不兼容的贝叶斯神经网络中的误差

Question

我是机器学习的新知识。我有一个有关贝叶斯神经网络的项目，以预测足球结果。然后，我遵循此链接中的指示。然后我制作这样的代码：

import sys
from math import floor
import edward as ed
import numpy as np
import pandas as pd
import tensorflow as tf
from edward.models import Normal, Categorical
from fancyimpute import KNN
from tqdm import tqdm
from sklearn.preprocessing import OneHotEncoder
data = pd.read_csv('features_dummies_with_label.csv', sep=',')

def impute_missing_values_by_KNN():
    home_data = data[[col for col in data.columns if 'hp' in col]]
    away_data = data[[col for col in data.columns if 'ap' in col]]
    label_data = data[[col for col in data.columns if 'label' in col]]
    home_filled = pd.DataFrame(KNN(3).complete(home_data))
    home_filled.columns = home_data.columns
    home_filled.index = home_data.index
    away_filled = pd.DataFrame(KNN(3).complete(away_data))
    away_filled.columns = away_data.columns
    away_filled.index = away_data.index
    data_frame_out = pd.concat([home_filled, away_filled, label_data], axis=1)
    return data_frame_out

dataset = impute_missing_values_by_KNN()
dataset = pd.DataFrame(data=dataset)
data_x = dataset.loc[:, dataset.columns != 'label'].as_matrix().astype(np.float32)
data_y_ = dataset.loc[:, 'label'].as_matrix().astype(np.float32)
enc = OneHotEncoder(sparse=False)
integer_encoded = np.array(data_y_).reshape(-1)
integer_encoded = integer_encoded.reshape(len(integer_encoded), 1)
onehot_encoded = enc.fit_transform(integer_encoded)
data_y = onehot_encoded
train_size = 0.9
train_cnt = floor(data_x.shape[0] * train_size)
N = int(train_cnt)
train_x, test_x = data_x[0:N], data_x[N:]
train_y, test_y = data_y[0:N], data_y[N:]
in_size = train_x.shape[1]
out_size = train_y.shape[1]
EPOCH_SUM = 5
BATCH_SIZE = 10
train_y2 = np.argmax(train_y, axis=1)
test_y2 = np.argmax(test_y, axis=1)
n_nodes_hl1 = 500
x_ = tf.placeholder(tf.float32, [None, in_size])
y_ = tf.placeholder(tf.float32)
# def neural_network_model(data):
w_h1 = Normal(loc=tf.zeros([in_size, out_size]), scale=tf.ones([in_size, out_size]))
b_h1 = Normal(loc=tf.zeros([out_size]), scale=tf.ones([out_size]))
y_pre = Normal(tf.matmul(x_, w_h1) + b_h1, scale=1.0)
qw_h1 = Normal(loc=tf.Variable(tf.random_normal([in_size, out_size])),
               scale=tf.Variable(tf.random_normal([in_size, out_size])))
qb_h1 = Normal(loc=tf.Variable(tf.random_normal([out_size])), scale=tf.Variable(tf.random_normal([out_size])))
y = Normal(tf.matmul(x_, qw_h1) + qb_h1, scale=1.0)
inference = ed.KLqp({w_h1: qw_h1, b_h1: qb_h1}, data={y_pre: y_})
inference.initialize()
sess = tf.Session()
sess.run(tf.global_variables_initializer())
with sess:
    samples_num = 100
    for epoch in tqdm(range(EPOCH_SUM), file=sys.stdout):
        perm = np.random.permutation(N)
        for i in range(0, N, BATCH_SIZE):
            batch_x = train_x[perm[i:i + BATCH_SIZE]]
            batch_y = train_y2[perm[i:i + BATCH_SIZE]]
            inference.update(feed_dict={x_: batch_x, y_: batch_y})
        y_samples = y.sample(samples_num).eval(feed_dict={x_: train_x})
        acc = (np.round(y_samples.sum(axis=0) / samples_num) == train_y2).mean()
        y_samples = y.sample(samples_num).eval(feed_dict={x_: test_x})
        tets_acc = (np.round(y_samples.sum(axis=0) / samples_num) == test_y2).mean()
        if (epoch + 1) % 1 == 0:
            tqdm.write('epoch:t{}taccuracy:t{}tvaridation accuracy:t{}'.format(epoch + 1, acc, tets_acc))

但是，当我调试时，有这样的错误：

InvalidArgumentError (see above for traceback): Incompatible shapes: [10] vs. [10,3]
     [[Node: inference/sample/Normal_2/log_prob/standardize/sub = Sub[T=DT_FLOAT, _device="/job:localhost/replica:0/task:0/cpu:0"](_arg_Placeholder_1_0_1, inference/sample/Normal_2/loc)]]

在此行中：

temperion.update(feed_dict = {x_：batch_x，y_：batch_y}(

错误的含义是什么？以及如何解决它？

Answer 1

而没有看到追溯，很难调试错误。但是，我认为您传递给推理的张量或数组。update是另一种形状，就像您在声明中所定义的形状一样。因此，我将检查例如：batch_x，train_x(适用于每个迭代(的形状，w_h1，qw_h1，...。将这些阵列/张量打印出或使用TFDBG进行比较。

请考虑这个答案不是最终的，而是评论。但是，由于我有＆lt；我无法评论的50分。但是，我想做出贡献，因为我认为我的帖子可以为解决方案做出贡献。