SimpleBlobDetector我正在尝试按照这个 SO 答案使用 OpenCV 检测大污点。
这是输入图像：
我首先尝试使用params.filterByArea = False，这样可以检测到大片黑色污渍：
然而，较小的斑点最终也被检测到了。因此我切换params.filterByArea = True希望强制执行对象区域的标准。
然而，当设置params.filterByArea = True时，params.minArea = 10最大的污渍不再被识别：
我尝试使用其他minArea参数但没有效果，甚至尝试了minArea0，这应该相当于根本没有过滤。
我在这里遗漏了什么？
考虑以下数据库：
from sqlalchemy import String, select, create_engine
from sqlalchemy.orm import DeclarativeBase, Mapped, mapped_column, Session
class Base(DeclarativeBase):
pass
class User(Base):
__tablename__ = "user_account"
name: Mapped[str] = mapped_column(String(30))
surname: Mapped[str] = mapped_column(String(30))
def __repr__(self):
return f"User(name={self.name!r}, surname={self.surname!r})"
engine = create_engine("sqlite+pysqlite:///test3.sqlite3", echo=True)
Base.metadata.create_all(engine)
with Session(engine) as session:
user1 = User(
name="Mario",
surname="Rossi"
)
user2 = User(
name="Mario",
surname="Bianchi",
)
user3 = User(
name="Giovanni",
surname="Bianchi",
)
session.add_all([user1, user2, user3])
session.commit()
现在假设我有一个想要查找的用户列表：
users = [("Mario", "Rossi"), ("Giovanni", "Bianchi")]
然后我会运行：
names = [name for name, _ in users]
surnames = [surname for _, surname in users]
with Session(engine) as session:
stmt = select(User).where(User.name.in_(names)).where(User.surname.in_(surnames))
print(session.execute(stmt).scalars().all())
返回：
[User(name='Mario', surname='Rossi'), User(name='Mario', surname='Bianchi'), User(name='Giovanni', surname='Bianchi')]
但“Mario Bianchi”不在我的输入用户列表中。
我如何连接 IN 语句以便仅选择成对对应关系？
也就是说，如果我有varA in (el1, el2) AND varB IN (el3, el4)，我不想选择带有varA==el1 AND varB==el4
我想用 Python 实现类似于这个 OCaml 的东西：
let example = fun v opt_n ->
let fltr = fun i -> i mod 2 = 0
let fltr = match opt_n with
| None -> fltr
| Some n -> fun i -> i mod n = 0 && fltr(n)
fltr v
这很容易组合/扩展，我可以在运行时添加任意数量的谓词。这当然是一个简化的示例，在现实生活中，我有许多可选的包含/排除集，以及用于成员资格的谓词检查。
在 Python 中以简单的方式执行此操作会失败：
def example(v: int, opt_n=None):
"""
doesn't work!
"""
# doesn't need to be a lambda, an explicitely defined function fails too
fltr = lambda i: i % 2 == 0
if opt_n is not None:
# fails miserably -> maximum recursion depth exceeded
fltr = lambda i: fltr(i) and i % opt_n == 0
return fltr(v)
example(10, 5)
这很烦人，因为似乎fltr只能在作业的左侧出现一次，所以我必须fltr在之后的每种情况下内联首字母：
def example(v: int, opt_n=None, opt_m=None):
"""annoying but works"""
fltr = None
# some inital filters
pred_0 = lambda _: True # do some real checks ...
pred_1 = lambda _: True # do some real checks ...
if opt_n is not None:
# fltr is inlined, only appears on left side, now it works
fltr = lambda i: pred_0(i) and pred_1(i) and opt_n % 2 == 0
if opt_m is not None:
# much repetition
fltr = lambda i: pred_0(i) and pred_1(i) and opt_n % 3 == 0
if fltr is None:
# inlined again
fltr = lambda i: pred_0(i) and pred_1(i)
return fltr(v)
有什么方法可以解决我的混乱，也许我遗漏了一些东西，以及/或者在 Python 中组合谓词的推荐方法是什么？
我正在运行这段代码，使用 PGP 公钥加密文件。
import gnupg
def pgp_encrypt(pub_file, out_file):
gpg = gnupg.GPG()
with open(pub_file, 'rb') as pgp_pub_key:
public_key_data = pgp_pub_key.read()
# import_keys_file() is NOT used as the key
# eventually will come from user-input
import_result = gpg.import_keys(public_key_data)
if import_result.count == 0:
print("Error: No keys imported. Make sure the public key file is correct.")
exit()
pgp_key_id = import_result.results[0]['fingerprint']
plaintext_data = b'This is the TEST data to encrypt'
encrypted_data = gpg.encrypt(
plaintext_data,
recipients=[pgp_key_id],
always_trust=True
)
if encrypted_data.ok:
print("Data encrypted successfully.")
print(encrypted_data.data)
with open(out_file, 'wb') as encrypted_file:
encrypted_file.write(encrypted_data.data)
else:
print("Encryption failed:")
print(encrypted_data.status)
## Apply
pgp_encrypt('pgp_pubkey.asc', 'pgp_encrypted_file')
因此，基本上它是读取公钥文件并将数据放入，然后在加密文件之前public_key_data使用它导入。gpg.import_keys(public_key_data)
print(encrypted_data.data)我的代码无法正确打印结果（在屏幕上，这是另一个需要解决的问题）但它可以正常工作：
santanu@mgtucrpi5:~/Scripts/gnupg $ python pgp_encrypt.py
Data encrypted successfully.
b'-----BEGIN PGP MESSAGE-----\n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\npROcEBIU2jtazibCiYD1uIy+lwS4w0en8ysFPrLnJuWcFQ==\n=UR1e\n-----END PGP MESSAGE-----\n'
但是当我从虚拟环境运行它时，出现以下错误：
santanu@mgtucrpi5:~/Scripts/gnupg $ source pgpenv/bin/activate
(pgpenv) santanu@mgtucrpi5:~/Scripts/gnupg $ python pgp_encrypt.py
Traceback (most recent call last):
File "/home/santanu/Scripts/gnupg/pgp_encrypt.py", line 34, in <module>
pgp_encrypt('pgp_pubkey.asc', 'pgp_encrypted_file')
File "/home/santanu/Scripts/gnupg/pgp_encrypt.py", line 11, in pgp_encrypt
if import_result.count == 0:
^^^^^^^^^^^^^^^^^^^
AttributeError: 'ImportResult' object has no attribute 'count'. Did you mean: 'counts'?
如果我将其更改为counts（只是尝试），我会收到不同的错误，而且看起来也不正确：
File "/home/santanu/Scripts/gnupg/pgpenv/lib/python3.11/site-packages/gnupg/gnupg.py", line 1064, in encrypt
result = self._encrypt(stream, recipients, **kwargs)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
TypeError: GPGBase._encrypt() got multiple values for argument 'recipients'
我该如何解决这个问题？
参考https://gnupg.readthedocs.io/en/latest/#importing-and-receiving-keys
=========================================
回答@fqlenos 的问题：
这是我创建虚拟环境的方式：
% cd ~/Scripts/gnupg
% python3 -m venv pgpenv
% source pgpenv/bin/activate
% pip install gnupg
我也尝试了pip install python-gnupg，但结果相同。我是否遗漏了什么或做错了什么？
目前，我想更改nRF Connect 应用程序上显示的Pico 2W设备的名称。我在 VsCode 中使用Micro Pico。我尝试了很多方法，但名称始终为“ N/A ”。
import bluetooth
from time import sleep
bluetooth.BLE().active(True)
ble = bluetooth.BLE()
device_name = 'PicoW_Device'
advertising_data = b'\x02\x01\x06\x03\x03\x0F\x18\x09' + bytes(device_name, 'utf-8')
ble.gap_advertise(100, adv_data=advertising_data)
print(f"The Name of the device is currently: {device_name}")
while True:
sleep(1)
我尝试更改名称，但始终显示“ N/A ”
我最近发现 Django 电子商务网站存在一个严重的安全问题，用户可以在将商品添加到购物车之前修改商品价格。
许多开发人员允许用户从前端发送价格数据，可以使用 Burp Suite 或浏览器开发人员工具轻松篡改这些数据。
考虑一个将商品添加到购物车的简单 Django 视图：
def add_item(request):
product_id = request.GET.get('product_id')
price = request.GET.get('price') # User-controlled value (security risk)
qty = int(request.GET.get('qty', 1))
cart_item = {
'product_id': product_id,
'qty': qty,
'price': price, # This price comes from the user, not the database!
}
request.session['cart'] = request.session.get('cart', {})
request.session['cart'][product_id] = cart_item
request.session.modified = True
return JsonResponse({'message': 'Added to cart'})
price字段更改为$1，并转发请求。很想听听你对此的想法！
我该如何编写代码以便当我收到“hello”时它运行一个函数？
import serial
ser = serial.Serial(port='/dev/tty.usbmodem11201',baudrate=9600)
while True:
value= ser.readline()
valueInString=str(value, 'UTF-8')
print(valueInString)
if ser.is_open == hello:
print("Hi")
这是我所看到的，但它不起作用。我该如何让它工作？谢谢！
https://open.kattis.com/problems/lyklagangriti?editresubmit=16843522 这个问题基本上就是你输入密码的地方。每次输入中出现 L，光标就会“移动”到左边，右边也是一样，当出现 B 时，你就删除你所在的字符。
text = input()
chars = list(text)
cursor = 0
password = []
for char in chars:
if char == 'L':
cursor -= 1
elif char == 'R':
cursor += 1
elif char == 'B':
password.pop(cursor - 1)
cursor -= 1
else:
password.insert(cursor, char)
cursor += 1
print(*password, sep='')
我正在 Kattis 上解决这个问题，到目前为止，我已经通过了很多测试用例，但我似乎超出了测试组 4 和 5 的时间限制。有没有什么方法可以加快这个代码的速度？
我需要暂停程序直到用户按下某个键，然后它应该返回所按下的键。
我看过的大多数地方都给出了以下两种解决方案之一：
# This doesn't work with what I need for obvious reasons
return input("Enter key here >>> ")
# but sometimes they'll use something like this:
import keyboard
while True:
event = keyboard.read_event()
if event.event_type == keyboard.KEY_DOWN:
return event.name
但是第二个例子，虽然它确实只在第一次按下时返回按键，正如预期的那样，但这种方法存在一个问题，即如果我在使用这段代码（key = input("Enter key here >>> ")）后进行常规输入，则按下的键就会输入到输入中。
我怎样才能仅在第一次按下（而不是按住）时返回按键，并且还用完该按键（不让按键执行任何其他操作）？
我正在阅读《用 Python 自动化无聊内容》的最后一章 - 第 20 章从安装和导入 pyautogui 开始，但我一直无法完成此操作。我已经能够在 Mac 上安装该模块，但我无法将其作为第三方模块添加到 Mu 编辑器中。这是我尝试将其作为模块添加到 Mu 编辑器时收到的错误：
Collecting pyautogui
Using cached PyAutoGUI-0.9.54.tar.gz (61 kB)
Installing build dependencies: started
Installing build dependencies: finished with status 'done'
Getting requirements to build wheel: started
Getting requirements to build wheel: finished with status 'done'
Preparing metadata (pyproject.toml): started
Preparing metadata (pyproject.toml): finished with status 'done'
Collecting pygetwindow>=0.0.5
Using cached PyGetWindow-0.0.9.tar.gz (9.7 kB)
Preparing metadata (setup.py): started
Preparing metadata (setup.py): finished with status 'done'
Collecting mouseinfo
Using cached MouseInfo-0.1.3.tar.gz (10 kB)
Preparing metadata (setup.py): started
Preparing metadata (setup.py): finished with status 'done'
Collecting pymsgbox
Using cached PyMsgBox-1.0.9.tar.gz (18 kB)
Installing build dependencies: started
Installing build dependencies: finished with status 'done'
Getting requirements to build wheel: started
Getting requirements to build wheel: finished with status 'done'
Preparing metadata (pyproject.toml): started
Preparing metadata (pyproject.toml): finished with status 'done'
Collecting pyobjc-framework-quartz
Using cached pyobjc_framework_Quartz-10.3.2-cp38-cp38-macosx_11_0_universal2.whl (211 kB)
Collecting pyobjc-core
Using cached pyobjc_core-11.0.tar.gz (994 kB)
Installing build dependencies: started
Installing build dependencies: finished with status 'done'
Getting requirements to build wheel: started
Getting requirements to build wheel: finished with status 'error'
error: subprocess-exited-with-error
× Getting requirements to build wheel did not run successfully.
│ exit code: 1
╰─> [2 lines of output]
<string>:18: DeprecationWarning: pkg_resources is deprecated as an API. See https://setuptools.pypa.io/en/latest/pkg_resources.html
PyObjC: Need at least Python 3.9
[end of output]
note: This error originates from a subprocess, and is likely not a problem with pip.
error: subprocess-exited-with-error
× Getting requirements to build wheel did not run successfully.
│ exit code: 1
╰─> See above for output.
note: This error originates from a subprocess, and is likely not a problem with pip.
FINISHED
我已经能够使用以下命令在我的 Mac 上成功安装 pyautogui：pip3 install pyautogui
我正在使用 python3.9 我在 macOS Sequia 15.3.1 上
但是，当我尝试将其作为第三方包添加到 Mu 编辑器时，它会退出并出现上述错误。
我搜索了互联网以寻找答案，并尝试过：
我希望安装能够顺利完成，以便我可以将其导入 REPL 并使用 pyautogui 进行练习。
我特别困惑为什么它说 PyObjC：至少需要 Python 3.9，即使我正在运行 Python 3.9.6。
请帮忙！