这是一个创建张量流图以简化虹膜数据的简单代码。它使用tf.estimator.inputs.numpy_input_fn
为tf.contrib.learn.KMeansClustering
k-means聚类器定义输入函数。
import os
import numpy as np
import tensorflow as tf
# Data set
IRIS = "iris.csv"
# Load datasets
iris = tf.contrib.learn.datasets.base.load_csv_with_header(
filename=IRIS,
target_dtype=np.int,
features_dtype=np.float32)
# Build KMeans Clustering model.
num_clusters = 4
estimator = tf.contrib.learn.KMeansClustering(
num_clusters,
model_dir="/tmp/iris_model",
initial_clusters='random',
distance_metric='squared_euclidean',
use_mini_batch=True)
# Define the training inputs
input_fn = tf.estimator.inputs.numpy_input_fn(
x={"x": np.array(iris.data)},
y=np.array(iris.target),
batch_size=4,
num_epochs=None,
shuffle=True)
# Fit model.
clusters = estimator.fit(input_fn=input_fn)
但Tensorflow会返回以下错误:
...
AssertionError: Tensor("random_shuffle_queue_DequeueMany:2", shape=(4,), dtype=int64)
你知道我为什么会收到这个错误以及如何调试它吗?
iris.csv:
150,4,setosa,versicolor,virginica
6.4,2.8,5.6,2.2,2
5.0,2.3,3.3,1.0,1
4.9,2.5,4.5,1.7,2
4.9,3.1,1.5,0.1,0
5.7,3.8,1.7,0.3,0
4.4,3.2,1.3,0.2,0
5.4,3.4,1.5,0.4,0
6.9,3.1,5.1,2.3,2
6.7,3.1,4.4,1.4,1
5.1,3.7,1.5,0.4,0
5.2,2.7,3.9,1.4,1
6.9,3.1,4.9,1.5,1
5.8,4.0,1.2,0.2,0
5.4,3.9,1.7,0.4,0
7.7,3.8,6.7,2.2,2
6.3,3.3,4.7,1.6,1
6.8,3.2,5.9,2.3,2
7.6,3.0,6.6,2.1,2
6.4,3.2,5.3,2.3,2
5.7,4.4,1.5,0.4,0
6.7,3.3,5.7,2.1,2
6.4,2.8,5.6,2.1,2
5.4,3.9,1.3,0.4,0
6.1,2.6,5.6,1.4,2
7.2,3.0,5.8,1.6,2
5.2,3.5,1.5,0.2,0
5.8,2.6,4.0,1.2,1
5.9,3.0,5.1,1.8,2
5.4,3.0,4.5,1.5,1
6.7,3.0,5.0,1.7,1
6.3,2.3,4.4,1.3,1
5.1,2.5,3.0,1.1,1
6.4,3.2,4.5,1.5,1
6.8,3.0,5.5,2.1,2
6.2,2.8,4.8,1.8,2
6.9,3.2,5.7,2.3,2
6.5,3.2,5.1,2.0,2
5.8,2.8,5.1,2.4,2
5.1,3.8,1.5,0.3,0
4.8,3.0,1.4,0.3,0
7.9,3.8,6.4,2.0,2
5.8,2.7,5.1,1.9,2
6.7,3.0,5.2,2.3,2
5.1,3.8,1.9,0.4,0
4.7,3.2,1.6,0.2,0
6.0,2.2,5.0,1.5,2
4.8,3.4,1.6,0.2,0
7.7,2.6,6.9,2.3,2
4.6,3.6,1.0,0.2,0
7.2,3.2,6.0,1.8,2
5.0,3.3,1.4,0.2,0
6.6,3.0,4.4,1.4,1
6.1,2.8,4.0,1.3,1
5.0,3.2,1.2,0.2,0
7.0,3.2,4.7,1.4,1
6.0,3.0,4.8,1.8,2
7.4,2.8,6.1,1.9,2
5.8,2.7,5.1,1.9,2
6.2,3.4,5.4,2.3,2
5.0,2.0,3.5,1.0,1
5.6,2.5,3.9,1.1,1
6.7,3.1,5.6,2.4,2
6.3,2.5,5.0,1.9,2
6.4,3.1,5.5,1.8,2
6.2,2.2,4.5,1.5,1
7.3,2.9,6.3,1.8,2
4.4,3.0,1.3,0.2,0
7.2,3.6,6.1,2.5,2
6.5,3.0,5.5,1.8,2
5.0,3.4,1.5,0.2,0
4.7,3.2,1.3,0.2,0
6.6,2.9,4.6,1.3,1
5.5,3.5,1.3,0.2,0
7.7,3.0,6.1,2.3,2
6.1,3.0,4.9,1.8,2
4.9,3.1,1.5,0.1,0
5.5,2.4,3.8,1.1,1
5.7,2.9,4.2,1.3,1
6.0,2.9,4.5,1.5,1
6.4,2.7,5.3,1.9,2
5.4,3.7,1.5,0.2,0
6.1,2.9,4.7,1.4,1
6.5,2.8,4.6,1.5,1
5.6,2.7,4.2,1.3,1
6.3,3.4,5.6,2.4,2
4.9,3.1,1.5,0.1,0
6.8,2.8,4.8,1.4,1
5.7,2.8,4.5,1.3,1
6.0,2.7,5.1,1.6,1
5.0,3.5,1.3,0.3,0
6.5,3.0,5.2,2.0,2
6.1,2.8,4.7,1.2,1
5.1,3.5,1.4,0.3,0
4.6,3.1,1.5,0.2,0
6.5,3.0,5.8,2.2,2
4.6,3.4,1.4,0.3,0
4.6,3.2,1.4,0.2,0
7.7,2.8,6.7,2.0,2
5.9,3.2,4.8,1.8,1
5.1,3.8,1.6,0.2,0
4.9,3.0,1.4,0.2,0
4.9,2.4,3.3,1.0,1
4.5,2.3,1.3,0.3,0
5.8,2.7,4.1,1.0,1
5.0,3.4,1.6,0.4,0
5.2,3.4,1.4,0.2,0
5.3,3.7,1.5,0.2,0
5.0,3.6,1.4,0.2,0
5.6,2.9,3.6,1.3,1
4.8,3.1,1.6,0.2,0
6.3,2.7,4.9,1.8,2
5.7,2.8,4.1,1.3,1
5.0,3.0,1.6,0.2,0
6.3,3.3,6.0,2.5,2
5.0,3.5,1.6,0.6,0
5.5,2.6,4.4,1.2,1
5.7,3.0,4.2,1.2,1
4.4,2.9,1.4,0.2,0
4.8,3.0,1.4,0.1,0
5.5,2.4,3.7,1.0,1
5.9,3.0,4.2,1.5,1
6.9,3.1,5.4,2.1,2
5.1,3.3,1.7,0.5,0
6.0,3.4,4.5,1.6,1
5.5,2.5,4.0,1.3,1
6.2,2.9,4.3,1.3,1
5.5,4.2,1.4,0.2,0
6.3,2.8,5.1,1.5,2
5.6,3.0,4.1,1.3,1
6.7,2.5,5.8,1.8,2
7.1,3.0,5.9,2.1,2
4.3,3.0,1.1,0.1,0
5.6,2.8,4.9,2.0,2
5.5,2.3,4.0,1.3,1
6.0,2.2,4.0,1.0,1
5.1,3.5,1.4,0.2,0
5.7,2.6,3.5,1.0,1
4.8,3.4,1.9,0.2,0
5.1,3.4,1.5,0.2,0
5.7,2.5,5.0,2.0,2
5.4,3.4,1.7,0.2,0
5.6,3.0,4.5,1.5,1
6.3,2.9,5.6,1.8,2
6.3,2.5,4.9,1.5,1
5.8,2.7,3.9,1.2,1
6.1,3.0,4.6,1.4,1
5.2,4.1,1.5,0.1,0
6.7,3.1,4.7,1.5,1
6.7,3.3,5.7,2.5,2
6.4,2.9,4.3,1.3,1
答案 0 :(得分:2)
您不能将tf.estimator.inputs.numpy_input_fn传递给任何tf.contrib类,因为它不返回功能,标签,而是包含两者的内容。
使用旧的contrib类时,最好的选择是自己编写输入函数。这是如何:
def make_numpy_input_fn(x, y, batch_size):
def input_fn():
features, labels = tf.train.shuffle_batch(
[tf.constant(x), tf.constant(y)],
batch_size=batch_size,
capacity=50*batch_size,
min_after_dequeue=20*batch_size,
enqueue_many=True)
features = {'x': features}
return features, labels
return input_fn