使用Keras基于图表预测Roc值时准确率为0的问题排查

我是神经网络领域新手，尝试用TensorFlow/Keras编写预测算法，基于提取的图表数据，通过Alt（高度）和Temp（温度）预测Roc值。此前模型准确率曾达到0.2至0.5，虽不理想但可优化，然而之后准确率骤降为0，无论调整参数都无法恢复。

附相关代码 #

# import tensorflow as tf
from tensorflow import keras
import numpy as np
import pandas as pd
import sklearn.model_selection

# Data collection
factor = 10
data = pd.read_csv("roc_6800_ibf.csv", sep=",")
data = data.apply(pd.to_numeric, errors='coerce')
data = (data / factor) + 5

predict = "Roc"

x = np.array(data.drop([predict], axis=1))
y = np.array(data[predict])

x_train, x_test, y_train, y_test = sklearn.model_selection.train_test_split(x, y, 
test_size=0.2)

x_shape = int(x.ndim)
y_shape = int(y.ndim)

# Model

model = keras.Sequential([
keras.layers.Dense(units=(2), input_shape=(2,), activation="relu"),
keras.layers.Dense(4, activation="relu"),
keras.layers.Dense(1, activation="relu")
])

model.compile(optimizer="adam", loss="MeanSquaredError", metrics=["accuracy"])

model.fit(x_train, y_train, epochs=20, batch_size=10, verbose=1)

results = model.evaluate(x_test, y_test)

print("- - - - - - - - - - - - - - - - - - - - - - - -")
print(results)

# Prediction

def dataPredict(inputvalues, outputvalues):
    print("- - - - - - - - - - - - - - - - - - - - - - - -")
    test_q = np.array([inputvalues])
    test_a = outputvalues
    prediction = model.predict((test_q / factor) + 5)

    print("Prediction " + str((prediction[0] - 5) * factor))
    print("Actual " + str(test_a[0]))
    print("Input " + str(test_q))


dataPredict([5.5,20.0],[3.6])
dataPredict([6.8,30.0],[0.4])

数据集（72条记录） #

Alt,Temp,Roc
-1.0,-40.0,9.6
0.0,-40.0,9.6
1.0,-40.0,9.6
2.0,-40.0,9.6
3.0,-40.0,9.6
4.0,-40.0,9.6
5.0,-40.0,9.6
6.0,-40.0,9.6
7.0,-40.0,8.1
8.0,-40.0,7.9
7.5,-40.0,9.1
-1.0,0.0,9.6
0.0,0.0,9.6
1.0,0.0,9.6
2.0,0.0,9.6
2.1,0.0,9.6
3.0,0.0,9.0
4.0,0.0,8.0
5.0,0.0,6.6
6.0,0.0,5.5
7.0,0.0,4.2
8.0,0.0,3.2
-1.0,20.0,9.6
0.0,20.0,9.6
0.5,20.0,9.0
1.0,20.0,8.6
2.0,20.0,7.8
3.0,20.0,6.2
4.0,20.0,5.2
5.0,20.0,4.0
6.0,20.0,2.9
7.0,20.0,1.8
8.0,20.0,0.5
-1.0,40.0,7.5
0.0,40.0,6.8
1.0,40.0,5.6
2.0,40.0,4.2
3.0,40.0,3.2
4.0,40.0,2.2
5.0,40.0,1.0
-1.0,50.0,5.4
0.0,50.0,4.2
-0.5,-40.0,9.5
0.5,-40.0,9.5
1.5,-40.0,9.5
2.5,-40.0,9.5
3.5,-40.0,9.5
4.5,-40.0,9.5
5.5,-40.0,9.5
6.5,-40.0,9.1
7.5,-40.0,8.1
-0.5,-10.0,9.5
0.5,-10.0,9.5
1.5,-10.0,9.5
2.5,-10.0,9.5
3.5,-10.0,9.5
4.5,-10.0,8.3
5.5,-10.0,7.1
6.5,-10.0,6.0
7.5,-10.0,5.0
-0.5,30.0,8.4
0.5,30.0,7.6
1.5,30.0,6.4
2.5,30.0,5.5
3.5,30.0,4.2
4.5,30.0,3.1
5.5,30.0,1.9
6.5,30.0,0.8
7.5,30.0,-0.5
5.2,10.0,5.3
6.8,10.0,4.0

异常训练日志 #

Epoch 20/20
6/6 [==============================] - 0s 2ms/step - loss: 32.5049 - accuracy: 0.0000e+00

1. 回归任务误用分类指标accuracy #

你做的是回归任务（预测连续型Roc值），但用了分类任务专属的accuracy指标。accuracy统计的是预测类别与真实类别完全匹配的比例，回归任务中预测值和真实值几乎不可能完全相等，因此准确率直接变为0。之前那0.2-0.5的数值也是无意义的巧合。

2. 输出层激活函数错误 #

输出层使用relu激活函数，但你的Roc值存在负数（如7.5,30.0,-0.5），relu会将所有负数截断为0，导致模型无法学习负向Roc的规律，最终训练崩溃。

3. 数据预处理逻辑冗余 #

自定义的(data / factor) + 5转换没必要，反而打乱了数据原始分布，增加模型学习难度。用标准化/归一化这类通用预处理方法更稳妥。

from tensorflow import keras
import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler

# 数据读取与预处理
data = pd.read_csv("roc_6800_ibf.csv", sep=",")
data = data.apply(pd.to_numeric, errors='coerce').dropna()  # 清理无效数据

predict = "Roc"
x = data.drop([predict], axis=1).values
y = data[predict].values

# 数据标准化（适配神经网络输入）
scaler_x = StandardScaler()
scaler_y = StandardScaler()
x_scaled = scaler_x.fit_transform(x)
y_scaled = scaler_y.fit_transform(y.reshape(-1, 1))

# 划分训练测试集
x_train, x_test, y_train, y_test = train_test_split(x_scaled, y_scaled, test_size=0.2, random_state=42)

# 构建回归模型
model = keras.Sequential([
    keras.layers.Dense(8, input_shape=(2,), activation="relu"),
    keras.layers.Dense(4, activation="relu"),
    keras.layers.Dense(1)  # 回归任务输出层无需激活函数
])

# 回归任务用MSE损失，监控MAE（平均绝对误差）指标
model.compile(optimizer="adam", loss="MeanSquaredError", metrics=["MeanAbsoluteError"])

# 训练模型（加入验证集监控过拟合）
history = model.fit(x_train, y_train, epochs=50, batch_size=8, verbose=1, validation_split=0.1)

# 评估模型
results = model.evaluate(x_test, y_test)
print("-"*50)
print(f"测试集损失: {results[0]:.4f}, 平均绝对误差: {results[1]:.4f}")

# 预测函数（适配标准化转换）
def dataPredict(inputvalues, outputvalues):
    print("-"*50)
    test_q_scaled = scaler_x.transform(np.array([inputvalues]))
    prediction_scaled = model.predict(test_q_scaled, verbose=0)
    prediction = scaler_y.inverse_transform(prediction_scaled)[0][0]
    
    print(f"预测值: {prediction:.2f}")
    print(f"真实值: {outputvalues[0]}")
    print(f"输入值: {inputvalues}")

dataPredict([5.5,20.0],[3.6])
dataPredict([6.8,30.0],[0.4])

• 数据集仅72条，数据量极小，易过拟合，可添加keras.layers.Dropout(0.2)层抑制过拟合，或使用更简单的模型结构。
• 训练时通过validation_split监控验证集损失，避免模型在训练集上过度拟合。
• 回归任务不要关注accuracy，重点看MeanSquaredError（均方误差）或MeanAbsoluteError（平均绝对误差），数值越小说明预测越准确。

内容的提问来源于stack exchange，提问作者Bengt B