Python Numpy diff倒置手术？

使用函数时，假设以下简单情况：

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

---

我怎样才能轻松地将x恢复到原始比例而不差分？我想其中有一个元素。

与第一个元素连接，然后使用-

对于串联，我们也可以使用

np.hstack

，如下所示-

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

或者使用

np.concatenate

进行连接-

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

绩效基准 Divakar提出了一些解决方案，我想知道，我应该采取什么措施，这里是性能基准。我还补充了答案

结果 长话短说-只需使用：

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

x：问题大小，y：1000次运行的计算时间

代码 输出

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

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')

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:])