我试着建立一个只有一层的CNN,但我有一些问题。事实上,编译器告诉我
ValueError:检查模型输入时的错误:期望conv1d_1_input具有三维,但得到形状为(569,30)的数组
这是密码
import numpy
from keras.models import Sequential
from keras.layers.convolutional import Conv1D
numpy.random.seed(7)
datasetTraining = numpy.loadtxt("CancerAdapter.csv",delimiter=",")
X = datasetTraining[:,1:31]
Y = datasetTraining[:,0]
datasetTesting = numpy.loadtxt("CancereEvaluation.csv",delimiter=",")
X_test = datasetTraining[:,1:31]
Y_test = datasetTraining[:,0]
model = Sequential()
model.add(Conv1D(2,2,activation='relu',input_shape=X.shape))
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
model.fit(X, Y, epochs=150, batch_size=5)
scores = model.evaluate(X_test, Y_test)
print("\n%s: %.2f%%" % (model.metrics_names[1], scores[1]*100))发布于 2017-04-13 18:27:28
td;lr您需要重塑您的数据才能使Conv1d有意义的空间维数:
X = np.expand_dims(X, axis=2) # reshape (569, 30) to (569, 30, 1)
# now input can be set as
model.add(Conv1D(2,2,activation='relu',input_shape=(30, 1))本质上重塑了如下所示的数据集:
features
.8, .1, .3
.2, .4, .6
.7, .2, .1 至:
[[.8
.1
.3],
[.2,
.4,
.6
],
[.7,
.2,
.1]]解释与示例
通常卷积作用于空间维数。核在产生张量的维数上“被下放”。在Conv1D的情况下,内核通过每个示例的“步骤”维度传递。
您将看到Conv1D在NLP中使用,其中steps是句子中的许多单词(填充到固定的最大长度)。这些单词将被编码为长度为4的向量。
下面是一个例句:
jack .1 .3 -.52 |
is .05 .8, -.7 |<--- kernel is `convolving` along this dimension.
a .5 .31 -.2 |
boy .5 .8 -.4 \|/在这种情况下,我们将输入设置为conv的方式:
maxlen = 4
input_dim = 3
model.add(Conv1D(2,2,activation='relu',input_shape=(maxlen, input_dim))在您的情况下,您将把这些特征作为空间维度来处理,每个特征的长度为1。
下面是您的数据集中的一个示例
att1 .04 |
att2 .05 | < -- kernel convolving along this dimension
att3 .1 | notice the features have length 1. each
att4 .5 \|/ example have these 4 featues.我们将Conv1D示例设置为:
maxlen = num_features = 4 # this would be 30 in your case
input_dim = 1 # since this is the length of _each_ feature (as shown above)
model.add(Conv1D(2,2,activation='relu',input_shape=(maxlen, input_dim))正如您所看到的,您的数据集必须重新调整为(569,30,1),使用:
X = np.expand_dims(X, axis=2) # reshape (569, 30, 1)
# now input can be set as
model.add(Conv1D(2,2,activation='relu',input_shape=(30, 1))下面是一个可以运行的完整示例(我将使用功能API)
from keras.models import Model
from keras.layers import Conv1D, Dense, MaxPool1D, Flatten, Input
import numpy as np
inp = Input(shape=(5, 1))
conv = Conv1D(filters=2, kernel_size=2)(inp)
pool = MaxPool1D(pool_size=2)(conv)
flat = Flatten()(pool)
dense = Dense(1)(flat)
model = Model(inp, dense)
model.compile(loss='mse', optimizer='adam')
print(model.summary())
# get some data
X = np.expand_dims(np.random.randn(10, 5), axis=2)
y = np.random.randn(10, 1)
# fit model
model.fit(X, y)发布于 2018-10-19 15:06:35
我亦曾在其他职位提及这点:
要将形状(nrows, ncols)的常用特征表数据输入到Keras的Conv1d,需要执行以下两个步骤:
xtrain.reshape(nrows, ncols, 1)
# For conv1d statement:
input_shape = (ncols, 1)例如,采用虹膜数据集的前4个特征:
看到通常的格式及其形状:
iris_array = np.array(irisdf.iloc[:,:4].values)
print(iris_array[:5])
print(iris_array.shape)输出显示通常的格式及其形状:
[[5.1 3.5 1.4 0.2]
[4.9 3. 1.4 0.2]
[4.7 3.2 1.3 0.2]
[4.6 3.1 1.5 0.2]
[5. 3.6 1.4 0.2]]
(150, 4)以下代码更改了格式:
nrows, ncols = iris_array.shape
iris_array = iris_array.reshape(nrows, ncols, 1)
print(iris_array[:5])
print(iris_array.shape)上述代码数据格式及其形状的输出:
[[[5.1]
[3.5]
[1.4]
[0.2]]
[[4.9]
[3. ]
[1.4]
[0.2]]
[[4.7]
[3.2]
[1.3]
[0.2]]
[[4.6]
[3.1]
[1.5]
[0.2]]
[[5. ]
[3.6]
[1.4]
[0.2]]]
(150, 4, 1)这对于Keras的Conv1d来说很好。对于input_shape (4,1)是必需的。
发布于 2019-06-05 11:55:25
我有一个稀疏矩阵作为输入,所以如果不将它转换成常规数组,我就无法重新塑造它。
解决办法是使用角面整形层:
from keras.layers.core import Reshape
...
model = Sequential()
model.add(Reshape((X.shape[1], 1), input_shape=(X.shape[1], )))
model.add(Conv1D(2,2,activation='relu'))
...https://stackoverflow.com/questions/43396572
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