如何从单列矩阵A提取元素生成新矩阵B？

Hey there! Let's figure out how to create your matrix B from A. The approach depends a bit on which programming language you're using—here are the most common solutions for tools people use for matrix work:

NumPy is the go-to library for matrix operations in Python, and it uses 0-based indexing (meaning the first element is index 0 instead of 1). We'll need to adjust your 1-based positions to match:

• Your requested A[2-6] (1-based) translates to A[1:6] in NumPy (since the end index is exclusive)
• A[11-16] (1-based) translates to A[10:16]

Here's the code to put it all together:

import numpy as np

# Example: Create a sample column matrix A with 16 elements (1 to 16)
A = np.arange(1, 17).reshape(-1, 1)

# Extract the desired elements and combine them into B
selected = np.concatenate([A[1:6], A[10:16]])
B = selected

This gives you a 10x1 column matrix B where B's first 5 elements are from A[2-6], and the last 5 are from A[11-16]—exactly filling B[1-10] as you wanted.

MATLAB and Octave use 1-based indexing, which aligns perfectly with your request. No need to adjust positions here:

% Example: Create a sample column matrix A with 16 elements
A = (1:16)';

% Combine the two ranges and assign to B
B = [A(2:6); A(11:16)];

The ; stacks the two subsets vertically, resulting in a 10x1 column matrix B. B(1-5) will hold A(2-6), and B(6-10) holds A(11-16).

R also uses 1-based indexing, making this straightforward. For a column matrix in R:

# Example: Create a sample column matrix A
A <- matrix(1:16, ncol = 1)

# Extract and stack the desired rows into B
B <- rbind(A[2:6, ], A[11:16, ])

rbind() combines the two row subsets into a new 10x1 matrix B, matching your requirements.

If you're working with a different language or tool, feel free to share more details and I can tweak the solution!

内容的提问来源于stack exchange，提问作者Sander