Python 在Keras中,如何通过预测方法获得与评价方法相同的精度值?
首先,我遵循中编码的示例 并制作以下代码:Python 在Keras中,如何通过预测方法获得与评价方法相同的精度值?,python,tensorflow,machine-learning,keras,tf.keras,Python,Tensorflow,Machine Learning,Keras,Tf.keras,首先,我遵循中编码的示例 并制作以下代码: import numpy as np import pandas as pd import tensorflow as tf from tensorflow import feature_column from tensorflow.keras import layers from sklearn.model_selection import train_test_split URL = 'https://storage.googleapis.co
import numpy as np
import pandas as pd
import tensorflow as tf
from tensorflow import feature_column
from tensorflow.keras import layers
from sklearn.model_selection import train_test_split
URL = 'https://storage.googleapis.com/applied-dl/heart.csv'
dataframe = pd.read_csv(URL)
dataframe.head()
train, test = train_test_split(dataframe, test_size=0.2)
train, val = train_test_split(train, test_size=0.2)
def df_to_dataset(dataframe, shuffle=True, batch_size=32):
dataframe = dataframe.copy()
labels = dataframe.pop('target')
ds = tf.data.Dataset.from_tensor_slices((dict(dataframe), labels))
if shuffle:
ds = ds.shuffle(buffer_size=len(dataframe))
ds = ds.batch(batch_size)
return ds
batch_size = 32
train_ds = df_to_dataset(train, batch_size=batch_size)
val_ds = df_to_dataset(val, shuffle=False, batch_size=batch_size)
test_ds = df_to_dataset(test, shuffle=False, batch_size=batch_size)
feature_columns = []
age = feature_column.numeric_column("age")
# numeric cols
for header in ['age', 'trestbps', 'chol', 'thalach', 'oldpeak', 'slope', 'ca']:
feature_columns.append(feature_column.numeric_column(header))
# bucketized cols
age_buckets = feature_column.bucketized_column(age, boundaries=[18, 25, 30, 35, 40, 45, 50, 55, 60, 65])
feature_columns.append(age_buckets)
# indicator cols
thal = feature_column.categorical_column_with_vocabulary_list(
'thal', ['fixed', 'normal', 'reversible'])
thal_one_hot = feature_column.indicator_column(thal)
feature_columns.append(thal_one_hot)
# embedding cols
thal_embedding = feature_column.embedding_column(thal, dimension=8)
feature_columns.append(thal_embedding)
# crossed cols
crossed_feature = feature_column.crossed_column([age_buckets, thal], hash_bucket_size=1000)
crossed_feature = feature_column.indicator_column(crossed_feature)
feature_columns.append(crossed_feature)
feature_layer = tf.keras.layers.DenseFeatures(feature_columns)
model = tf.keras.Sequential([
feature_layer,
layers.Dense(128, activation='relu'),
layers.Dense(128, activation='relu'),
layers.Dense(1)
])
model.compile(optimizer='adam',
loss=tf.keras.losses.BinaryCrossentropy(from_logits=True),
metrics=['accuracy'])
model.fit(train_ds,
validation_data=val_ds,
epochs=5)
loss, accuracy = model.evaluate(test_ds)
print("Accuracy", accuracy)
# Try to use predict to get the same accuracy
predictions = model.predict(test_ds)
for i, p in enumerate(predictions):
print(p, test.iloc[i,-1])
执行后,我得到的精度为0.6885246
然后,我尝试使用predict
方法获得评估数据集的预测,但在print(p,test.iloc[I,-1]
中得到的结果是:
[-1.7059733] 0
[-0.914219] 0
[2.6422875] 1
[-0.50430596] 1
[-1.2348572] 0
[-0.57301724] 0
[-2.1014583] 0
[-4.370711] 0
[0.21761642] 0
[-2.8065221] 0
[-3.2469923] 0
[-0.25715744] 1
[0.05394493] 1
[1.2391514] 0
[-3.7102253] 1
[-4.0611124] 0
[1.36385] 0
[-1.1096503] 0
[3.4140522] 1
[0.6951326] 0
[-3.232728] 0
[0.98346126] 0
[0.04960524] 0
[-0.90004027] 0
[1.918218] 0
[-0.02936329] 0
[-0.55671084] 1
[-2.1650188] 1
[-4.8975983] 0
[-1.5514184] 1
[-2.1743653] 0
[0.56928] 0
[-2.8607953] 0
[2.4095147] 0
[0.5155109] 1
[0.7517127] 0
[-1.6738821] 0
[-3.733505] 0
[2.2426589] 1
[-2.6165645] 0
[-2.1079547] 0
[-1.8746301] 0
[-4.116344] 0
[0.33854234] 1
[-2.3230617] 0
[-0.02075209] 1
[-0.33064234] 0
[1.6755556] 1
[1.1898655] 1
[0.40846193] 0
[-0.33131325] 0
[-0.63726294] 0
[-2.7144134] 0
[-0.48318636] 0
[1.516653] 1
[2.5299337] 1
[-2.1182806] 0
[-2.5583768] 1
[-0.65298045] 1
[-1.4936553] 0
[-0.7257029] 0
我的问题是应该使用什么方法将浮点结果转换为二进制(0或1)并比较目标?我的最终目标是通过评估方法获得精度值0.6885246
获得解决方案后编辑:
Accuracy 0.6885246
0.6885245901639344
我对Tensorflow教程中最近的这种方式感到非常惊讶,它在模型最后一层(
Dense(1)
)中使用线性激活函数来解决分类问题,然后在损失函数中要求使用from_logits=True
。我想原因是它可能会导致更好的数值稳定性,如以下所述:
来自\u logits
:是否将y\u pred
解释为值的张量。
默认情况下,我们假设y_pred
包含概率(即值
在[0,1])中。注意:使用from_logits=True
可能更符合数字
马厩
其中“by defaul”表示此处损失函数参数的默认值为from\u logits=False
在任何情况下,你最终得到的都是logits的预测,而不是类似教程(和实践)中通常出现的概率。logits的问题恰恰是,与概率预测相比,它们缺乏直观的解释
您应该做的是通过sigmoid函数传递您的logit,将它们转换为概率:
import numpy as np
def sigmoid(x):
return 1 / (1 + np.exp(-x))
您的前四个预测示例:
preds = np.array([-1.7059733, -0.914219, 2.6422875, -0.50430596])
sigmoid(preds)
# array([0.15368673, 0.28613728, 0.93353404, 0.37652929])
然后将其转换为阈值为0.5的“硬”预测:
final_preds = [1 if x>0.5 else 0 for x in preds]
final_preds
# [0, 0, 1, 0]
在这种形式下,你可以将它们与基本事实相比较
但是我建议你考虑的是,为了避免这一点,把你的最后一层改为
Dense(1, activation='sigmoid')
并从损失定义中删除
(from_logits=True)
参数。这样model.predict
应该返回硬预测(未经测试)。我对Tensorflow教程中最近的这种在模型最后一层中使用线性激活函数的方式感到非常惊讶(密集(1)
)对于分类问题,然后在损失函数中从_logits=True中询问。我猜原因是它可能会导致更好的数值稳定性,如以下所述:
来自\u logits
:是否将y\u pred
解释为值的张量。
默认情况下,我们假设y_pred
包含概率(即值
在[0,1])中。注意:使用from_logits=True
可能更符合数字
马厩
其中“by defaul”表示此处损失函数参数的默认值为from\u logits=False
在任何情况下,你最终得到的都是logits的预测,而不是类似教程(和实践)中通常出现的概率。logits的问题恰恰是,与概率预测相比,它们缺乏直观的解释
您应该做的是通过sigmoid函数传递您的logit,将它们转换为概率:
import numpy as np
def sigmoid(x):
return 1 / (1 + np.exp(-x))
您的前四个预测示例:
preds = np.array([-1.7059733, -0.914219, 2.6422875, -0.50430596])
sigmoid(preds)
# array([0.15368673, 0.28613728, 0.93353404, 0.37652929])
然后将其转换为阈值为0.5的“硬”预测:
final_preds = [1 if x>0.5 else 0 for x in preds]
final_preds
# [0, 0, 1, 0]
在这种形式下,你可以将它们与基本事实相比较
但是我建议你考虑的是,为了避免这一点,把你的最后一层改为
Dense(1, activation='sigmoid')
并从损失定义中删除(from_logits=True)
参数。这样model.predict
应该返回硬预测(未测试)