在Keras中，如何通过predict方法获得与evaluate方法相同的accuracy值？

Question

首先，我遵循了Tutorials

中编码的示例

并编写以下代码：

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.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])

执行后得到Accuracy = 0.6885246.

然后，我尝试使用 predict 方法来获得评估数据集的预测，但我在 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

我的问题是我应该使用什么方法将浮点数结果转换为二进制（0或1）并比较目标？我的最终目标是通过evaluate方法得到的精度值0.6885246。

得到解后编辑：

改为"from_logits=false"
改变输出层"layers.Dense(1, activation='sigmoid')"
在model.predict

final_preds = [1 if x>0.5 else 0 for x in predictions]
m = 0
for i, p in enumerate(final_preds):
   if p == test.iloc[i, -1]
      m += 1

print(m / len(final_preds))

在运行之后，我得到：

Accuracy 0.6885246
0.6885245901639344

Answer 1

我对 Tensorflow 教程中最近流行的在模型最后一层 (Dense(1)) 中使用线性激活函数来解决分类问题感到非常惊讶，并且然后在损失函数中求 from_logits=True 。我想原因是它可能会导致更好的数值稳定性，如 documentation:

中所述

from_logits: Whether to interpret y_pred as a tensor of logit values. By default, we assume that y_pred contains probabilities (i.e., values in [0, 1]). Note: Using from_logits=True may be more numerically stable.

其中"by defaul"表示这里的损失函数参数默认值为from_logits=False.

无论如何，您最终得到的是 logits 的预测，而不是迄今为止在类似教程（和实践中）中通常情况下的概率。与概率预测相比，logits 的问题恰恰在于它们缺乏直观的解释。

你应该做的是从 sigmoid 函数传递你的 logits 以将它们转换为概率：

import numpy as np

def sigmoid(x):
    return 1 / (1 + np.exp(-x))

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])

然后将它们转换为 "hard" 阈值为 0.5 的预测：

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，您可以将它们与基本事实进行比较。

但是为了避免这种情况，我建议您考虑将最后一层更改为

Dense(1, activation='sigmoid')

并从损失定义中删除 (from_logits=True) 参数。这样 model.predict 应该 return 硬预测（未测试）。

在Keras中，如何通过predict方法获得与evaluate方法相同的accuracy值？

In Keras, how obtain the same accuracy value of evaluate method by predict method?

python

machine-learning

keras

tensorflow

tf.keras