如何利用np.argwhere生成的位置数组批量修改目标数组值（避免循环）

Hey there! Congrats on hitting your first month with Python—numpy’s vectorized operations can feel a bit counterintuitive at first, but you’re already asking the right questions by trying to avoid slow for loops. Let’s fix this for you!

Your goal is to modify elements in c that line up with positions where a has values greater than 3. The key here is to use numpy’s built-in indexing capabilities, which are optimized for exactly this kind of task (no loops needed!).

First, note that your c is a Python list of lists—we’ll need to convert it to a numpy array first, since native lists don’t support numpy’s advanced indexing. Here’s how to do it properly:

import numpy as np

# Your original arrays
a = np.arange(6).reshape(2,3)
b = np.argwhere(a > 3)
# Convert c to a numpy array (critical for vectorized indexing)
c = np.array([[7,8,9],[10,11,12]])

Now you have two straightforward options to update c:

Option 1: Use a boolean mask (most concise) #

Instead of using np.argwhere, you can directly use the boolean array generated by a > 3 to index into c. This is the most efficient and readable approach:

# Assign new values to the matching positions
c[a > 3] = [99, 100]  # Replace with your desired values

Option 2: Use the argwhere result (if you need to work with b explicitly) #

If you already have b from np.argwhere, you can use its rows as indices. Since b is a 2D array where each row is [row, column], we slice it to get separate row and column indices:

c[b[:, 0], b[:, 1]] = [99, 100]

Either way, running this will give you the updated c:

[[ 7  8  9]
 [10 99 100]]

Both methods use numpy’s vectorized operations, which are executed in optimized C code under the hood. This will be drastically faster than a Python for loop, especially with large datasets—exactly what you’re looking for!

内容的提问来源于stack exchange，提问作者Vitor Pereira Vaz