如何替换numpy数组中的值列表？

Question

我有一个未排序的数字数组。

我需要用特定的替代品（也在相应的列表中给出）替换某些数字（在列表中给出）

我编写了以下代码（似乎有效）：

import numpy as np

numbers = np.arange(0,40)
np.random.shuffle(numbers)
problem_numbers = [33, 23, 15]  # table, night_stand, plant
alternative_numbers = [12, 14, 26]  # desk, dresser, flower_pot

for i in range(len(problem_numbers)):
    idx = numbers == problem_numbers[i]
    numbers[idx] = alternative_numbers[i]

但是，这似乎非常低效（对于更大的数组，这需要完成数百万次）。

我发现this 问题回答了类似的问题，但在我的情况下，数字没有排序，它们需要保持原来的位置。

注意：numbers 可能在problem_numbers 中包含多个或不包含多个元素

Answer 1

编辑：我在 this answer 中实现了一个 TensorFlow 版本（几乎完全相同，除了替换是一个字典）。

---

这是一个简单的方法：

import numpy as np

numbers = np.arange(0,40)
np.random.shuffle(numbers)
problem_numbers = [33, 23, 15]  # table, night_stand, plant
alternative_numbers = [12, 14, 26]  # desk, dresser, flower_pot

# Replace values
problem_numbers = np.asarray(problem_numbers)
alternative_numbers = np.asarray(alternative_numbers)
n_min, n_max = numbers.min(), numbers.max()
replacer = np.arange(n_min, n_max + 1)
# Mask replacements out of range
mask = (problem_numbers >= n_min) & (problem_numbers <= n_max)
replacer[problem_numbers[mask] - n_min] = alternative_numbers[mask]
numbers = replacer[numbers - n_min]

只要numbers 中的值范围（最小和最大之间的差异）不是很大（例如，您没有类似1、@987654325 @ 和 10000000000)。

基准测试

我已经将 OP 中的代码与三个（截至目前）建议的解决方案进行了比较：

import numpy as np

def method_itzik(numbers, problem_numbers, alternative_numbers):
    numbers = np.asarray(numbers)
    for i in range(len(problem_numbers)):
        idx = numbers == problem_numbers[i]
        numbers[idx] = alternative_numbers[i]
    return numbers

def method_mseifert(numbers, problem_numbers, alternative_numbers):
    numbers = np.asarray(numbers)
    replacer = dict(zip(problem_numbers, alternative_numbers))
    numbers_list = numbers.tolist()
    numbers = np.array(list(map(replacer.get, numbers_list, numbers_list)))
    return numbers

def method_divakar(numbers, problem_numbers, alternative_numbers):
    numbers = np.asarray(numbers)
    problem_numbers = np.asarray(problem_numbers)
    problem_numbers = np.asarray(alternative_numbers)
    # Pre-process problem_numbers and correspondingly alternative_numbers
    # such that repeats and no matches are taken care of
    sidx_pn = problem_numbers.argsort()
    pn = problem_numbers[sidx_pn]
    mask = np.concatenate(([True],pn[1:] != pn[:-1]))
    an = alternative_numbers[sidx_pn]

    minN, maxN = numbers.min(), numbers.max()
    mask &= (pn >= minN) & (pn <= maxN)

    pn = pn[mask]
    an = an[mask]

    # Pre-pocessing done. Now, we need to use pn and an in place of
    # problem_numbers and alternative_numbers repectively. Map, index and assign.
    sidx = numbers.argsort()
    idx = sidx[np.searchsorted(numbers, pn, sorter=sidx)]
    valid_mask = numbers[idx] == pn
    numbers[idx[valid_mask]] = an[valid_mask]

def method_jdehesa(numbers, problem_numbers, alternative_numbers):
    numbers = np.asarray(numbers)
    problem_numbers = np.asarray(problem_numbers)
    alternative_numbers = np.asarray(alternative_numbers)
    n_min, n_max = numbers.min(), numbers.max()
    replacer = np.arange(n_min, n_max + 1)
    # Mask replacements out of range
    mask = (problem_numbers >= n_min) & (problem_numbers <= n_max)
    replacer[problem_numbers[mask] - n_min] = alternative_numbers[mask]
    numbers = replacer[numbers - n_min]
    return numbers

结果：

import numpy as np

np.random.seed(100)

MAX_NUM = 100000
numbers = np.random.randint(0, MAX_NUM, size=100000)
problem_numbers = np.unique(np.random.randint(0, MAX_NUM, size=500))
alternative_numbers = np.random.randint(0, MAX_NUM, size=len(problem_numbers))

%timeit method_itzik(numbers, problem_numbers, alternative_numbers)
10 loops, best of 3: 63.3 ms per loop

# This method expects lists
problem_numbers_l = list(problem_numbers)
alternative_numbers_l = list(alternative_numbers)
%timeit method_mseifert(numbers, problem_numbers_l, alternative_numbers_l)
10 loops, best of 3: 20.5 ms per loop

%timeit method_divakar(numbers, problem_numbers, alternative_numbers)
100 loops, best of 3: 9.45 ms per loop

%timeit method_jdehesa(numbers, problem_numbers, alternative_numbers)
1000 loops, best of 3: 822 µs per loop

Answer 2

如果不是所有problem_values 都在numbers 中，甚至可能出现多次：

在这种情况下，我将只使用 dict 来保留要替换的值并使用 dict.get 来翻译有问题的数字：

replacer = dict(zip(problem_numbers, alternative_numbers))
numbers_list = numbers.tolist()
numbers = np.array(list(map(replacer.get, numbers_list, numbers_list)))

即使它必须“通过 Python”，这几乎是不言自明的，它并不比 NumPy 解决方案慢多少（可能）。

如果每个problem_value 都实际存在于numbers 数组中并且仅一次：

如果你有numpy_indexed 包，你可以简单地使用numpy_indexed.indices：

>>> import numpy_indexed as ni
>>> numbers[ni.indices(numbers, problem_numbers)] = alternative_numbers

即使对于大型数组，这也应该非常有效。