给定一个docprep.pyx简单的文件
from spacy.structs cimport TokenC
print("loading")
Run Code Online (Sandbox Code Playgroud)
并试图通过
cythonize -3 -i docprep.pyx
Run Code Online (Sandbox Code Playgroud)
我收到以下错误消息
docprep.c:613:10: fatal error: ios: No such file or directory
#include "ios"
^~~~~
compilation terminated
Run Code Online (Sandbox Code Playgroud)
从路径中可以看出,该系统安装了 Python 3.7 的 Anaconda。numpy,spacy并且cython都是通过conda.
如果我有这样的MainActivity:
public class MainActivity extends AppCompatActivity
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
// Set up database
RealmConfiguration realmConfiguration = new RealmConfiguration.Builder(this).build();
Realm.deleteRealm(realmConfiguration); // Clean slate
Realm.setDefaultConfiguration(realmConfiguration); // Make this Realm the default
realm = Realm.getDefaultInstance();
}
@Override
public void onDestroy() {
realm.close();
super.onDestroy();
}
}
Run Code Online (Sandbox Code Playgroud)
我realm.getDefaultInstance()在另一个类(相同的线程)中使用这样的:
public class ViewBookActivity extends Activity {
private Realm realm;
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_scan_result);
realm = Realm.getDefaultInstance();
}
}
Run Code Online (Sandbox Code Playgroud)
我应该再调用realm.close()中onDestroy()的ViewBookActivity?或者关闭它是否足够MainActivity? …
假设我有一个排序算法和两个自定义比较器:
mySort :: [[Int]] -> ([Int] -> [Int] -> Ordering) -> [[Int]]
mySort = undefined
myCmp1 :: [Int] -> [Int] -> Ordering
myCmp1 xs ys
| a0 < b0 = LT
| a0 > b0 = GT
| a1 < b1 = LT
| a1 > b1 = GT
| a1 == b1 = EQ where
a0 = head xs
b0 = head ys
a1 = last xs
b1 = last ys
myCmp2 :: [Int] -> [Int] -> Ordering
myCmp2 …Run Code Online (Sandbox Code Playgroud) 如何使用MATLAB将下图中的两个连接圆分开?我尝试过使用imerode,但这并没有给出好的结果.腐蚀不起作用,因为为了腐蚀足以分离圆圈,线条消失或变得严重.在其他起始图片中,圆和线重叠,因此隔离重叠的对象也不起作用.
图像显示标识的bwboundaries对象,每个对象绘制不同的颜色.正如你所看到的那样,两个淡蓝色圆圈相连,我想将它们分开,产生两个独立的圆圈.谢谢

我正在使用FileCookieStore它需要一个 cookiefile 的路径,并且该文件已经创建。我想使用fs.writeFile,如果它不存在创建此文件,但如果它来覆盖它确实存在。但是,在下面的实现中,即使 cookiefile 存在,它也会覆盖 cookiefile 的内容,即使wx-flag 应该抛出错误并且如果文件已经存在,则不会覆盖该文件。
var cookiePath = "cookie.json"
fs.writeFile(cookiePath, null, { flags: 'wx' }, function (err) {
if (err) throw err;
});
var j = request.jar(new FileCookieStore(cookiePath));
request = request.defaults({ jar : j });
function login(callback) {
// puts data into j which automatically writes the content to cookiePath
}
Run Code Online (Sandbox Code Playgroud)
我是否误解writeFile了 -wx标志的正确使用?如果我运行login()cookie的内容自动保存到cookie.json,但如果我不调用再次运行该文件login,fs.writeFile清空cookiefile。
给定 5 万个网站 url 的列表,我的任务是找出其中哪些是可用的/可访问的。这个想法只是向HEAD每个 URL发送一个请求并查看状态响应。从我听到一个异步方法是要去的地方,现在我使用的是asyncio用aiohttp。
我想出了以下代码,但速度非常糟糕。在我的 10 兆位连接上,1000 个 URL 大约需要 200 秒。我不知道期望的速度是多少,但我是 Python 异步编程的新手,所以我想我在某个地方走错了地方。如您所见,我已尝试将允许的同时连接数增加到 1000(从默认值 100 增加)以及 DNS 解析保留在缓存中的持续时间;都没有什么大的影响。环境有 Python 3.6 和 aiohttp3.5.4。
与问题无关的代码审查也受到赞赏。
import asyncio
import time
from socket import gaierror
from typing import List, Tuple
import aiohttp
from aiohttp.client_exceptions import TooManyRedirects
# Using a non-default user-agent seems to avoid lots of 403 (Forbidden) errors
HEADERS = {
'user-agent': ('Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_5) '
'AppleWebKit/537.36 (KHTML, like …Run Code Online (Sandbox Code Playgroud) 使用 Pandas 1.0.1,如果
df = df.merge(df2, on=some_column)
Run Code Online (Sandbox Code Playgroud)
产量
File /home/torstein/code/fintechdb/Sheets/sheets/gild.py, line 42, in gild
df = df.merge(df2, on=some_column)
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/frame.py, line 7297, in merge
validate=validate,
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/reshape/merge.py, line 88, in merge
return op.get_result()
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/reshape/merge.py, line 643, in get_result
join_index, left_indexer, right_indexer = self._get_join_info()
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/reshape/merge.py, line 862, in _get_join_info
(left_indexer, right_indexer) = self._get_join_indexers()
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/reshape/merge.py, line 841, in _get_join_indexers
self.left_join_keys, self.right_join_keys, sort=self.sort, how=self.how
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/reshape/merge.py, line 1311, in _get_join_indexers
zipped = zip(*mapped)
File /home/torstein/anaconda3/lib/python3.7/site-packages/pandas/core/reshape/merge.py, line 1309, in <genexpr>
for …Run Code Online (Sandbox Code Playgroud) 想象一下,我有一些看起来像这样的类:
class Car {
private Image carImage;
public Car(int imageIndex) {
switch (imageIndex) {
case 1: carImage = generateCarImage(1); break;
# and so forth
}
}
}
class Audi extends Car {
private int numberOfSeats;
public Audi(int imageIndex, int numberOfSeats) {
super(imageIndex);
this.numberOfSeats = numberOfSeats;
}
}
Run Code Online (Sandbox Code Playgroud)
现在假设我使用相同的图像创建多个奥迪:
Audi car1 = new Audi(1,2);
Audi car2 = new Audi(1,3);
Run Code Online (Sandbox Code Playgroud)
car1和car2会延伸相同的物体吗?我假设没有,但有没有办法可以做到这一点?我问,因为我想避免两次生成和存储相同的图像.
编辑:
这两个奥迪会参考同一辆车吗,例如图像只生成和存储一次,对一个的任何变化都会影响另一个吗?
class Car {
private Image carImage;
public Car(int imageIndex) {
switch (imageIndex) {
case 1: # carImage = readfile(1.jpeg) …Run Code Online (Sandbox Code Playgroud) 我conda在 bash 终端中有一个环境,带有 Intel Python Distribution 解释器。但是,在导入包时,它们是从系统默认 Python 的用户目录中导入的,而不是从环境中导入的。查看版本差异和包的__spec__来源pandas。
~ ? $ ? conda activate idp
~ ? $ ? which python
~/anaconda3/envs/idp/bin/python
~ ? $ ? python
Python 3.6.8 |Intel Corporation| (default, Mar 1 2019, 00:10:45)
[GCC 4.8.2 20140120 (Red Hat 4.8.2-15)] on linux
Type "help", "copyright", "credits" or "license" for more information.
Intel(R) Distribution for Python is brought to you by Intel Corporation.
Please check out: https://software.intel.com/en-us/python-distribution
>>> import pandas
>>> pandas.__version__ …Run Code Online (Sandbox Code Playgroud)