我有一个数据是这样的:
它可以查看here,并且已经包含在下面的代码中。 实际上,我有大约7000个样本(行),downloadable too。在
任务是给抗原,预测相应的表位。 所以抗原表位始终是抗原的一个精确的亚序列。这相当于 Sequence to Sequence Learning。下面是我在Keras下运行的递归神经网络代码。它是根据example建模的。在
我的问题是:
这是我的运行代码,它给出了非常糟糕的准确分数。在
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from __future__ import print_function
import sys
import json
import pandas as pd
from keras.models import Sequential
from keras.engine.training import slice_X
from keras.layers.core import Activation, RepeatVector, Dense
from keras.layers import recurrent, TimeDistributed
import numpy as np
from six.moves import range
class CharacterTable(object):
'''
Given a set of characters:
+ Encode them to a one hot integer representation
+ Decode the one hot integer representation to their character output
+ Decode a vector of probabilties to their character output
'''
def __init__(self, chars, maxlen):
self.chars = sorted(set(chars))
self.char_indices = dict((c, i) for i, c in enumerate(self.chars))
self.indices_char = dict((i, c) for i, c in enumerate(self.chars))
self.maxlen = maxlen
def encode(self, C, maxlen=None):
maxlen = maxlen if maxlen else self.maxlen
X = np.zeros((maxlen, len(self.chars)))
for i, c in enumerate(C):
X[i, self.char_indices[c]] = 1
return X
def decode(self, X, calc_argmax=True):
if calc_argmax:
X = X.argmax(axis=-1)
return ''.join(self.indices_char[x] for x in X)
class colors:
ok = '\033[92m'
fail = '\033[91m'
close = '\033[0m'
INVERT = True
HIDDEN_SIZE = 128
BATCH_SIZE = 64
LAYERS = 3
# Try replacing GRU, or SimpleRNN
RNN = recurrent.LSTM
def main():
"""
Epitope_core = answers
Antigen = questions
"""
epi_antigen_df = pd.io.parsers.read_table("http://dpaste.com/2PZ9WH6.txt")
antigens = epi_antigen_df["Antigen"].tolist()
epitopes = epi_antigen_df["Epitope Core"].tolist()
if INVERT:
antigens = [ x[::-1] for x in antigens]
allchars = "".join(antigens+epitopes)
allchars = list(set(allchars))
aa_chars = "".join(allchars)
sys.stderr.write(aa_chars + "\n")
max_antigen_len = len(max(antigens, key=len))
max_epitope_len = len(max(epitopes, key=len))
X = np.zeros((len(antigens),max_antigen_len, len(aa_chars)),dtype=np.bool)
y = np.zeros((len(epitopes),max_epitope_len, len(aa_chars)),dtype=np.bool)
ctable = CharacterTable(aa_chars, max_antigen_len)
sys.stderr.write("Begin vectorization\n")
for i, antigen in enumerate(antigens):
X[i] = ctable.encode(antigen, maxlen=max_antigen_len)
for i, epitope in enumerate(epitopes):
y[i] = ctable.encode(epitope, maxlen=max_epitope_len)
# Shuffle (X, y) in unison as the later parts of X will almost all be larger digits
indices = np.arange(len(y))
np.random.shuffle(indices)
X = X[indices]
y = y[indices]
# Explicitly set apart 10% for validation data that we never train over
split_at = len(X) - len(X) / 10
(X_train, X_val) = (slice_X(X, 0, split_at), slice_X(X, split_at))
(y_train, y_val) = (y[:split_at], y[split_at:])
sys.stderr.write("Build model\n")
model = Sequential()
# "Encode" the input sequence using an RNN, producing an output of HIDDEN_SIZE
# note: in a situation where your input sequences have a variable length,
# use input_shape=(None, nb_feature).
model.add(RNN(HIDDEN_SIZE, input_shape=(max_antigen_len, len(aa_chars))))
# For the decoder's input, we repeat the encoded input for each time step
model.add(RepeatVector(max_epitope_len))
# The decoder RNN could be multiple layers stacked or a single layer
for _ in range(LAYERS):
model.add(RNN(HIDDEN_SIZE, return_sequences=True))
# For each of step of the output sequence, decide which character should be chosen
model.add(TimeDistributed(Dense(len(aa_chars))))
model.add(Activation('softmax'))
model.compile(loss='categorical_crossentropy',
optimizer='adam',
metrics=['accuracy'])
# Train the model each generation and show predictions against the validation dataset
for iteration in range(1, 200):
print()
print('-' * 50)
print('Iteration', iteration)
model.fit(X_train, y_train, batch_size=BATCH_SIZE, nb_epoch=5,
validation_data=(X_val, y_val))
###
# Select 10 samples from the validation set at random so we can visualize errors
for i in range(10):
ind = np.random.randint(0, len(X_val))
rowX, rowy = X_val[np.array([ind])], y_val[np.array([ind])]
preds = model.predict_classes(rowX, verbose=0)
q = ctable.decode(rowX[0])
correct = ctable.decode(rowy[0])
guess = ctable.decode(preds[0], calc_argmax=False)
# print('Q', q[::-1] if INVERT else q)
print('T', correct)
print(colors.ok + '☑' + colors.close if correct == guess else colors.fail + '☒' + colors.close, guess)
print('---')
if __name__ == '__main__':
main()
是的,这些你都可以用。LSTMs和gru是RNN的类型;如果你所说的RNN是指a fully-connected RNN,那么由于梯度消失问题(1,2),它们已经不再受欢迎了。由于数据集中的示例相对较少,GRU可能比LSTM更好,因为它的体系结构更简单。在
你提到培训和验证错误都是不好的。一般来说,这可能是由于以下几个因素之一:
另一种选择是,由于表位是输入的子串,因此预测抗原序列内表位的起始和结束指数(可能由抗原序列的长度标准化),而不是一次预测一个字符的子串。这将是两个任务的回归问题。例如,如果抗原是FSKIAGLTVT(10个字母长),它的表位是KIAGL(位置3到7,一个基),那么输入就是FSKIAGLTVT,输出是0.3(第一个任务)和0.7(第二个任务)。在
或者,如果您可以使所有抗原的长度相同(通过删除带有短抗原的数据集的部分和/或切掉长抗原的末端,假设您知道表位不在末端附近,则将长抗原的末端切掉),你可以用两个任务(开始和结束)和序列长度类将它定义为一个分类问题,在这个问题中,你试图为抗原分配一个概率,从每个位置开始和结束。在
减少层的数量将显著提高代码的速度。此外,由于GRU的架构更简单,GRU将比LSTM更快。然而,这两种类型的递归网络都比卷积网络慢。在
如果你对合作感兴趣,可以给我发一封电子邮件(我个人资料中的地址)。在
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