我想在django中使用yolov5模型,但我遇到了麻烦。
我想要做的是,如果用户上传一个图像到django服务器,用yolov5模型进行对象检测,然后在web上显示结果。这个过程本身很简单。,但我不知道如何附加yolov5 api和django.。
正如你们中使用yolo的人所知道的,yolo基本上运行基于命令的apis。
!python train.py --img 512 --batch 16 --epochs 100 --data ~~ # for training
!python detect.py --weights'/content/yolov5/runs/~~ # for detection有在Django视图中运行此操作的任何方法吗?
我尝试了python函数,比如execfile()和os.system(),它们以解释器的方式在Python中执行命令,但是它不能正常工作。(我认为Django和Yolo之间的道路是扭曲的。)
实际上,如果可能的话,最好将yolo作为一个像Keras这样的模块加载,然后像函数一样运行它,而不是命令方法。但是我找不到像模块一样使用yolov5的方法或者类似的方法。
我该如何解决这个问题?请给我一些建议。
发布于 2021-07-21 17:12:45
基于Django和Yolov5的目标检测
使用yolov5配置Django项目:
我上传了django-对象检测这里,以便于参考.
Models.py:
import os
from django.db import models
from django.utils.translation import gettext_lazy as _
class ImageModel(models.Model):
image = models.ImageField(_("image"), upload_to='images')
class Meta:
verbose_name = "Image"
verbose_name_plural = "Images"
def __str__(self):
return str(os.path.split(self.image.path)[-1])Views.py:
import io
from PIL import Image as im
import torch
from django.shortcuts import render
from django.views.generic.edit import CreateView
from .models import ImageModel
from .forms import ImageUploadForm
class UploadImage(CreateView):
model = ImageModel
template_name = 'image/imagemodel_form.html'
fields = ["image"]
def post(self, request, *args, **kwargs):
form = ImageUploadForm(request.POST, request.FILES)
if form.is_valid():
img = request.FILES.get('image')
img_instance = ImageModel(
image=img
)
img_instance.save()
uploaded_img_qs = ImageModel.objects.filter().last()
img_bytes = uploaded_img_qs.image.read()
img = im.open(io.BytesIO(img_bytes))
# Change this to the correct path
path_hubconfig = "absolute/path/to/yolov5_code"
path_weightfile = "absolute/path/to/yolov5s.pt" # or any custom trained model
model = torch.hub.load(path_hubconfig, 'custom',
path=path_weightfile, source='local')
results = model(img, size=640)
results.render()
for img in results.imgs:
img_base64 = im.fromarray(img)
img_base64.save("media/yolo_out/image0.jpg", format="JPEG")
inference_img = "/media/yolo_out/image0.jpg"
form = ImageUploadForm()
context = {
"form": form,
"inference_img": inference_img
}
return render(request, 'image/imagemodel_form.html', context)
else:
form = ImageUploadForm()
context = {
"form": form
}
return render(request, 'image/imagemodel_form.html', context)在这个views.py中可以修改很多东西,比如添加更多的函数和逻辑,但是这里的目标是连接yolov5 & Django。views.py内部的配置是最重要的,因为它是yolov5 hubconf.py文件的网关。
Forms.py
from django import forms
from .models import ImageModel
class ImageUploadForm(forms.ModelForm):
class Meta:
model = ImageModel
fields = ['image']Imagemodel_form.html
{% extends "base.html" %}
{% load crispy_forms_tags %}
{% block leftbar %}
<div class="col-sm-3">
</div>
{% endblock leftbar %}
{% block content %}
<div class="col-sm-9">
<div id="uploadedImage"></div>
<div class="mt-4">
<form action="" enctype="multipart/form-data" id="imageUploadForm" method="post">
{% csrf_token %}
{{ form|crispy }}
<button class="btn btn-outline-success" type="submit">Submit</button>
</form>
</div>
</div>
<div class="mt-4">
{% if inference_img %}
<img src="{{inference_img}}" class="img-fluid" />
{% else %}
The inferenced image will be displayed here.
{% endif %}
</div>
</div>
{% endblock content %}最初的简单网页:

发现后:

https://stackoverflow.com/questions/64002542
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