Find entry exit on intersect

up vote
0
down vote

favorite

Im using the following code to find entry and exit count of the people.

 # check to see if the object has been counted or not

            if not to.counted:

                # if the direction is negative (indicating the object

                # is moving up) AND the centroid is above the center

                # line, count the object

                if direction < 0 and centroid[1] < H // 2:

                    totalUp += 1

                    to.counted = True



                # if the direction is positive (indicating the object

                # is moving down) AND the centroid is below the

                # center line, count the object

                elif direction > 0 and centroid[1] > H // 2:

                    totalDown += 1

                    to.counted = True

As per this code if the same person comes back and enters again, the entry count is still the same as the person has been counted already. I want to find the entry and exit count everytime when the person intersects the line. How do I sort it out?

asked Nov 7 at 14:40

RockKil

1

What is the to object? If it's a class you define, you could keep track of counted as an int and increment counted with to.counted+=1. This would also require you get rid of the if statement, otherwise you'll only count at the instance counted=0 or in your case counted=False
– C.Nivs
Nov 7 at 14:44

object is the person..
– RockKil
Nov 7 at 14:45

Add the code used to define to (including any class definition) to your question, that might make things a touch more clear for us to help and we can see exactly what you're working with
– C.Nivs
Nov 7 at 14:47

add a comment |

up vote
0
down vote

favorite

Im using the following code to find entry and exit count of the people.

 # check to see if the object has been counted or not

            if not to.counted:

                # if the direction is negative (indicating the object

                # is moving up) AND the centroid is above the center

                # line, count the object

                if direction < 0 and centroid[1] < H // 2:

                    totalUp += 1

                    to.counted = True



                # if the direction is positive (indicating the object

                # is moving down) AND the centroid is below the

                # center line, count the object

                elif direction > 0 and centroid[1] > H // 2:

                    totalDown += 1

                    to.counted = True

asked Nov 7 at 14:40

RockKil

1

What is the to object? If it's a class you define, you could keep track of counted as an int and increment counted with to.counted+=1. This would also require you get rid of the if statement, otherwise you'll only count at the instance counted=0 or in your case counted=False
– C.Nivs
Nov 7 at 14:44

object is the person..
– RockKil
Nov 7 at 14:45

Add the code used to define to (including any class definition) to your question, that might make things a touch more clear for us to help and we can see exactly what you're working with
– C.Nivs
Nov 7 at 14:47

add a comment |

up vote
0
down vote

favorite

Im using the following code to find entry and exit count of the people.

 # check to see if the object has been counted or not

            if not to.counted:

                # if the direction is negative (indicating the object

                # is moving up) AND the centroid is above the center

                # line, count the object

                if direction < 0 and centroid[1] < H // 2:

                    totalUp += 1

                    to.counted = True



                # if the direction is positive (indicating the object

                # is moving down) AND the centroid is below the

                # center line, count the object

                elif direction > 0 and centroid[1] > H // 2:

                    totalDown += 1

                    to.counted = True

asked Nov 7 at 14:40

RockKil

Im using the following code to find entry and exit count of the people.

 # check to see if the object has been counted or not

            if not to.counted:

                # if the direction is negative (indicating the object

                # is moving up) AND the centroid is above the center

                # line, count the object

                if direction < 0 and centroid[1] < H // 2:

                    totalUp += 1

                    to.counted = True



                # if the direction is positive (indicating the object

                # is moving down) AND the centroid is below the

                # center line, count the object

                elif direction > 0 and centroid[1] > H // 2:

                    totalDown += 1

                    to.counted = True

python opencv image-processing tracking dlib

asked Nov 7 at 14:40

RockKil

asked Nov 7 at 14:40

RockKil

asked Nov 7 at 14:40

RockKil

asked Nov 7 at 14:40

RockKil

asked Nov 7 at 14:40

RockKil

1

What is the to object? If it's a class you define, you could keep track of counted as an int and increment counted with to.counted+=1. This would also require you get rid of the if statement, otherwise you'll only count at the instance counted=0 or in your case counted=False
– C.Nivs
Nov 7 at 14:44

object is the person..
– RockKil
Nov 7 at 14:45

Add the code used to define to (including any class definition) to your question, that might make things a touch more clear for us to help and we can see exactly what you're working with
– C.Nivs
Nov 7 at 14:47

add a comment |

1

What is the to object? If it's a class you define, you could keep track of counted as an int and increment counted with to.counted+=1. This would also require you get rid of the if statement, otherwise you'll only count at the instance counted=0 or in your case counted=False
– C.Nivs
Nov 7 at 14:44

object is the person..
– RockKil
Nov 7 at 14:45

Add the code used to define to (including any class definition) to your question, that might make things a touch more clear for us to help and we can see exactly what you're working with
– C.Nivs
Nov 7 at 14:47

What is the to object? If it's a class you define, you could keep track of counted as an int and increment counted with to.counted+=1. This would also require you get rid of the if statement, otherwise you'll only count at the instance counted=0 or in your case counted=False
– C.Nivs
Nov 7 at 14:44

object is the person..
– RockKil
Nov 7 at 14:45

Add the code used to define to (including any class definition) to your question, that might make things a touch more clear for us to help and we can see exactly what you're working with
– C.Nivs
Nov 7 at 14:47

add a comment |

1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

A quick way to do that would be to ignore the counted attribute entirely:

# if the direction is negative (indicating the object

# is moving up) AND the centroid is above the center

# line, count the object

if direction < 0 and centroid[1] < H // 2:

      totalUp += 1



# if the direction is positive (indicating the object

# is moving down) AND the centroid is below the

# center line, count the object

elif direction > 0 and centroid[1] > H // 2:

    totalDown += 1

This is assuming that your total counts are not per person but a grand total of occurrences. If that's the case, ignore the if to.counted because you don't care if they've been counted already, you just care if the conditions you've set have been satisfied

answered Nov 7 at 14:53

C.Nivs

1,6661414

I tried doing so. It increments the count dramatically fast.
– RockKil
Nov 7 at 14:56

Is that expected behavior? If not, why not?
– C.Nivs
Nov 7 at 14:58

it is not the expected behaviour. Everytime when the person hits the line, either the entry or exit count should increase only once until he hits it again... if he hits it again, entry or exit count should increment once more... it shall not increment constantly.
– RockKil
Nov 7 at 15:08

based on the if logic and available code, the only way the entry or exit values are incremented is if they satisfy one of the two conditions. Are you sure that the program is not finding those matches at a fairly high rate and so it looks like the count is increasing constantly? I would need to see your entire code to tell why the values are increasing at an unacceptable rate
– C.Nivs
Nov 7 at 22:28

add a comment |

Your Answer

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: "1"
};
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',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53191680%2ffind-entry-exit-on-intersect%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

A quick way to do that would be to ignore the counted attribute entirely:

# if the direction is negative (indicating the object

# is moving up) AND the centroid is above the center

# line, count the object

if direction < 0 and centroid[1] < H // 2:

      totalUp += 1



# if the direction is positive (indicating the object

# is moving down) AND the centroid is below the

# center line, count the object

elif direction > 0 and centroid[1] > H // 2:

    totalDown += 1

answered Nov 7 at 14:53

C.Nivs

1,6661414

I tried doing so. It increments the count dramatically fast.
– RockKil
Nov 7 at 14:56

Is that expected behavior? If not, why not?
– C.Nivs
Nov 7 at 14:58

it is not the expected behaviour. Everytime when the person hits the line, either the entry or exit count should increase only once until he hits it again... if he hits it again, entry or exit count should increment once more... it shall not increment constantly.
– RockKil
Nov 7 at 15:08

based on the if logic and available code, the only way the entry or exit values are incremented is if they satisfy one of the two conditions. Are you sure that the program is not finding those matches at a fairly high rate and so it looks like the count is increasing constantly? I would need to see your entire code to tell why the values are increasing at an unacceptable rate
– C.Nivs
Nov 7 at 22:28

add a comment |

up vote
0
down vote

accepted

A quick way to do that would be to ignore the counted attribute entirely:

# if the direction is negative (indicating the object

# is moving up) AND the centroid is above the center

# line, count the object

if direction < 0 and centroid[1] < H // 2:

      totalUp += 1



# if the direction is positive (indicating the object

# is moving down) AND the centroid is below the

# center line, count the object

elif direction > 0 and centroid[1] > H // 2:

    totalDown += 1

answered Nov 7 at 14:53

C.Nivs

1,6661414

I tried doing so. It increments the count dramatically fast.
– RockKil
Nov 7 at 14:56

Is that expected behavior? If not, why not?
– C.Nivs
Nov 7 at 14:58

it is not the expected behaviour. Everytime when the person hits the line, either the entry or exit count should increase only once until he hits it again... if he hits it again, entry or exit count should increment once more... it shall not increment constantly.
– RockKil
Nov 7 at 15:08

based on the if logic and available code, the only way the entry or exit values are incremented is if they satisfy one of the two conditions. Are you sure that the program is not finding those matches at a fairly high rate and so it looks like the count is increasing constantly? I would need to see your entire code to tell why the values are increasing at an unacceptable rate
– C.Nivs
Nov 7 at 22:28

add a comment |

up vote
0
down vote

accepted

A quick way to do that would be to ignore the counted attribute entirely:

# if the direction is negative (indicating the object

# is moving up) AND the centroid is above the center

# line, count the object

if direction < 0 and centroid[1] < H // 2:

      totalUp += 1



# if the direction is positive (indicating the object

# is moving down) AND the centroid is below the

# center line, count the object

elif direction > 0 and centroid[1] > H // 2:

    totalDown += 1

answered Nov 7 at 14:53

C.Nivs

1,6661414

A quick way to do that would be to ignore the counted attribute entirely:

# if the direction is negative (indicating the object

# is moving up) AND the centroid is above the center

# line, count the object

if direction < 0 and centroid[1] < H // 2:

      totalUp += 1



# if the direction is positive (indicating the object

# is moving down) AND the centroid is below the

# center line, count the object

elif direction > 0 and centroid[1] > H // 2:

    totalDown += 1

answered Nov 7 at 14:53

C.Nivs

1,6661414

answered Nov 7 at 14:53

C.Nivs

1,6661414

answered Nov 7 at 14:53

C.Nivs

1,6661414

answered Nov 7 at 14:53

C.Nivs

1,6661414

I tried doing so. It increments the count dramatically fast.
– RockKil
Nov 7 at 14:56

Is that expected behavior? If not, why not?
– C.Nivs
Nov 7 at 14:58

it is not the expected behaviour. Everytime when the person hits the line, either the entry or exit count should increase only once until he hits it again... if he hits it again, entry or exit count should increment once more... it shall not increment constantly.
– RockKil
Nov 7 at 15:08

based on the if logic and available code, the only way the entry or exit values are incremented is if they satisfy one of the two conditions. Are you sure that the program is not finding those matches at a fairly high rate and so it looks like the count is increasing constantly? I would need to see your entire code to tell why the values are increasing at an unacceptable rate
– C.Nivs
Nov 7 at 22:28

add a comment |

I tried doing so. It increments the count dramatically fast.
– RockKil
Nov 7 at 14:56

Is that expected behavior? If not, why not?
– C.Nivs
Nov 7 at 14:58

it is not the expected behaviour. Everytime when the person hits the line, either the entry or exit count should increase only once until he hits it again... if he hits it again, entry or exit count should increment once more... it shall not increment constantly.
– RockKil
Nov 7 at 15:08

based on the if logic and available code, the only way the entry or exit values are incremented is if they satisfy one of the two conditions. Are you sure that the program is not finding those matches at a fairly high rate and so it looks like the count is increasing constantly? I would need to see your entire code to tell why the values are increasing at an unacceptable rate
– C.Nivs
Nov 7 at 22:28

I tried doing so. It increments the count dramatically fast.
– RockKil
Nov 7 at 14:56

Is that expected behavior? If not, why not?
– C.Nivs
Nov 7 at 14:58

it is not the expected behaviour. Everytime when the person hits the line, either the entry or exit count should increase only once until he hits it again... if he hits it again, entry or exit count should increment once more... it shall not increment constantly.
– RockKil
Nov 7 at 15:08

based on the if logic and available code, the only way the entry or exit values are incremented is if they satisfy one of the two conditions. Are you sure that the program is not finding those matches at a fairly high rate and so it looks like the count is increasing constantly? I would need to see your entire code to tell why the values are increasing at an unacceptable rate
– C.Nivs
Nov 7 at 22:28

add a comment |

draft saved

draft discarded

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Wsrtjtyk