英文:
Finding elements of a list in an array in Python (Numpy, Pytorch)
问题
我正在处理以下问题：我有一个列表，假设 a=[1, 2, 3] 和一个数组 b=[[2, 4, 6],[3, 2, 5],[4, 1, 3]]，列表中的元素数量等于数组中的行数，输出应该是一个列表 c=[-1, 1, 2]，其中 -1 表示数字 a[0] 没有在数组 b 的第0行中找到，其他数字告诉了元素 a[i] 在行 b[i, :] 中的索引，索引从0开始。
我使用循环解决了这个问题，但我正在寻找一种更优化的方法，可以使用 np.where 或 torch 的替代方法来解决，我会感谢任何帮助。
英文:
I am dealing with the following problem: I have a list, let's say a=[1, 2, 3] and an array b=[[2, 4, 6],[3, 2, 5],[4, 1, 3]], the number of elements in the list is equal to the number of rows in the array, the output should be a list c=[-1, 1, 2], where -1 indicates that the number a[0] was not found in the 0-th row of the array b, the other numbers tells the index where the element a[i] is in the row b[i, :], the index enumeration starts at 0.
I solved it using loops but I am searching for a more optimal way to do it using np.where or the torch alternative, I would appreciate any help.
答案1
得分: 2
import numpy as npa = [1, 2, 3]b = np.array([[2, 4, 6], [3, 2, 5], [4, 1, 3]])# 通过将每个元素与a的对应元素进行比较来创建布尔掩码mask = b == np.array(a)[:, np.newaxis]# 找到掩码每行中第一个True的索引indices = np.argmax(mask, axis=1)# 使用np.where条件地分配匹配发生的索引，未匹配的地方分配-1c = np.where(np.any(mask, axis=1), indices, -1).tolist()print(c)
英文:
Try this
import numpy as npa = [1, 2, 3]b = np.array([[2, 4, 6], [3, 2, 5], [4, 1, 3]])# Create a boolean mask by comparing each element of b with the corresponding element of amask = b == np.array(a)[:, np.newaxis]# Find the index of the first occurrence of True along each row of the maskindices = np.argmax(mask, axis=1)# Use np.where to conditionally assign the indices where matches were found and -1 where no matches occurredc = np.where(np.any(mask, axis=1), indices, -1).tolist()print(c)
Or
import numpy as npa = [1, 2, 3]b = np.array([[2, 4, 6], [3, 2, 5], [4, 1, 3]])c = []for i in range(len(a)):# Find the indices where the i-th element of a matches the elements in the i-th row of bindices = np.where(b[i, :] == a[i])[0]if len(indices) > 0:# If indices are found, append the first index to the list cc.append(indices[0])else:# If no indices are found, append -1 to the list cc.append(-1)print(c)
Output
[-1, 1, 2]
答案2
得分: 0
假设每行最多只有一个匹配项，你可以使用以下代码：
out = np.full_like(a, -1)idx, vals = np.where(b == a[:,None])out[idx] = vals
输出结果：array([-1, 1, 2])
可重现的输入数据：
a = np.array([1, 2, 3])b = np.array([[2, 4, 6],[3, 2, 5],[4, 1, 3]])
英文:
Assuming at most one match per row, you could use:
out = np.full_like(a, -1)idx, vals = np.where(b == a[:,None])out[idx] = vals
Output: array([-1, 1, 2])
Reproducible input:
a = np.array([1, 2, 3])b = np.array([[2, 4, 6],[3, 2, 5],[4, 1, 3]])
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