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I’m having trouble getting VectorContinuousCallback to work as desired and I’m not sure what I’m doing wrong. I have a large system of equations, and essentially, any time any of the values cross some threshold value (in my system it’s 10e-30 but in this reprex 0.05), I want the value to go to zero.
That is, if at any point the values of u go below 0.05, I want the callback to take the value to zero, but right now, the solver seems to just almost ignore the callback? Not any of the crosses of the threshold are recognized.
A reprex:
using DifferentialEquations, Plots
function biomass_sim!(du, u, p, t)
# change = growth + gain from eating - loss from eating - loss
du[1] = 0.2*u[1] + (0.1*u[2] + 0.15*u[3]) - (0.2*u[4]) - 0.9*u[1]
du[2] = 0.2*u[2] + (0.1*u[1] + 0.05*u[3]) - (0.1*u[1] + 0.4*u[4]) - 0.5*u[2]
du[3] = 1.2*u[3] + 0 - (0.15*u[1] + 0.005*u[2]) - 1.3*u[3]
du[4] = 0.2*u[4] + (0.2*u[1] + 0.4*u[2]) - 1.9*u[1]
end
# set up extinction callback
function extinction_threshold(out,u,t,integrator)
# loop through all species to make the condition check all of them
for i in 1:4
out[i] = 0.05 - u[i]
end
end
function extinction_affect!(integrator, event_idx)
# loop again through all species
for i in 1:4
if event_idx == i
integrator.u[i] = 0
end
end
end
extinction_callback =
VectorContinuousCallback(extinction_threshold,
extinction_affect!,
4,
save_positions = (true, true),
interp_points = 1000
)
tspan = (0.0, 10.0)
u0 = [10, 10, 10, 10]
prob = ODEProblem(biomass_sim!,
u0,
tspan)
sol= solve(prob,
Tsit5(),
abstol = 1e-15,
reltol = 1e-10,
callback = extinction_callback,
progress = true,
progress_steps = 1)
plot(sol)
What I want to see here is that the two values that DO cross the threshold to be < 0.05 at least visually (u3 and u4 clearly go below zero), I want those values to become zero.
The application of this is an extinction to a species, so if they go below some threshold, I want to consider them extinct and therefore not able to be consumed by other species.
I’ve tried changing the tolerances, using different solvers (I’m not married to Tsit5()), but have yet to find a way to do this.
Any help much appreciated!!
The full output:
retcode: Success
Interpolation: specialized 4th order "free" interpolation
t: 165-element Vector{Float64}:
0.0
0.004242012928189618
0.01597661154828718
0.03189297583643294
0.050808376324350105
⋮
9.758563212772982
9.850431863368996
9.94240515017787
10.0
u: 165-element Vector{Vector{Float64}}:
[10.0, 10.0, 10.0, 10.0]
[9.972478301795496, 9.97248284719235, 9.98919421326857, 9.953394015118882]
[9.89687871881005, 9.896943262844019, 9.95942008072302, 9.825050883392004]
[9.795579619066798, 9.795837189358107, 9.919309541156514, 9.652323759097303]
[9.677029343447844, 9.67768414040866, 9.872046236050455, 9.449045554530718]
⋮
[43.86029800419986, 110.54225328286441, -12.173991695732434, -186.40702483057268]
[45.33660997599057, 114.35164541304869, -12.725800474246844, -192.72257104623995]
[46.86398454218351, 118.2922142830212, -13.295579652115606, -199.25572621838901]
[47.84633546050675, 120.82634905853745, -13.661479860003494, -203.45720035095707]
Answered in https://github.com/SciML/DifferentialEquations.jl/issues/843. This was a "user error". When you check the callback:
function extinction_affect!(integrator, event_idx)
# loop again through all species
#show integrator.u,event_idx
for i in 1:4
if event_idx == i
integrator.u[i] = 0
end
end
biomass_sim!(get_tmp_cache(integrator)[1], integrator.u, integrator.p, integrator.t)
#show get_tmp_cache(integrator)[1]
end
It is definitely called and does exactly as intended.
(integrator.u, event_idx) = ([5.347462662161639, 5.731062469090074, 7.64667777801325, 0.05000000000000008], 4)
(get_tmp_cache(integrator))[1] = [-2.0231159499021523, -1.3369848518263594, -1.5954424894710222, -6.798261538038756]
(integrator.u, event_idx) = ([12.968499097445866, 30.506371944743357, 0.050000000000001314, -53.521085634736835], 3)
(get_tmp_cache(integrator))[1] = [4.676904953209599, 12.256522670471728, -2.0978067243405967, -20.548116814707996]
but what this also shows is that, even if u[4] = 0, du[4] < 0 and so it's clear why it goes negative: that's due to how the ODE is defined. You should flip a parameter or something to make the derivative = 0 if you want to keep it at zero past the callback point.
please I want to create a function that computes the Ordinal Pooling neural network like the following figure:
this is my function :
def Ordinal_Pooling_NN(x):
wights = torch.tensor([0.6, 0.25, 0.10, 0.05])
top = torch.topk(x, 4, dim = 1)
wights = wights.repeat(x.shape[0], 1)
result = torch.sum(wights * (top.values), dim = 1 )
return result
but as a result, I get the following error:
<ipython-input-112-ddf99c812d56> in Ordinal_Pooling_NN(x)
9 top = torch.topk(x, 4, dim = 1)
10 wights = wights.repeat(x.shape[0], 1)
---> 11 result = torch.sum(wights * (top.values), dim = 1 )
12 return result
RuntimeError: The size of tensor a (4) must match the size of tensor b (16) at non-singleton dimension 2
Your implementation is actually correct, I believe you did not feed the function with a 2D tensor, the input must have a batch axis. For instance, the code below will run:
>>> Ordinal_Pooling_NN(torch.tensor([[1.9, 0.4, 1.3, 0.8]]))
tensor([1.5650])
Do note you are not required to repeat the weights tensor, it will be broadcasted automatically when computing the point-wise multiplication. You only need the following:
def Ordinal_Pooling_NN(x):
w = torch.tensor([0.6, 0.25, 0.10, 0.05])
top = torch.topk(x, k=4, dim=1)
result = torch.sum(w*top.values, dim=1)
return result
Python version: 3.8
Pytorch version: 1.9.0+cpu
Platform: Anaconda Spyder5.0
To reproduce this problem, just copy every code below to a single file.
The ILSVRC2012_val_00000293.jpg file used in this code is shown below, you also need to download it and then change its destination in the code.
Some background of this problem:
I am now working on a project that aims to develop a hardware accelerator to complete the inference process of the MobileNet V2 network. I used pretrained quantized Pytorch model to simulate the outcome, and the result comes out very well.
In order to use hardware to complete this task, I wish to know every inputs and outputs as well as intermidiate variables during runing this piece of pytorch code. I used a package named torchextractor to fetch the outcomes of first layer, which in this case, is a 3*3 convolution layer.
import numpy as np
import torchvision
import torch
from torchvision import transforms, datasets
from PIL import Image
from torchvision import transforms
import torchextractor as tx
import math
#########################################################################################
##### Processing of input image
#########################################################################################
normalize = transforms.Normalize(mean=[0.485, 0.456, 0.406],
std=[0.229, 0.224, 0.225])
test_transform = transforms.Compose([
transforms.Resize(256),
transforms.CenterCrop(224),
transforms.ToTensor(),
normalize,])
preprocess = transforms.Compose([
transforms.Resize(256),
transforms.CenterCrop(224),
transforms.ToTensor(),
transforms.Normalize(mean=[0.485, 0.456, 0.406], std=[0.229, 0.224, 0.225]),
])
#image file destination
filename = "D:\Project_UM\MobileNet_VC709\MobileNet_pytorch\ILSVRC2012_val_00000293.jpg"
input_image = Image.open(filename)
input_tensor = preprocess(input_image)
input_batch = input_tensor.unsqueeze(0)
#########################################################################################
#########################################################################################
#########################################################################################
#----First verify that the torchextractor class should not influent the inference outcome
# ofmp of layer1 before putting into torchextractor
a,b,c = quantize_tensor(input_batch)# to quantize the input tensor and return an int8 tensor, scale and zero point
input_qa = torch.quantize_per_tensor(torch.tensor(input_batch.clone().detach()), b, c, torch.quint8)# Using quantize_per_tensor method of torch
# Load a quantized mobilenet_v2 model
model_quantized = torchvision.models.quantization.mobilenet_v2(pretrained=True, quantize=True)
model_quantized.eval()
with torch.no_grad():
output = model_quantized.features[0][0](input_qa)# Ofmp of layer1, datatype : quantized_tensor
# print("FM of layer1 before tx_extractor:\n",output.int_repr())# Ofmp of layer1, datatype : int8 tensor
output1_clone = output.int_repr().detach().numpy()# Clone ofmp of layer1, datatype : ndarray
#########################################################################################
#########################################################################################
#########################################################################################
# ofmp of layer1 after adding torchextractor
model_quantized_ex = tx.Extractor(model_quantized, ["features.0.0"])#Capture of the module inside first layer
model_output, features = model_quantized_ex(input_batch)# Forward propagation
# feature_shapes = {name: f.shape for name, f in features.items()}
# print(features['features.0.0']) # Ofmp of layer1, datatype : quantized_tensor
out1_clone = features['features.0.0'].int_repr().numpy() # Clone ofmp of layer1, datatype : ndarray
if(out1_clone.all() == output1_clone.all()):
print('Model with torchextractor attached output the same value as the original model')
else:
print('Torchextractor method influence the outcome')
Here I define a numpy quantization scheme based on the quantization scheme proposed by
Quantization and Training of Neural Networks for Efficient
Integer-Arithmetic-Only Inference
# Convert a normal regular tensor to a quantized tensor with scale and zero_point
def quantize_tensor(x, num_bits=8):# to quantize the input tensor and return an int8 tensor, scale and zero point
qmin = 0.
qmax = 2.**num_bits - 1.
min_val, max_val = x.min(), x.max()
scale = (max_val - min_val) / (qmax - qmin)
initial_zero_point = qmin - min_val / scale
zero_point = 0
if initial_zero_point < qmin:
zero_point = qmin
elif initial_zero_point > qmax:
zero_point = qmax
else:
zero_point = initial_zero_point
# print(zero_point)
zero_point = int(zero_point)
q_x = zero_point + x / scale
q_x.clamp_(qmin, qmax).round_()
q_x = q_x.round().byte()
return q_x, scale, zero_point
#%%
# #############################################################################################
# --------- Simulate the inference process of layer0: conv33 using numpy
# #############################################################################################
# get the input_batch quantized buffer data
input_scale = b.item()
input_zero = c
input_quantized = a[0].detach().numpy()
# get the layer0 output scale and zero_point
output_scale = model_quantized.features[0][0].state_dict()['scale'].item()
output_zero = model_quantized.features[0][0].state_dict()['zero_point'].item()
# get the quantized weight with scale and zero_point
weight_scale = model_quantized.features[0][0].state_dict()["weight"].q_scale()
weight_zero = model_quantized.features[0][0].state_dict()["weight"].q_zero_point()
weight_quantized = model_quantized.features[0][0].state_dict()["weight"].int_repr().numpy()
# print(weight_quantized)
# print(weight_quantized.shape)
# bias_quantized,bias_scale,bias_zero= quantize_tensor(model_quantized.features[0][0].state_dict()["bias"])# to quantize the input tensor and return an int8 tensor, scale and zero point
# print(bias_quantized.shape)
bias = model_quantized.features[0][0].state_dict()["bias"].detach().numpy()
# print(input_quantized)
print(type(input_scale))
print(type(output_scale))
print(type(weight_scale))
Then I write a quantized 2D convolution using numpy, hope to figure out every details in pytorch data flow during the inference.
#%% numpy simulated layer0 convolution function define
def conv_cal(input_quantized, weight_quantized, kernel_size, stride, out_i, out_j, out_k):
weight = weight_quantized[out_i]
input = np.zeros((input_quantized.shape[0], kernel_size, kernel_size))
for i in range(weight.shape[0]):
for j in range(weight.shape[1]):
for k in range(weight.shape[2]):
input[i][j][k] = input_quantized[i][stride*out_j+j][stride*out_k+k]
# print(np.dot(weight,input))
# print(input,"\n")
# print(weight)
return np.multiply(weight,input).sum()
def QuantizedConv2D(input_scale, input_zero, input_quantized, output_scale, output_zero, weight_scale, weight_zero, weight_quantized, bias, kernel_size, stride, padding, ofm_size):
output = np.zeros((weight_quantized.shape[0],ofm_size,ofm_size))
input_quantized_padding = np.full((input_quantized.shape[0],input_quantized.shape[1]+2*padding,input_quantized.shape[2]+2*padding),0)
zero_temp = np.full(input_quantized.shape,input_zero)
input_quantized = input_quantized - zero_temp
for i in range(input_quantized.shape[0]):
for j in range(padding,padding + input_quantized.shape[1]):
for k in range(padding,padding + input_quantized.shape[2]):
input_quantized_padding[i][j][k] = input_quantized[i][j-padding][k-padding]
zero_temp = np.full(weight_quantized.shape, weight_zero)
weight_quantized = weight_quantized - zero_temp
for i in range(output.shape[0]):
for j in range(output.shape[1]):
for k in range(output.shape[2]):
# output[i][j][k] = (weight_scale*input_scale)*conv_cal(input_quantized_padding, weight_quantized, kernel_size, stride, i, j, k) + bias[i] #floating_output
output[i][j][k] = weight_scale*input_scale/output_scale*conv_cal(input_quantized_padding, weight_quantized, kernel_size, stride, i, j, k) + bias[i]/output_scale + output_zero
output[i][j][k] = round(output[i][j][k])
# int_output
return output
Here I input the same image, weight, and bias together with their zero_point and scale, then compare this "numpy simulated" result to the PyTorch calculated one.
quantized_model_out1_int8 = np.squeeze(features['features.0.0'].int_repr().numpy())
print(quantized_model_out1_int8.shape)
print(quantized_model_out1_int8)
out1_np = QuantizedConv2D(input_scale, input_zero, input_quantized, output_scale, output_zero, weight_scale, weight_zero, weight_quantized, bias, 3, 2, 1, 112)
np.save("out1_np.npy",out1_np)
for i in range(quantized_model_out1_int8.shape[0]):
for j in range(quantized_model_out1_int8.shape[1]):
for k in range(quantized_model_out1_int8.shape[2]):
if(out1_np[i][j][k] < 0):
out1_np[i][j][k] = 0
print(out1_np)
flag = np.zeros(quantized_model_out1_int8.shape)
for i in range(quantized_model_out1_int8.shape[0]):
for j in range(quantized_model_out1_int8.shape[1]):
for k in range(quantized_model_out1_int8.shape[2]):
if(quantized_model_out1_int8[i][j][k] == out1_np[i][j][k]):
flag[i][j][k] = 1
out1_np[i][j][k] = 0
quantized_model_out1_int8[i][j][k] = 0
# Compare the simulated result to extractor fetched result, gain the total hit rate
print(flag.sum()/(112*112*32)*100,'%')
If the "numpy simulated" results are the same as the extracted one, call it a hit. Print the total hit rate, it shows that numpy gets 92% of the values right. Now the problem is, I have no idea why the rest 8% of values come out wrong.
Comparison of two outcomes:
The picture below shows the different values between Numpy one and PyTorch one, the sample channel is index[1]. The left upper corner is Numpy one, and the upright corner is PyTorch one, I have set all values that are the same between them to 0, as you can see, most of the values just have a difference of 1(This can be view as the error brought by the precision loss of fixed point arithmetics), but some have large differences, e.g. the value[1][4], 121 vs. 76 (I don't know why)
Focus on one strange value:
This code is used to replay the calculation process of the value[1][4], originally I was expecting a trial and error process could lead me to solve this problem, to get my wanted number of 76, but no matter how I tried, it didn't output 76. If you want to try this, I paste this code for your convenience.
#%% A test code to check the calculation process
weight_quantized_sample = weight_quantized[2]
M_t = input_scale * weight_scale / output_scale
ifmap_t = np.int32(input_quantized[:,1:4,7:10])
weight_t = np.int32(weight_quantized_sample)
bias_t = bias[2]
bias_q = bias_t/output_scale
res_t = 0
for ch in range(3):
ifmap_offset = ifmap_t[ch]-np.int32(input_zero)
weight_offset = weight_t[ch]-np.int32(weight_zero)
res_ch = np.multiply(ifmap_offset, weight_offset)
res_ch = res_ch.sum()
res_t = res_t + res_ch
res_mul = M_t*res_t
# for n in range(1, 30):
# res_mul = multiply(n, M_t, res_t)
res_t = round(res_mul + output_zero + bias_q)
print(res_t)
Could you help me out of this, have been stuck here for a long time.
I implemented my own version of quantized convolution and got from 99.999% to 100% hitrate (and mismatch of a single value is by 1 that I can consider to be a rounding issue). The link on the paper in the question helped a lot.
But I found that your formulas are the same as mine. So I don't know what was your issue. As I understand quantization in pytorch is hardware dependent.
Here is my code:
def my_Conv2dRelu_b2(input_q, conv_layer, output_shape):
'''
Args:
input_q: quantized tensor
conv_layer: quantized tensor
output_shape: the pre-computed shape of the result
Returns:
'''
output = np.zeros(output_shape)
# extract needed float numbers from quantized operations
weights_scale = conv_layer.weight().q_per_channel_scales()
input_scale = input_q.q_scale()
weights_zp = conv_layer.weight().q_per_channel_zero_points()
input_zp = input_q.q_zero_point()
# extract needed convolution parameters
padding = conv_layer.padding
stride = conv_layer.stride
# extract float numbers for results
output_zp = conv_layer.zero_point
output_scale = conv_layer.scale
conv_weights_int = conv_layer.weight().int_repr()
input_int = input_q.int_repr()
biases = conv_layer.bias().numpy()
for k in range(input_q.shape[0]):
for i in range(conv_weights_int.shape[0]):
output[k][i] = manual_convolution_quant(
input_int[k].numpy(),
conv_weights_int[i].numpy(),
biases[i],
padding=padding,
stride=stride,
image_zp=input_zp, image_scale=input_scale,
kernel_zp=weights_zp[i].item(), kernel_scale=weights_scale[i].item(),
result_zp=output_zp, result_scale=output_scale
)
return output
def manual_convolution_quant(image, kernel, b, padding, stride, image_zp, image_scale, kernel_zp, kernel_scale,
result_zp, result_scale):
H = image.shape[1]
W = image.shape[2]
new_H = H // stride[0]
new_W = W // stride[1]
results = np.zeros([new_H, new_W])
M = image_scale * kernel_scale / result_scale
bias = b / result_scale
paddedIm = np.pad(
image,
[(0, 0), (padding[0], padding[0]), (padding[1], padding[1])],
mode="constant",
constant_values=image_zp,
)
s = kernel.shape[1]
for i in range(new_H):
for j in range(new_W):
patch = paddedIm[
:, i * stride[0]: i * stride[0] + s, j * stride[1]: j * stride[1] + s
]
res = M * ((kernel - kernel_zp) * (patch - image_zp)).sum() + result_zp + bias
if res < 0:
res = 0
results[i, j] = round(res)
return results
Code to compare pytorch and my own version.
def calc_hit_rate(array1, array2):
good = (array1 == array2).astype(np.int).sum()
all = array1.size
return good / all
# during inference
y2 = model.conv1(y1)
y2_int = torch.int_repr(y2)
y2_int_manual = my_Conv2dRelu_b2(y1, model.conv1, y2.shape)
print(f'y2 hit rate= {calc_hit_rate(y2.int_repr().numpy(), y2_int_manual)}') #hit_rate=1.0
I have the following neural network model.
nn_classifier = Sequential()
nn_classifier.add(Dense(output_dim = 16 ,activation='relu',input_dim = 13))
nn_classifier.add(Dense(output_dim = 16,activation='relu'))
nn_classifier.add(Dense(output_dim = 1, activation = 'sigmoid'))
nn_classifier.compile(optimizer = 'sgd', loss = 'binary_crossentropy', metrics=[tf.keras.metrics.BinaryAccuracy(threshold=0.5)])
model=nn_classifier.fit(X_train, Y_train ,validation_split=0.33, batch_size = 10, nb_epoch = 100)
Y_pred = nn_classifier.predict(X_test)
As I have used the sigmoid function in my output layer, I was expecting the predicted values (Y_pred) are either 0 or 1. But I get some decimal values. Is my understanding wrong?
Sigmoid always gives a value in [0,1] you need to round the value that means fix a threshold if it is higher than threshold then 1 else 0.
I am reading through the documentation of PyTorch and found an example where they write
gradients = torch.FloatTensor([0.1, 1.0, 0.0001])
y.backward(gradients)
print(x.grad)
where x was an initial variable, from which y was constructed (a 3-vector). The question is, what are the 0.1, 1.0 and 0.0001 arguments of the gradients tensor ? The documentation is not very clear on that.
Explanation
For neural networks, we usually use loss to assess how well the network has learned to classify the input image (or other tasks). The loss term is usually a scalar value. In order to update the parameters of the network, we need to calculate the gradient of loss w.r.t to the parameters, which is actually leaf node in the computation graph (by the way, these parameters are mostly the weight and bias of various layers such Convolution, Linear and so on).
According to chain rule, in order to calculate gradient of loss w.r.t to a leaf node, we can compute derivative of loss w.r.t some intermediate variable, and gradient of intermediate variable w.r.t to the leaf variable, do a dot product and sum all these up.
The gradient arguments of a Variable's backward() method is used to calculate a weighted sum of each element of a Variable w.r.t the leaf Variable. These weight is just the derivate of final loss w.r.t each element of the intermediate variable.
A concrete example
Let's take a concrete and simple example to understand this.
from torch.autograd import Variable
import torch
x = Variable(torch.FloatTensor([[1, 2, 3, 4]]), requires_grad=True)
z = 2*x
loss = z.sum(dim=1)
# do backward for first element of z
z.backward(torch.FloatTensor([[1, 0, 0, 0]]), retain_graph=True)
print(x.grad.data)
x.grad.data.zero_() #remove gradient in x.grad, or it will be accumulated
# do backward for second element of z
z.backward(torch.FloatTensor([[0, 1, 0, 0]]), retain_graph=True)
print(x.grad.data)
x.grad.data.zero_()
# do backward for all elements of z, with weight equal to the derivative of
# loss w.r.t z_1, z_2, z_3 and z_4
z.backward(torch.FloatTensor([[1, 1, 1, 1]]), retain_graph=True)
print(x.grad.data)
x.grad.data.zero_()
# or we can directly backprop using loss
loss.backward() # equivalent to loss.backward(torch.FloatTensor([1.0]))
print(x.grad.data)
In the above example, the outcome of first print is
2 0 0 0
[torch.FloatTensor of size 1x4]
which is exactly the derivative of z_1 w.r.t to x.
The outcome of second print is :
0 2 0 0
[torch.FloatTensor of size 1x4]
which is the derivative of z_2 w.r.t to x.
Now if use a weight of [1, 1, 1, 1] to calculate the derivative of z w.r.t to x, the outcome is 1*dz_1/dx + 1*dz_2/dx + 1*dz_3/dx + 1*dz_4/dx. So no surprisingly, the output of 3rd print is:
2 2 2 2
[torch.FloatTensor of size 1x4]
It should be noted that weight vector [1, 1, 1, 1] is exactly derivative of loss w.r.t to z_1, z_2, z_3 and z_4. The derivative of loss w.r.t to x is calculated as:
d(loss)/dx = d(loss)/dz_1 * dz_1/dx + d(loss)/dz_2 * dz_2/dx + d(loss)/dz_3 * dz_3/dx + d(loss)/dz_4 * dz_4/dx
So the output of 4th print is the same as the 3rd print:
2 2 2 2
[torch.FloatTensor of size 1x4]
Typically, your computational graph has one scalar output says loss. Then you can compute the gradient of loss w.r.t. the weights (w) by loss.backward(). Where the default argument of backward() is 1.0.
If your output has multiple values (e.g. loss=[loss1, loss2, loss3]), you can compute the gradients of loss w.r.t. the weights by loss.backward(torch.FloatTensor([1.0, 1.0, 1.0])).
Furthermore, if you want to add weights or importances to different losses, you can use loss.backward(torch.FloatTensor([-0.1, 1.0, 0.0001])).
This means to calculate -0.1*d(loss1)/dw, d(loss2)/dw, 0.0001*d(loss3)/dw simultaneously.
Here, the output of forward(), i.e. y is a a 3-vector.
The three values are the gradients at the output of the network. They are usually set to 1.0 if y is the final output, but can have other values as well, especially if y is part of a bigger network.
For eg. if x is the input, y = [y1, y2, y3] is an intermediate output which is used to compute the final output z,
Then,
dz/dx = dz/dy1 * dy1/dx + dz/dy2 * dy2/dx + dz/dy3 * dy3/dx
So here, the three values to backward are
[dz/dy1, dz/dy2, dz/dy3]
and then backward() computes dz/dx
The original code I haven't found on PyTorch website anymore.
gradients = torch.FloatTensor([0.1, 1.0, 0.0001])
y.backward(gradients)
print(x.grad)
The problem with the code above is there is no function based on how to calculate the gradients. This means we don't know how many parameters (arguments the function takes) and the dimension of parameters.
To fully understand this I created an example close to the original:
Example 1:
a = torch.tensor([1.0, 2.0, 3.0], requires_grad = True)
b = torch.tensor([3.0, 4.0, 5.0], requires_grad = True)
c = torch.tensor([6.0, 7.0, 8.0], requires_grad = True)
y=3*a + 2*b*b + torch.log(c)
gradients = torch.FloatTensor([0.1, 1.0, 0.0001])
y.backward(gradients,retain_graph=True)
print(a.grad) # tensor([3.0000e-01, 3.0000e+00, 3.0000e-04])
print(b.grad) # tensor([1.2000e+00, 1.6000e+01, 2.0000e-03])
print(c.grad) # tensor([1.6667e-02, 1.4286e-01, 1.2500e-05])
I assumed our function is y=3*a + 2*b*b + torch.log(c) and the parameters are tensors with three elements inside.
You can think of the gradients = torch.FloatTensor([0.1, 1.0, 0.0001]) like this is the accumulator.
As you may hear, PyTorch autograd system calculation is equivalent to Jacobian product.
In case you have a function, like we did:
y=3*a + 2*b*b + torch.log(c)
Jacobian would be [3, 4*b, 1/c]. However, this Jacobian is not how PyTorch is doing things to calculate the gradients at a certain point.
PyTorch uses forward pass and backward mode automatic differentiation (AD) in tandem.
There is no symbolic math involved and no numerical differentiation.
Numerical differentiation would be to calculate δy/δb, for b=1 and b=1+ε where ε is small.
If you don't use gradients in y.backward():
Example 2
a = torch.tensor(0.1, requires_grad = True)
b = torch.tensor(1.0, requires_grad = True)
c = torch.tensor(0.1, requires_grad = True)
y=3*a + 2*b*b + torch.log(c)
y.backward()
print(a.grad) # tensor(3.)
print(b.grad) # tensor(4.)
print(c.grad) # tensor(10.)
You will simply get the result at a point, based on how you set your a, b, c tensors initially.
Be careful how you initialize your a, b, c:
Example 3:
a = torch.empty(1, requires_grad = True, pin_memory=True)
b = torch.empty(1, requires_grad = True, pin_memory=True)
c = torch.empty(1, requires_grad = True, pin_memory=True)
y=3*a + 2*b*b + torch.log(c)
gradients = torch.FloatTensor([0.1, 1.0, 0.0001])
y.backward(gradients)
print(a.grad) # tensor([3.3003])
print(b.grad) # tensor([0.])
print(c.grad) # tensor([inf])
If you use torch.empty() and don't use pin_memory=True you may have different results each time.
Also, note gradients are like accumulators so zero them when needed.
Example 4:
a = torch.tensor(1.0, requires_grad = True)
b = torch.tensor(1.0, requires_grad = True)
c = torch.tensor(1.0, requires_grad = True)
y=3*a + 2*b*b + torch.log(c)
y.backward(retain_graph=True)
y.backward()
print(a.grad) # tensor(6.)
print(b.grad) # tensor(8.)
print(c.grad) # tensor(2.)
Lastly few tips on terms PyTorch uses:
PyTorch creates a dynamic computational graph when calculating the gradients in forward pass. This looks much like a tree.
So you will often hear the leaves of this tree are input tensors and the root is output tensor.
Gradients are calculated by tracing the graph from the root to the leaf and multiplying every gradient in the way using the chain rule. This multiplying occurs in the backward pass.
Back some time I created PyTorch Automatic Differentiation tutorial that you may check interesting explaining all the tiny details about AD.