如何一次建立一列矩阵 [英] How to build a matrix one column at a time

问题描述

我想根据下面的代码一次建立一列矩阵，并使用现有矩阵中的列.

I am wanting to build some matrices one column at a time, with columns from an existing matrix, as per the code below.

# x is an existing matrix, y is an array, pos_classes is a set

for i in range(len(y)):
    if y[i] in pos_classes:
        x_pos = np.append(x_pos, x[i])
        y_pos_actual = np.append(y_pos_actual, y[i])
    else:
        x_neg =  np.append(x_neg, x[i])
        y_neg_actual = np.append(y_neg_actual, y[i])

我的问题是，应如何初始化x_pos和x_neg，以便每个x[i]都按列追加?我尝试如下，并且每个追加将x_pos变成一维数组.

My question is, what should I initialise x_pos and x_neg as so that each x[i] gets appended column-wise? I tried as below, and each append turns x_pos into a 1D array.

x_pos = np.empty((0,x.shape[1]))

我对python还是很陌生，可能缺少明显的东西.

I'm fairly new to python and am probably missing something obvious.

推荐答案

您正在以错误的方式进行操作.在numpy中以矢量化的方式执行此类操作几乎总是更好.

You're going about this the wrong way. It's almost always better to do such things in a vectorized way in numpy.

首先，构建索引数组y_pos_actual.然后，只需

First, build your index array y_pos_actual. Then, simply do

x_pos = x[:, y_pos_actual]

这会将索引y_pos_actual给定的所有列都选择到新的矩阵x_pos中.您可以使用

This will select all columns given by the indices y_pos_actual into a new matrix x_pos. You can do the same row-wise using

x_pos = x[y_pos_actual, :]

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