)
ax.legend()
ax.set_xscale('log')
ax.set_yscale('log')

输出

array size: 1
time for 1000x "np.r_[x[0], x_diff].cumsum()": 0.011935
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 0.006159
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 0.003221
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 0.003482
array size: 10
time for 1000x "np.r_[x[0], x_diff].cumsum()": 0.009031
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 0.006170
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 0.003082
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 0.003467
array size: 100
time for 1000x "np.r_[x[0], x_diff].cumsum()": 0.009754
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 0.006332
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 0.003296
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 0.004249
array size: 1000
time for 1000x "np.r_[x[0], x_diff].cumsum()": 0.010550
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 0.008595
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 0.005414
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 0.006916
array size: 10000
time for 1000x "np.r_[x[0], x_diff].cumsum()": 0.029658
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 0.028389
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 0.024410
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 0.034652
array size: 100000
time for 1000x "np.r_[x[0], x_diff].cumsum()": 0.221405
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 0.219564
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 0.215796
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 0.310225
array size: 1000000
time for 1000x "np.r_[x[0], x_diff].cumsum()": 2.660822
time for 1000x "np.hstack((x[0], x_diff)).cumsum()": 2.664244
time for 1000x "np.concatenate(([x[0]], x_diff)).cumsum()": 2.636382
time for 1000x "csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])": 3.770557
np.r_[x[0], x_diff].cumsum()
np.hstack((x[0], x_diff)).cumsum()
np.concatenate(([x[0]], x_diff)).cumsum()
csp0 = np.zeros(shape=(len(x) + 1,)); np.cumsum(x, out=csp0[1:])

- Markus Dutschke