我已经在谷歌硬盘上有一个(2K图像)数据集的拉链.我必须在ML训练算法中使用它.Code下面以字符串格式提取内容:
from pydrive.auth import GoogleAuth
from pydrive.drive import GoogleDrive
from google.colab import auth
from oauth2client.client import GoogleCredentials
import io
import zipfile
# Authenticate and create the PyDrive client.
# This only needs to be done once per notebook.
auth.authenticate_user()
gauth = GoogleAuth()
gauth.credentials = GoogleCredentials.get_application_default()
drive = GoogleDrive(gauth)
# Download a file based on its file ID.
#
# A file ID looks like: laggVyWshwcyP6kEI-y_W3P8D26sz
file_id = '1T80o3Jh3tHPO7hI5FBxcX-jFnxEuUE9K' #-- Updated File ID for my zip
downloaded = drive.CreateFile({'id': file_id})
#print('Downloaded …
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当我在14:20 - 14:30 通过Google Python Class Day 1 Part 2时,Guy说"不要使用list.sort
".他还提到"恐龙使用它!" (即它是一种旧的排序方式).但他没有提到原因.
谁能告诉我为什么我们不应该使用list.sort
?
我最近开始使用机器学习,我正在学习CNN,我计划在这个Keras博客和这个github repo的帮助下编写一个Car Damage严重性检测应用程序.
这是汽车数据集的样子:
F:\WORKSPACE\ML\CAR_DAMAGE_DETECTOR\DATASET\DATA3A
????training (979 Images for all 3 categories of training set)
? ????01-minor
? ????02-moderate
? ????03-severe
????validation (171 Images for all 3 categories of validation set)
????01-minor
????02-moderate
????03-severe
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以下代码只给我32%的准确性.
from keras.preprocessing.image import ImageDataGenerator
from keras.models import Sequential
from keras.layers import Conv2D, MaxPooling2D
from keras.layers import Activation, Dropout, Flatten, Dense
from keras import backend as K
# dimensions of our images.
img_width, img_height = 150, 150
train_data_dir = 'dataset/data3a/training'
validation_data_dir = …
Run Code Online (Sandbox Code Playgroud) 我是一名python开发人员,我的公司想用PHP -laravel做一个项目.他也想让我开始Vagrant
.所以我遇到了这个简单的教程,并试图从它开始.但是我仍然收到错误没有指定输入文件.
我今天一整天都在搜索这个问题,我发现了许多帖子,其中包含相同的查询.我尝试了laravel.io的几乎所有解决方案和一些堆栈溢出查询但是没有得到确切的解决方案.
目录结构:
Git Cloned Homestead - /home/laxmikant/Work/PHPWORK/Homestead
映射位置 - /home/laxmikant/Work/PHPWORK/codebase
(完全空目录)
映射的文件夹to
:( /home/vagrant/codebase
包含Laravel
项目)
这是我的 Homestead.yaml
---
ip: "192.168.10.10"
memory: 2048
cpus: 1
provider: virtualbox
authorize: ~/.ssh/id_rsa.pub
keys:
- ~/.ssh/id_rsa
folders:
- map: ~/Work/PHPWORK/codebase #-- The path on my local machine
to: /home/vagrant/codebase #-- The path from the vagrant box
sites:
- map: homestead.app
to: /home/vagrant/codebase/Laravel/public #-- The path from the vagrant box which is mapped with …
Run Code Online (Sandbox Code Playgroud) 我们需要在当前作业执行时动态调度多个作业。
大概的场景是:
resume_dttime
resume_dttime
所以我的代码是:
from apscheduler.schedulers.blocking import BlockingScheduler
sched = BlockingScheduler()
@sched.scheduled_job('cron', day_of_week='mon-fri', hour=6)
def scheduled_job():
"""
"""
liveusers = todays_userslist() #Get users from table with todays resume_dttime
for u in liveusers:
user_job = get_userjob(u.id)
runtime = u.resume_dttime #eg u.resume_dttime is datetime(2015, 12, 13, 16, 30, 5)
sched.add_job(user_job, 'date', run_date=runtime, args=[u.name])
if __name__ == "__main__":
sched.start()
sched.shutdown(wait=True)
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查询是:
根据我们客户的要求,我必须开发一个应该能够处理大量CSV文件的应用程序.文件大小可以在10 MB - 2GB的范围内.
根据大小,模块决定是Multiprocessing pool
使用普通文件还是使用普通文件来读取文件CSV reader
.但是从观察开始,在测试大小为100 MB的文件的两种模式时,multi processing
需要比平时更长的时间CSV reading
.
这是正确的行为吗?或者我做错了什么?
这是我的代码:
def set_file_processing_mode(self, fpath):
""" """
fsize = self.get_file_size(fpath)
if fsize > FILE_SIZE_200MB:
self.read_in_async_mode = True
else:
self.read_in_async_mode = False
def read_line_by_line(self, filepath):
"""Reads CSV line by line"""
with open(filepath, 'rb') as csvin:
csvin = csv.reader(csvin, delimiter=',')
for row in iter(csvin):
yield row
def read_huge_file(self, filepath):
"""Read file in chunks"""
pool = mp.Pool(1)
for chunk_number in range(self.chunks): #self.chunks = 20
proc …
Run Code Online (Sandbox Code Playgroud) 我只是在检查 django,并试图通过id
作为参数传递给 URL来列出书籍的视图books/urls.py
。但是获取 404 页面未找到错误。当我在浏览器中输入这个 url 时,我没有发现 url 有什么问题:
http://192.168.0.106:8000/books/list/21/
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书店/urls.py
urlpatterns = [
path('admin/', admin.site.urls),
path('books/', include("books.urls"))
]
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设置.py
INSTALLED_APPS = [
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
'books'
]
...
...
...
ROOT_URLCONF = 'bookstore.urls'
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书籍/网址.py
urlpatterns = [
path('home/', create),
path('list/(?P<id>\d+)', list_view),
]
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书籍/views.py
def create(request):
form = CreateForm(request.POST or None, request.FILES or None)
if form.is_valid():
instance = form.save(commit=False)
instance.save()
messages.success(request, "Book Created")
return redirect('/books/list', kwargs={"id":instance.id})
return render(request, "home.html", {"form":form})
def list_view(request, id=None): …
Run Code Online (Sandbox Code Playgroud) 我是Python的新手,目前正在学习lambda
表达.我正在解决一个辅导课程
定义一个函数
max_of_three()
,它将三个数字作为参数并返回其中最大的一个.
我已经阅读了这篇旧文章并尝试过没有成功:
>>> max_of_three = lambda x, y, z : x if x > y else (y if y>z else z)
>>> max_of_three(91,2,322)
91
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为什么不回Z?这是X.
我有一个查询,因为我想准备一个JSFiddle.但它不适用于小型点击程序.其中只有两行代码.
HTML
<body>
<button onclick="javascript:launchdialog();">Click Me</button>
</body>
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JavaScript的
function launchdialog() {
alert('test');
}
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我只是在两行代码中找不到任何错误.请看看这个JSFiddle - http://jsfiddle.net/g47qqzra/
我必须通过 JavaScript 或 jQuery 在嵌套的 JSON 中搜索一个键。在我的 JSON 对象中,所有键都是唯一的。我自己尝试了一些解决方案,但没有奏效。这是我的代码:
json = {
"app": [{
"Garden": {
"Flowers": {
"Red flower": "Rose",
"White Flower": "Jasmine",
"Yellow Flower": "Marigold"
}
},
"Fruits": {
"Yellow fruit 1": "Mango",
"Green fruit 2": "Guava",
"White Flower 3": "groovy"
},
"Trees": {
"label": {
"Yellow fruit 2": [{"type a":"Pumpkin", "type b": "Banana",..}],
"White Flower 2": ["Bogan 1", "Bogan 2", ...]
}
}],...
}
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如何在给定对象中搜索特定键?
如果我通过lookup(json, "type a")
它应该返回"Pumpkin"
,或者如果我搜索"White Flower 2"
它应该返回 ["Bogan …
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django ×1
django-2.0 ×1
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jquery ×1
jsfiddle ×1
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