Python 解决cvxpy优化问题时的自动终端输出

在尝试使用cvxpy解决逻辑回归问题时，我在调用solve函数时得到了一堆终端输出，即使没有编程打印输出。此外，即使将verbose设置为true且无法访问最佳值，也没有向终端打印有关问题的信息

我猜我在问题的表述上做错了什么，但我不太明白它是什么

在最小代码示例中，问题定义如下：

import numpy as np
import cvxpy as cp

y_vec = np.random.choice([0, 1], size=(728,), p=[9./10, 1./10])
M_mat = np.random.choice([0, 1], size=(728,801), p=[9./10, 1./10])
beta = cp.Variable(M_mat.shape[0])
objective = 0
for i in range(400):
    objective += y_vec[i] * M_mat[:, i].T @ beta - \
        cp.log(1 + cp.exp((M_mat[:, i].T @ beta)))

prob = cp.Problem(cp.Maximize(objective))
prob.solve(verbose=True)
print("Optimal var reached",  beta.value)

y_vec和M_mat都是数据类型为int64的numpy数组。两者都是仅由0和1组成的分类问题的选择矩阵。出于最小代码示例的目的，随机生成它们以再现错误。此外，还检查了M_mat[：，i].T@beta，以获得预期的标量

当我执行代码时，我会得到很多这样的打印输出，程序在某个数字后终止

这里显示的只是程序终止时打印输出的结束。但是有许多log1.0+exp[0.0…….0.]*var0形式的块，其中该输出序列的长度与变量beta相同

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我觉得这个结果很令人困惑。如何获得优化参数beta的单个向量？非常感谢您的帮助

经过一些尝试和错误，我发现使用cvxpy.logisticsfunction可以成功地计算出具有所需输出向量的解

这是通过如下重新制定目标函数来实现的：

objective = 0
for i in range(400):
    objective += y_vec[i] * M_mat[:, i].T @ beta - cp.logistic(M_mat[:, i].T @ beta)

---

尽管这两种实现在数学上应该是相同的，但结果却截然不同。我不知道为什么会这样。我希望这个解决方案可能对某些人有用，我很想知道，如果有人知道的更多，为什么行为会如此不同

经过一些尝试和错误，我发现使用cvxpy.logisticsfunction可以成功地计算出具有所需输出向量的解

这是通过如下重新制定目标函数来实现的：

objective = 0
for i in range(400):
    objective += y_vec[i] * M_mat[:, i].T @ beta - cp.logistic(M_mat[:, i].T @ beta)

---

尽管这两种实现在数学上应该是相同的，但结果却截然不同。我不知道为什么会这样。我希望这个解决方案可能对某些人有用，我很想知道，如果有人知道的更多，为什么行为会如此不同

您对问题的原始表述不正确。我承认产出远不理想；CVXPY正在打印不符合DCP的表达式，但是它们太多了，所以输出是无用的。将for循环中的400替换为1，您将得到以下输出，这更有用，因为它适合您的终端窗口

The objective is not DCP. Its following subexpressions are not:
log(1.0 + exp([0. 0. 1. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
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 0. 1. 0. 0. 1. 0. 1. 0. 0. 1. 1. 0. 1. 0. 0. 0. 1. 0. 0. 0. 1. 0. 1. 0.
 0. 0. 1. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
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 0. 0. 0. 0. 1. 1. 0. 0.] @ var6400))

---

你最初对这个问题的表述是不正确的。我承认产出远不理想；CVXPY正在打印不符合DCP的表达式，但是它们太多了，所以输出是无用的。将for循环中的400替换为1，您将得到以下输出，这更有用，因为它适合您的终端窗口

The objective is not DCP. Its following subexpressions are not:
log(1.0 + exp([0. 0. 1. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 1. 0. 0. 0. 0. 0. 1. 0. 1. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.
 1. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 1. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 1. 0.
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 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0.
 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 1. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 0.
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 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 1. 0. 0.
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 0. 0. 0. 1. 0. 0. 1. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0.
 0. 1. 0. 0. 1. 0. 1. 0. 0. 1. 1. 0. 1. 0. 0. 0. 1. 0. 0. 0. 1. 0. 1. 0.
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 0. 0. 0. 0. 1. 1. 0. 0.] @ var6400))

---

请提供一个可以运行的最小完整示例。感谢您的反馈。该问题现已调整。请提供一个可以运行的最小完整示例。感谢您的反馈。问题现在已经解决了，它告诉我明天只能接受我自己的答案。所以我明天会这么做：太好了。谢谢你的关注和发布，很多人都不介意。欢迎！我想问一下，您是否认为值得做一个后续问题，询问是否有人知道这两种不同行为存在的原因？即使我现在可以继续我的工作，我想这也不是cvxpy想要的方式。也许它甚至值得一个bug报告？正如原子函数文档中提到的，CVXPY知道逻辑函数的曲率是凸的。它不知道cp.log1+cp.expx是凸的，因为该表达式不符合DCP规则。它告诉我明天只能接受我自己的答案。所以我明天会这么做：太好了。谢谢你的关注和发布，很多人都不介意。欢迎！我想问一下，您是否认为值得做一个后续问题，询问是否有人知道这两种不同行为存在的原因？即使我现在可以继续我的工作，我想这也不是cvxpy想要的方式。也许它甚至值得一个bug报告？正如原子函数文档中提到的，CVXPY知道逻辑函数的曲率是凸的。它不知道cp.log1+cp.expx是凸的，因为该表达式不符合DCP规则。感谢您的回复。是的，那句话会有很大帮助的！谢谢你的回复。是的，那句话会有很大帮助的！