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Conv1D with kernel_size=1 vs Linear layer

I'm working on very sparse vectors as input. I started working with simple Linear (dense/fully connected layers) and my network yielded pretty good results (let's take accuracy as my metric here, 95.8%).

I later tried to use a Conv1d with a kernel_size=1 and a MaxPool1d, and this network works slightly better (96.4% accuracy).

Question: How are these two implementation different ? Shouldn't a Conv1d with a unit kernel_size do the same as a Linear layer?

I've tried multiple runs, the CNN always yields slightly better results.

like image 854
RobinFrcd Avatar asked Apr 08 '19 14:04

RobinFrcd


People also ask

Is Conv1d same as linear?

Figure-3: PyTorch code to showcase that Conv1d and Linear layer operations are equivalent.

What is Conv1d layer?

1D convolution layer (e.g. temporal convolution). This layer creates a convolution kernel that is convolved with the layer input over a single spatial (or temporal) dimension to produce a tensor of outputs. If use_bias is True, a bias vector is created and added to the outputs.

What is kernel size in Conv1d?

The kernel size is the size of the sequential window of the input. If the kernel size is set at 1, then each time interval will have its kernel and therefore, the output shape won't change from the (8, 16)[16 filters as above example].


2 Answers

nn.Conv1d with a kernel size of 1 and nn.Linear give essentially the same results. The only differences are the initialization procedure and how the operations are applied (which has some effect on the speed). Note that using a linear layer should be faster as it is implemented as a simple matrix multiplication (+ adding a broadcasted bias vector)

@RobinFrcd your answers are either different due to MaxPool1d or due to the different initialization procedure.

Here are a few experiments to prove my claims:

def count_parameters(model):
    """Count the number of parameters in a model."""
    return sum([p.numel() for p in model.parameters()])

conv = torch.nn.Conv1d(8,32,1)
print(count_parameters(conv))
# 288

linear = torch.nn.Linear(8,32)
print(count_parameters(linear))
# 288

print(conv.weight.shape)
# torch.Size([32, 8, 1])
print(linear.weight.shape)
# torch.Size([32, 8])

# use same initialization
linear.weight = torch.nn.Parameter(conv.weight.squeeze(2))
linear.bias = torch.nn.Parameter(conv.bias)

tensor = torch.randn(128,256,8)
permuted_tensor = tensor.permute(0,2,1).clone().contiguous()

out_linear = linear(tensor)
print(out_linear.mean())
# tensor(0.0067, grad_fn=<MeanBackward0>)

out_conv = conv(permuted_tensor)
print(out_conv.mean())
# tensor(0.0067, grad_fn=<MeanBackward0>)

Speed test:

%%timeit
_ = linear(tensor)
# 151 µs ± 297 ns per loop

%%timeit
_ = conv(permuted_tensor)
# 1.43 ms ± 6.33 µs per loop

As Hanchen's answer show, the results can differ very slightly due to numerical precision.

like image 109
Yann Dubois Avatar answered Oct 03 '22 03:10

Yann Dubois


I've encountered similar issues when working with 3d point clouds with models such as PointNet (CVPR'17). Therefore I've made a few more interpretations based on Yann Dubois's answers. We first define a few utility functions and then report our findings:

import torch, timeit, torch.nn as nn, matplotlib.pyplot as plt


def count_params(model):
    """Count the number of parameters in a module."""
    return sum([p.numel() for p in model.parameters()])


def compare_params(linear, conv1d):
    """Compare whether two modules have identical parameters."""
    return (linear.weight.detach().numpy() == conv1d.weight.detach().numpy().squeeze()).all() and \
           (linear.bias.detach().numpy() == conv1d.bias.detach().numpy()).all()


def compare_tensors(out_linear, out_conv1d):
    """Compare whether two tensors are identical."""
    return (out_linear.detach().numpy() == out_conv1d.permute(0, 2, 1).detach().numpy()).all()
  1. With the same input and parameters, nn.Conv1d and nn.Linear are expected to produce same forward results arithmetically, but experiments show that there are different. We show this by plotting the histogram of the numerical differences. Note that this numerical difference will increase as the network goes deep.
conv1d, linear = nn.Conv1d(8, 32, 1), nn.Linear(8, 32)

# same input tensor
tensor = torch.randn(128, 256, 8)
permuted_tensor = tensor.permute(0, 2, 1).clone().contiguous()

# same weights and bias
linear.weight = nn.Parameter(conv1d.weight.squeeze(2))
linear.bias = nn.Parameter(conv1d.bias)
print(compare_params(linear, conv1d))  # True

# check on the forward tensor
out_linear = linear(tensor)  # torch.Size([128, 256, 32])
out_conv1d = conv1d(permuted_tensor)  # torch.Size([128, 32, 256])
print(compare_tensors(out_linear, out_conv1d))  # False
plt.hist((out_linear.detach().numpy() - out_conv1d.permute(0, 2, 1).detach().numpy()).ravel())

Fig.1 Histogram of forward tensor between nn.Conv1d and nn.Linear

  1. Gradient updates in back propagation will also be numerically different.
target = torch.randn(out_linear.shape)
permuted_target = target.permute(0, 2, 1).clone().contiguous()

loss_linear = nn.MSELoss()(target, out_linear)
loss_linear.backward()
loss_conv1d = nn.MSELoss()(permuted_target, out_conv1d)
loss_conv1d.backward()

plt.hist((linear.weight.grad.detach().numpy() - 
    conv1d.weight.grad.permute(0, 2, 1).detach().numpy()).ravel())

Fig.2 Histogram of backward tensor between nn.Conv1d and nn.Linear

  1. Computing Speed on GPU. nn.Linear is a bit faster than nn.Conv1d
# test execution speed on CPUs
print(timeit.timeit("_ = linear(tensor)", number=10000, setup="from __main__ import tensor, linear"))
print(timeit.timeit("_ = conv1d(permuted_tensor)", number=10000, setup="from __main__ import conv1d, permuted_tensor"))

# change everything in *.cuda(), then test speed on GPUs
like image 29
Hanchen Avatar answered Oct 03 '22 03:10

Hanchen