Loss function for a linear classifier
$begingroup$
I'm writing a loss function for a linear classifier. The function takes in several arrays and outputs the loss. s is a score array which contains 10 entries that are the scores for each class. y is the array that contains the correct class assignments. My loss function takes the max of 0 and the calculation between the correct score and all the other scores to see if there is a higher value in s than y[i].
import numpy as np
import math
def model_loss(W, b, x, y):
n = x.shape[0]
lossSize = 1/n
Li = 0.0
loss = 0.0
for i in range(n):
s = (np.dot(W.transpose(), x[i])) + b
for j in range (W.shape[1]):
if (j != y[i]):
Li += max(0.0, (s[j] - s[y[i]] + 1.0))
loss += Li
Li = 0.0
loss *= LossSize
return loss
This is a loss function for a linear classifier so W is an array of weights with the shape (16, 10) and x is the data I want to test. The shape of x is (10,000, 16) because it stores about 10,000 images. Each image contains a histogram of 16 entries. s is an array which contains 10 entries. The highest value of s is the class prediction. So if entry 3 on s has the highest value then 3 is the prediction class (these numbers could correspond to things like boat, car, building etc). y is the array that contains the correct class assignment. So y = [3, 8, 8 ...]. For each y value, it's value is treated as an index for s. So if s[y[i]] is the highest value then it means the linear classifier assigned the image to the right class. If it's not then it means the linear classifier assigned the wrong class to the image so the values from all the other entries in s will be compared with the y entry to see just how much loss was taken. Hence it's called a loss function.
This produces what I want but now I need to optimize it. My question is, is there a way to get rid of the two for loops since for loops can be rather slow and I feel like there is a more efficient way of doing this.
python performance numpy machine-learning clustering
edited 17 mins ago
200_success
129k15153415
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm writing a loss function for a linear classifier. The function takes in several arrays and outputs the loss. s is a score array which contains 10 entries that are the scores for each class. y is the array that contains the correct class assignments. My loss function takes the max of 0 and the calculation between the correct score and all the other scores to see if there is a higher value in s than y[i].
import numpy as np
import math
def model_loss(W, b, x, y):
n = x.shape[0]
lossSize = 1/n
Li = 0.0
loss = 0.0
for i in range(n):
s = (np.dot(W.transpose(), x[i])) + b
for j in range (W.shape[1]):
if (j != y[i]):
Li += max(0.0, (s[j] - s[y[i]] + 1.0))
loss += Li
Li = 0.0
loss *= LossSize
return loss
This is a loss function for a linear classifier so W is an array of weights with the shape (16, 10) and x is the data I want to test. The shape of x is (10,000, 16) because it stores about 10,000 images. Each image contains a histogram of 16 entries. s is an array which contains 10 entries. The highest value of s is the class prediction. So if entry 3 on s has the highest value then 3 is the prediction class (these numbers could correspond to things like boat, car, building etc). y is the array that contains the correct class assignment. So y = [3, 8, 8 ...]. For each y value, it's value is treated as an index for s. So if s[y[i]] is the highest value then it means the linear classifier assigned the image to the right class. If it's not then it means the linear classifier assigned the wrong class to the image so the values from all the other entries in s will be compared with the y entry to see just how much loss was taken. Hence it's called a loss function.
This produces what I want but now I need to optimize it. My question is, is there a way to get rid of the two for loops since for loops can be rather slow and I feel like there is a more efficient way of doing this.
python performance numpy machine-learning clustering
edited 17 mins ago
200_success
129k15153415
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
That is the entire function.
$endgroup$
– Brandon MacLeod
39 mins ago
$begingroup$
Ok I included the def.
$endgroup$
– Brandon MacLeod
37 mins ago
$begingroup$
Ok I updated my question.
$endgroup$
– Brandon MacLeod
22 mins ago
add a comment |
$begingroup$
I'm writing a loss function for a linear classifier. The function takes in several arrays and outputs the loss. s is a score array which contains 10 entries that are the scores for each class. y is the array that contains the correct class assignments. My loss function takes the max of 0 and the calculation between the correct score and all the other scores to see if there is a higher value in s than y[i].
import numpy as np
import math
def model_loss(W, b, x, y):
n = x.shape[0]
lossSize = 1/n
Li = 0.0
loss = 0.0
for i in range(n):
s = (np.dot(W.transpose(), x[i])) + b
for j in range (W.shape[1]):
if (j != y[i]):
Li += max(0.0, (s[j] - s[y[i]] + 1.0))
loss += Li
Li = 0.0
loss *= LossSize
return loss
This is a loss function for a linear classifier so W is an array of weights with the shape (16, 10) and x is the data I want to test. The shape of x is (10,000, 16) because it stores about 10,000 images. Each image contains a histogram of 16 entries. s is an array which contains 10 entries. The highest value of s is the class prediction. So if entry 3 on s has the highest value then 3 is the prediction class (these numbers could correspond to things like boat, car, building etc). y is the array that contains the correct class assignment. So y = [3, 8, 8 ...]. For each y value, it's value is treated as an index for s. So if s[y[i]] is the highest value then it means the linear classifier assigned the image to the right class. If it's not then it means the linear classifier assigned the wrong class to the image so the values from all the other entries in s will be compared with the y entry to see just how much loss was taken. Hence it's called a loss function.
This produces what I want but now I need to optimize it. My question is, is there a way to get rid of the two for loops since for loops can be rather slow and I feel like there is a more efficient way of doing this.
python performance numpy machine-learning clustering
edited 17 mins ago
200_success
129k15153415
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm writing a loss function for a linear classifier. The function takes in several arrays and outputs the loss. s is a score array which contains 10 entries that are the scores for each class. y is the array that contains the correct class assignments. My loss function takes the max of 0 and the calculation between the correct score and all the other scores to see if there is a higher value in s than y[i].
import numpy as np
import math
def model_loss(W, b, x, y):
n = x.shape[0]
lossSize = 1/n
Li = 0.0
loss = 0.0
for i in range(n):
s = (np.dot(W.transpose(), x[i])) + b
for j in range (W.shape[1]):
if (j != y[i]):
Li += max(0.0, (s[j] - s[y[i]] + 1.0))
loss += Li
Li = 0.0
loss *= LossSize
return loss
This is a loss function for a linear classifier so W is an array of weights with the shape (16, 10) and x is the data I want to test. The shape of x is (10,000, 16) because it stores about 10,000 images. Each image contains a histogram of 16 entries. s is an array which contains 10 entries. The highest value of s is the class prediction. So if entry 3 on s has the highest value then 3 is the prediction class (these numbers could correspond to things like boat, car, building etc). y is the array that contains the correct class assignment. So y = [3, 8, 8 ...]. For each y value, it's value is treated as an index for s. So if s[y[i]] is the highest value then it means the linear classifier assigned the image to the right class. If it's not then it means the linear classifier assigned the wrong class to the image so the values from all the other entries in s will be compared with the y entry to see just how much loss was taken. Hence it's called a loss function.
This produces what I want but now I need to optimize it. My question is, is there a way to get rid of the two for loops since for loops can be rather slow and I feel like there is a more efficient way of doing this.
python performance numpy machine-learning clustering
python performance numpy machine-learning clustering
edited 17 mins ago
200_success
129k15153415
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 mins ago
200_success
129k15153415
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 mins ago
200_success
129k15153415
edited 17 mins ago
200_success
129k15153415
edited 17 mins ago
200_success
129k15153415
129k15153415
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
asked 2 hours ago
Brandon MacLeodBrandon MacLeod
112
112
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Brandon MacLeod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
That is the entire function.
$endgroup$
– Brandon MacLeod
39 mins ago
$begingroup$
Ok I included the def.
$endgroup$
– Brandon MacLeod
37 mins ago
$begingroup$
Ok I updated my question.
$endgroup$
– Brandon MacLeod
22 mins ago
add a comment |
$begingroup$
That is the entire function.
$endgroup$
– Brandon MacLeod
39 mins ago
$begingroup$
Ok I included the def.
$endgroup$
– Brandon MacLeod
37 mins ago
$begingroup$
Ok I updated my question.
$endgroup$
– Brandon MacLeod
22 mins ago
$begingroup$
That is the entire function.
$endgroup$
– Brandon MacLeod
39 mins ago
$begingroup$
That is the entire function.
$endgroup$
– Brandon MacLeod
39 mins ago
$begingroup$
Ok I included the def.
$endgroup$
– Brandon MacLeod
37 mins ago
$begingroup$
Ok I included the def.
$endgroup$
– Brandon MacLeod
37 mins ago
$begingroup$
Ok I updated my question.
$endgroup$
– Brandon MacLeod
22 mins ago
$begingroup$
Ok I updated my question.
$endgroup$
– Brandon MacLeod
22 mins ago
add a comment |
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$begingroup$
That is the entire function.
$endgroup$
– Brandon MacLeod
39 mins ago
$begingroup$
Ok I included the def.
$endgroup$
– Brandon MacLeod
37 mins ago
$begingroup$
Ok I updated my question.
$endgroup$
– Brandon MacLeod
22 mins ago