在python中，根据列表重新排列数组元素

Question

我有一个大小为2, 1403的2D数组a和一个有2个列表的列表b。

a.shape = (2, 1403) # a is 2D array, each row has got unique elements.

len(b) = 2 # b is list

len(b[0]), len(b[1]) = 415, 452 # here also both the list inside b has got unique elements

b[0] and b[1]中存在的所有元素分别存在于a[0] and a[1]中

现在我想在b的元素的基础上重新排列a的元素。我想重新排列，使得b[0]中的所有元素(也存在于a[0]中(都应该出现在a[0]的末尾，这意味着新的a应该是a[0][:-len(b[0])] = b[0]，类似于a[1][:-len(b[1])] = b[1]。

玩具示例

a具有类似[[1,2,3,4,5,6,7,8,9,10,11,12],[1,2,3,4,5,6,7,8,9,10,11,12]的元素

b具有类似[[5, 9, 10], [2, 6, 8, 9, 11]]的元素

new_a变为[[1,2,3,4,6,7,8,11,12,5,9,10], [1,3,4,5,7,10,12,2,6,8,9,11]]

我写了一个代码，它在所有元素上循环，变得非常慢，如下面所示

a_temp = []
remove_temp = []
for i, array in enumerate(a):
a_temp_inner = []
remove_temp_inner = []
for element in array:
if element not in b[i]:
a_temp_inner.append(element) # get all elements first which are not present in b
else:
remove_temp_inner.append(element) #if any element present in b, remove it from main array
a_temp.append(a_temp_inner)
remove_temp.append(b_temp_inner)
a_temp = torch.tensor(a_temp)
remove_temp = torch.tensor(remove_temp)
a = torch.cat((a_temp, remove_temp), dim = 1) 

有人能帮我做一些比这个更好的更快的实现吗

Answer 1

假设a是np.array，b为list，则可以使用

np.array([np.concatenate((i[~np.in1d(i, j)], j)) for i, j in zip(a,b)])

输出

array([[ 1,  2,  3,  4,  6,  7,  8, 11, 12,  5,  9, 10],
[ 1,  3,  4,  5,  7, 10, 12,  2,  6,  8,  9, 11]])

如果b包含空lists，则可以进行微优化

np.array([np.concatenate((i[~np.in1d(i, j)], j)) if j else i for i, j in zip(a,b)])

在我的基准测试中，对于少于~100个元素的np.arrays，转换.tolist()比np.concatenate更快

np.array([i[~np.in1d(i, j)].tolist() + j for i, j in zip(a,b)])

---

此解决方案的数据示例和导入

import numpy as np
a = np.array([
[1,2,3,4,5,6,7,8,9,10,11,12],
[1,2,3,4,5,6,7,8,9,10,11,12]
])
b = [[5, 9, 10],
[2, 6, 8, 9, 11]]

Answer 2

这是我的方法：

index_ = np.array([[False if i in d else True for i in c] for c, d in zip(a,b)])
arr_filtered =[[np.extract(ind, c) for c, d, ind in zip(a,b,index_)], [np.extract(np.logical_not(ind), c) for c, d, ind in zip(a,b, index_)]]
arr_final = ar = np.array([np.concatenate((i, j)) for i, j in zip(*arr_filtered)])