Minimal character-level language model with a Vanilla Recurrent Neural Network, in Python/numpy

min-char-rnn.py

"""
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 range

Why am I getting this error?

Can somebody explain why we need to multiply 0.01 when initializing the weights ? Arent we already using np.random.randn to sample from a normal distribution?
I am talking about this specific line of code
Wxh = np.random.randn(hidden_size, vocab_size)*0.01

@MistryWoman Its essential to break the symmetry here. Since the outputs of the succeeding weights depends on the sum of inputs multiplied by the corresponding weight. If all weights are assigned to zero, every hidden unit will get same the type of signal ie zero in this case. No matter what was the input - if all weights are the same, all units in hidden layer will be the same too. This is not something we desire because we want different hidden units to compute different functions. However, this is not possible if you initialize all to the same value.

Wonderful code，but there are somecodes confused me.

# 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# 梯度的累加
        param += -learning_rate * dparam / np.sqrt(mem + 1e-8) # adagrad update

Wxh,..mby were defined as global variables，while param，dparam，mem are just local variables.How could Adagrad update change the value of global variables?I tried to test my thought by code like below.

import numpy as np
Wxh, Whh, Why, bh, by=1,2,3,4,5
dWxh, dWhh, dWhy, dbh, dby=1,2,3,4,5
mWxh, mWhh, mWhy, mbh, mby=1,2,3,4,5
while True:
    for param, dparam, mem in zip([Wxh, Whh, Why, bh, by], 
                                  [dWxh, dWhh, dWhy, dbh, dby], 
                                [mWxh, mWhh, mWhy, mbh, mby]):
        mem += dparam * dparam# 梯度的累加
        param += dparam / np.sqrt(mem + 1e-8) 
    print(Wxh)#Output never change!

It's might be a very simple problem.Just puzzled me!Someone helps?Thanks a lot.

Not sure if anyone ever addressed this further down the thread but I found an answer on stack overflow that addresses modifying mutable global variables like Wxh and the like within a local scope.

'param' is referencing all the weight arrays rather than copying their values in their original locations. It is not copying them into a new storage location. So when adding to all the values in 'param' with adagrad we are actually modifying the global values of Wxh, etc... not the values in param. I think?

https://stackoverflow.com/questions/28566563/global-scope-when-accessing-array-element-inside-function

@MistryWoman We are trying to push our weights closer to zero, so that gradient doesn't vanish. For large weights , gradients become very small while using tanh. Please take a look at https://gist.github.com/karpathy/d4dee566867f8291f086#gistcomment-2180482

@karpathy Thank you for posting this. I had a small doubt in line 58. Why do we perform Whh.T in np.dot(Whh.T,dhraw)? Since Whh has (hidden_size,hidden_size) as shape, we can directly multiply it with dhraw. Or am I missing something here?

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 range

Why 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!

I have a problem in the code of line 58: dhnext = np.dot(Whh.T, dhraw). Could anyone tell me what it means?

The expression for forward propagation is:

$$ \begin{align} z_t & = W_{hh} \cdot h_{t-1} + W_{xh} \cdot x_t + b_h \tag{1}\\ h_t & = Tanh(z_t) \tag{2}\\ y_t & = W_{hy} \cdot h_t + b_y \tag{3}\\ o_t & = Softmax(y_t) \tag{4}\\ \end{align} $$

Here is the gradients expression of weights that I derived:

$$ \begin{align} \frac{\partial L}{\partial W_{hy}} & = \frac{\partial L}{\partial y_t} \otimes {h_t}^T \tag{a}\\ \frac{\partial L}{\partial W_{hh}} & = {\color{red} {W_{hy}}^T \cdot \frac{\partial L}{\partial y_t} \odot Tanh'(z_t)} \otimes {h_{t-1}}^T \tag{b}\\ \frac{\partial L}{\partial W_{xh}} & = {\color{red} {W_{hy}}^T \cdot \frac{\partial L}{\partial y_t} \odot Tanh'(z_t)} \otimes x_t^T \tag{c} \end{align} $$

It is easy to see that the red part is what dhraw represents in the code. And we can get dWxh and dWhh from formula (b) and formula (c) without dhnext. So what does dhnext mean?

Stupid question : In which lines does the vanishing gradient problem manifest itself ?

It uses adaGrad optimization in the last few lines. Hence the gradient is always divided by the accumulate sum of each gradients' scaler .
mem += dparam*dparam
param += -learning_rate * dparam / np.sqrt(mem+1e-8)

However , I find that the mem and param are all local variables in the loop. Don't know if the implementation of adaGrad is correct.

I have a problem in the code of line 58: dhnext = np.dot(Whh.T, dhraw). Could anyone tell me what it means?

The expression for forward propagation is:

(1)zt=Whh⋅ht−1+Wxh⋅xt+bh(2)ht=Tanh(zt)(3)yt=Why⋅ht+by(4)ot=Softmax(yt)

Here is the gradients expression of weights that I derived:

(a)∂L∂Whh=∂L∂yt⊗htT(b)∂L∂Whh=WhyT⋅∂L∂yt⊙Tanh′(zt)⊗ht−1T(c)∂L∂Wxh=WhyT⋅∂L∂yt⊙Tanh′(zt)⊗xtT

It is easy to see that the red part is what dhraw represents in the code. And we can get dWxh and dWhh from formula (b) and formula (c) without dhnext. So what does dhnext mean?

Since dh means the gradient of loss wrt the hidden state, there are two ways the gradients flow(back propogation). From the equation and rnn structure we learn that the hidden state feeds forward to a output and next hidden state. The dhnext represents the gradient update for current state from the next hidden state. Note that dhnext is zero at first iteration because for the last-layer (unrolled) of rnn , there 's not a gradient update flow from the next hidden state.
Hope this will help you .

I have a problem in the code of line 58: dhnext = np.dot(Whh.T, dhraw). Could anyone tell me what it means?
The expression for forward propagation is:
(1)zt=Whh⋅ht−1+Wxh⋅xt+bh(2)ht=Tanh(zt)(3)yt=Why⋅ht+by(4)ot=Softmax(yt)
Here is the gradients expression of weights that I derived:
(a)∂L∂Whh=∂L∂yt⊗htT(b)∂L∂Whh=WhyT⋅∂L∂yt⊙Tanh′(zt)⊗ht−1T(c)∂L∂Wxh=WhyT⋅∂L∂yt⊙Tanh′(zt)⊗xtT
It is easy to see that the red part is what dhraw represents in the code. And we can get dWxh and dWhh from formula (b) and formula (c) without dhnext. So what does dhnext mean?

Since dh means the gradient of loss wrt the hidden state, there are two ways the gradients flow(back propogation). From the equation and rnn structure we learn that the hidden state feeds forward to a output and next hidden state. The dhnext represents the gradient update for current state from the next hidden state. Note that dhnext is zero at first iteration because for the last-layer (unrolled) of rnn , there 's not a gradient update flow from the next hidden state. Hope this will help you .

Thank you so much for your replying. I missed the partial derivative wrt the next hidden state.

@logeek404

However , I find that the mem and param are all local variables in the loop. Don't know if the implementation of adaGrad is correct.

mem and param point to the numpy ndarrays from the zip, and += updates their values in-place.

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Can somebody explain why we need to multiply 0.01 when initializing the weights ? Arent we already using np.random.randn to sample from a normal distribution? I am talking about this specific line of code Wxh = np.random.randn(hidden_size, vocab_size)*0.01

Thia is to keep weight matrix parameter to be small.