""" | |
Minimal character-level Vanilla RNN model. Written by Andrej Karpathy (@karpathy) | |
BSD License | |
""" | |
import numpy as np | |
# data I/O | |
data = open('input.txt', 'r').read() # should be simple plain text file | |
chars = list(set(data)) | |
data_size, vocab_size = len(data), len(chars) | |
print 'data has %d characters, %d unique.' % (data_size, vocab_size) | |
char_to_ix = { ch:i for i,ch in enumerate(chars) } | |
ix_to_char = { i:ch for i,ch in enumerate(chars) } | |
# hyperparameters | |
hidden_size = 100 # size of hidden layer of neurons | |
seq_length = 25 # number of steps to unroll the RNN for | |
learning_rate = 1e-1 | |
# model parameters | |
Wxh = np.random.randn(hidden_size, vocab_size)*0.01 # input to hidden | |
Whh = np.random.randn(hidden_size, hidden_size)*0.01 # hidden to hidden | |
Why = np.random.randn(vocab_size, hidden_size)*0.01 # hidden to output | |
bh = np.zeros((hidden_size, 1)) # hidden bias | |
by = np.zeros((vocab_size, 1)) # output bias | |
def lossFun(inputs, targets, hprev): | |
""" | |
inputs,targets are both list of integers. | |
hprev is Hx1 array of initial hidden state | |
returns the loss, gradients on model parameters, and last hidden state | |
""" | |
xs, hs, ys, ps = {}, {}, {}, {} | |
hs[-1] = np.copy(hprev) | |
loss = 0 | |
# forward pass | |
for t in xrange(len(inputs)): | |
xs[t] = np.zeros((vocab_size,1)) # encode in 1-of-k representation | |
xs[t][inputs[t]] = 1 | |
hs[t] = np.tanh(np.dot(Wxh, xs[t]) + np.dot(Whh, hs[t-1]) + bh) # hidden state | |
ys[t] = np.dot(Why, hs[t]) + by # unnormalized log probabilities for next chars | |
ps[t] = np.exp(ys[t]) / np.sum(np.exp(ys[t])) # probabilities for next chars | |
loss += -np.log(ps[t][targets[t],0]) # softmax (cross-entropy loss) | |
# backward pass: compute gradients going backwards | |
dWxh, dWhh, dWhy = np.zeros_like(Wxh), np.zeros_like(Whh), np.zeros_like(Why) | |
dbh, dby = np.zeros_like(bh), np.zeros_like(by) | |
dhnext = np.zeros_like(hs[0]) | |
for t in reversed(xrange(len(inputs))): | |
dy = np.copy(ps[t]) | |
dy[targets[t]] -= 1 # backprop into y. see http://cs231n.github.io/neural-networks-case-study/#grad if confused here | |
dWhy += np.dot(dy, hs[t].T) | |
dby += dy | |
dh = np.dot(Why.T, dy) + dhnext # backprop into h | |
dhraw = (1 - hs[t] * hs[t]) * dh # backprop through tanh nonlinearity | |
dbh += dhraw | |
dWxh += np.dot(dhraw, xs[t].T) | |
dWhh += np.dot(dhraw, hs[t-1].T) | |
dhnext = np.dot(Whh.T, dhraw) | |
for dparam in [dWxh, dWhh, dWhy, dbh, dby]: | |
np.clip(dparam, -5, 5, out=dparam) # clip to mitigate exploding gradients | |
return loss, dWxh, dWhh, dWhy, dbh, dby, hs[len(inputs)-1] | |
def sample(h, seed_ix, n): | |
""" | |
sample a sequence of integers from the model | |
h is memory state, seed_ix is seed letter for first time step | |
""" | |
x = np.zeros((vocab_size, 1)) | |
x[seed_ix] = 1 | |
ixes = [] | |
for t in xrange(n): | |
h = np.tanh(np.dot(Wxh, x) + np.dot(Whh, h) + bh) | |
y = np.dot(Why, h) + by | |
p = np.exp(y) / np.sum(np.exp(y)) | |
ix = np.random.choice(range(vocab_size), p=p.ravel()) | |
x = np.zeros((vocab_size, 1)) | |
x[ix] = 1 | |
ixes.append(ix) | |
return ixes | |
n, p = 0, 0 | |
mWxh, mWhh, mWhy = np.zeros_like(Wxh), np.zeros_like(Whh), np.zeros_like(Why) | |
mbh, mby = np.zeros_like(bh), np.zeros_like(by) # memory variables for Adagrad | |
smooth_loss = -np.log(1.0/vocab_size)*seq_length # loss at iteration 0 | |
while True: | |
# prepare inputs (we're sweeping from left to right in steps seq_length long) | |
if p+seq_length+1 >= len(data) or n == 0: | |
hprev = np.zeros((hidden_size,1)) # reset RNN memory | |
p = 0 # go from start of data | |
inputs = [char_to_ix[ch] for ch in data[p:p+seq_length]] | |
targets = [char_to_ix[ch] for ch in data[p+1:p+seq_length+1]] | |
# sample from the model now and then | |
if n % 100 == 0: | |
sample_ix = sample(hprev, inputs[0], 200) | |
txt = ''.join(ix_to_char[ix] for ix in sample_ix) | |
print '----\n %s \n----' % (txt, ) | |
# forward seq_length characters through the net and fetch gradient | |
loss, dWxh, dWhh, dWhy, dbh, dby, hprev = lossFun(inputs, targets, hprev) | |
smooth_loss = smooth_loss * 0.999 + loss * 0.001 | |
if n % 100 == 0: print 'iter %d, loss: %f' % (n, smooth_loss) # print progress | |
# perform parameter update with Adagrad | |
for param, dparam, mem in zip([Wxh, Whh, Why, bh, by], | |
[dWxh, dWhh, dWhy, dbh, dby], | |
[mWxh, mWhh, mWhy, mbh, mby]): | |
mem += dparam * dparam | |
param += -learning_rate * dparam / np.sqrt(mem + 1e-8) # adagrad update | |
p += seq_length # move data pointer | |
n += 1 # iteration counter |
Traceback (most recent call last):
File "minimal_rnn.py", line 100, in
loss, dWxh, dWhh, dWhy, dbh, dby, hprev = lossFun(inputs, targets, hprev)
File "minimal_rnn.py", line 43, in lossFun
loss += -np.log(ps[t][targets[t],0]) # softmax (cross-entropy loss)
IndexError: list index out of rangeWhy am I getting this error?
I was getting the same error but it seems like the reason is that your input file is too small. Try writing more stuff in your input.txt
I made a two layer recurrent neural network based off of this and I am not sure why it does not work. So, if anyone could check out and make a PR if you find a problem?
https://github.com/lanttu1243/vanilla_recurrent_neural_network.git
Hi, can anyone explain the line where we pass in the gradient through the softmax function " dy [targets[t]] -=1 ". Why are we doing this operation ??
To anyone finding the code hard to understand, I provide detailed explanations here: https://mkffl.github.io/ Hope it helps
Good explanation for second part of lossFunction.
Could someone explain how to use this? I can't generate text. It is running though.
@pjoer, do you have an input.txt
file (really any text file) in the same directory of this file which you should be running like python min-char-rnn.py
It seems the above was written with python2 in mind, but I just updated a few lines and got it working on Python 3.9.6: https://gist.github.com/ijkilchenko/84be862a5e18240c59b4505177c9c34c
Good luck!
@karpathy Thank you for posting this. I had a small doubt in line 58. Why do we perform
Whh.T
innp.dot(Whh.T,dhraw)
? SinceWhh
has(hidden_size,hidden_size)
as shape, we can directly multiply it withdhraw
. Or am I missing something here?