How to eliminate for loops in Python?
$begingroup$
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]. Here's my code.
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 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
asked 3 mins ago
Brandon MacLeodBrandon MacLeod
1
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]. Here's my code.
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 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
asked 3 mins ago
Brandon MacLeodBrandon MacLeod
1
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]. Here's my code.
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 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
asked 3 mins ago
Brandon MacLeodBrandon MacLeod
1
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]. Here's my code.
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 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
python performance numpy
asked 3 mins ago
Brandon MacLeodBrandon MacLeod
1
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 3 mins ago
Brandon MacLeodBrandon MacLeod
1
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 3 mins ago
Brandon MacLeodBrandon MacLeod
1
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 3 mins ago
Brandon MacLeodBrandon MacLeod
1
asked 3 mins ago
Brandon MacLeodBrandon MacLeod
1
1
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.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Brandon MacLeod is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212186%2fhow-to-eliminate-for-loops-in-python%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Brandon MacLeod is a new contributor. Be nice, and check out our Code of Conduct.
Brandon MacLeod is a new contributor. Be nice, and check out our Code of Conduct.
Brandon MacLeod is a new contributor. Be nice, and check out our Code of Conduct.
Brandon MacLeod is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212186%2fhow-to-eliminate-for-loops-in-python%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
