在VS C++代码中,如果我没有选择任何内容或选择完整行并按下注释选择(Ctrl + K + Ctrl + C),那么它将使用//注释整行
int x = 5;
按Ctrl + K + Ctrl + C后不选择任何内容或选择全行.
// int x = 5;
现在,如果我选择该行的某些部分并再次按下注释按钮,则仅评论所选文本(粗体表示已选中)
int x = 5 ;
按下Ctrl + K + Ctrl + C并选择x = 5.
int /*x = 5*/;
包含多条线
int x = 5;
int y = 2;
int z = x*5;
评论后快捷方式
int/* x = 5;
int y = 2;
int z =*/ x * 5;
我想要的是
//int x = 5;
//int y = 2; …我安装了多个版本的Boost(Windows 7/MinGW).我需要使用一个特定的(1.53.0).
我在CMakeFiles.txt文件中定义了BOOST_ROOT: SET(BOOST_ROOT C:/boost_1_53_0/)但是我一直收到这个错误:
> cmake .
BOOST_ROOT=C:/boost_1_53_0/
CMake Error at C:/Program Files (x86)/CMake 2.8/share/cmake-2.8/Modules/FindBoost.cmake:1191 (message):
Unable to find the requested Boost libraries.
Boost version: 1.48.0
Boost include path: C:/Boost/include/boost-1_48
Detected version of Boost is too old. Requested version was 1.53 (or
newer).
The following Boost libraries could not be found:
boost_filesystem
No Boost libraries were found. You may need to set BOOST_LIBRARYDIR to the
directory containing Boost libraries or BOOST_ROOT to the location of
Boost.
我还将BOOST_ROOT定义为环境变量,但结果相同.
为什么cmake还在寻找旧版本？
当我没有明确指定应该使用密钥时,对元组列表(字典键,密钥是随机字符串的值对)进行排序更快(编辑:在@Chepner的评论中添加了operator.itemgetter(0)和密钥版现在更快!):
import timeit
setup ="""
import random
import string
random.seed('slartibartfast')
d={}
for i in range(1000):
d[''.join(random.choice(string.ascii_uppercase) for _ in range(16))] = 0
"""
print min(timeit.Timer('for k,v in sorted(d.iteritems()): pass',
setup=setup).repeat(7, 1000))
print min(timeit.Timer('for k,v in sorted(d.iteritems(),key=lambda x: x[0]): pass',
setup=setup).repeat(7, 1000))
print min(timeit.Timer('for k,v in sorted(d.iteritems(),key=operator.itemgetter(0)): pass',
setup=setup).repeat(7, 1000))
得到:
0.575334150664
0.579534521128
0.523808984422 (the itemgetter version!)
但是,如果我创建一个自定义对象key=lambda x: x[0]显式传递sorted使其更快:
setup ="""
import random
import string
random.seed('slartibartfast')
d={}
class A(object):
def __init__(self):
self.s …这个:
import numpy as np
a = np.array([1, 2, 1])
w = np.array([[.5, .6], [.7, .8], [.7, .8]])
print(np.dot(a, w))
# [ 2.6 3. ] # plain nice old matrix multiplication n x (n, m) -> m
import tensorflow as tf
a = tf.constant(a, dtype=tf.float64)
w = tf.constant(w)
with tf.Session() as sess:
print(tf.matmul(a, w).eval())
结果是:
C:\_\Python35\python.exe C:/Users/MrD/.PyCharm2017.1/config/scratches/scratch_31.py
[ 2.6 3. ]
# bunch of errors in windows...
Traceback (most recent call last):
File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\common_shapes.py", line 671, in _call_cpp_shape_fn_impl
input_tensors_as_shapes, status) …我遇到的一件事是服务类(比如JBoss服务)由于帮助器内部类而变得过大.我还没有找到一个打破课堂的好方法.这些助手通常是线程.这是一个例子:
/** Asset service keeps track of the metadata about assets that live on other
* systems. Complications include the fact the assets have a lifecycle and their
* physical representation lives on other systems that have to be polled to find
* out if the Asset is still there. */
public class AssetService
{
//...various private variables
//...various methods
public AssetService()
{
Job pollerJob = jobService.schedule( new AssetPoller() );
Job lifeCycleJob = jobService.schedule( AssetLifecycleMonitor() );
}
class AssetPoller
{ …我知道,有几个这种类型的问题,但我尝试了所有这些,但它仍然无效.
好的,对于我的应用; 我有一个活动.在此活动中,有4个选项卡,第四个包含列表和记录按钮.当我推送记录时,有一个GPS监听器启动.获取新的gps值后,它会将其推送到列表中.
到目前为止这个工作!
如果我单击主页按钮它仍然有效,如果我长按它.它恢复了具有特定Tab的Activity,并且列表仍然保持listitems并且gps监听器仍处于活动状态.
这个工作也没关系!
现在我想添加一个通知,它将列表的gps值计数显示为.number.在每个新的gps信号上,它会使用新号码更新通知图标.这没问题,但点击通知的动作完全搞砸了我的申请.
实际代码如下所示:
public void callNotify(String text) {
notif = new Notification();
Intent contentIntent = new Intent(this, tabactivity.class);
contentIntent.putExtra("fid", this.specialvalue);
contentIntent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP
| Intent.FLAG_ACTIVITY_SINGLE_TOP);
notif.icon = R.drawable.icon;
notif.setLatestEventInfo(this, getString(R.string.app_name), text,
PendingIntent.getActivity(this.getBaseContext(), 0, contentIntent,
PendingIntent.FLAG_UPDATE_CURRENT));
notif.ledARGB = Color.RED;
notif.ledOnMS = 200;
notif.ledOffMS = 200;
notif.flags = Notification.FLAG_SHOW_LIGHTS
| Notification.FLAG_ONGOING_EVENT
| Notification.FLAG_ONLY_ALERT_ONCE;
notif.number = notifyNumber;
mNotificationManager.notify(notifyNumber, notif);
}
public void updateNotify(String text) {
notifyNumber++;
notif.number = (int) (notifyNumber / 2);
Intent contentIntent = new Intent(this, tabactivity.class);
contentIntent.putExtra("fid", this.specialvalue); …我试图有一个恒定的gps监听器,每隔x分钟将其位置(长和lat坐标)发送到Web服务器.在按钮上单击它还会将其位置发送到Web服务器.我知道要获取gps信号,你可以输入寻找位置的频率,但是如何编写一个程序可以获得gps位置并每x分钟发送一次坐标(即使在背景中也没有按下按钮) ？
//在点击中
LocationManager lm = (LocationManager) getSystemService(Context.LOCATION_SERVICE);
locationManager.requestLocationUpdates(
LocationManager.GPS_PROVIDER, whatshouldIputhere?, 0, this);
和
public void onLocationChanged(Location location) {
if (location != null) {
double lat = location.getLatitude();
double lng = location.getLongitude();
}
}
我有一个数据帧
df = pd.DataFrame(columns = ["AA", "BB", "CC"])
df.loc[0]= ["a", "b", "c1"]
df.loc[1]= ["a", "b", "c2"]
df.loc[2]= ["a", "b", "c3"]
我需要在标题中添加secod行
df.columns = pd.MultiIndex.from_tuples(zip(df.columns, ["DD", "EE", "FF"]))
我的df现在
AA BB CC
DD EE FF
0 a b c1
1 a b c2
2 a b c3
但是当我将这个数据帧写入csv文件时
df.to_csv("test.csv", index = False)
我按预期更多地获得一排
AA,BB,CC
DD,EE,FF
,,
a,b,c1
a,b,c2
a,b,c3
我正在尝试实现以下内容:
$ prog.py -h
usage: prog.py [-h] [-s | -m] [[-y [year]] | [[-1 | -3] [month] [year]]]
但是,无论我如何玩add_argument_group和add_mutually_exclusive_group,
#!/usr/bin/env python
import argparse
def main(opt):
print(opt)
if __name__ == '__main__':
parser = argparse.ArgumentParser()
bar = parser.add_mutually_exclusive_group()
bar.add_argument('-s', action='store_true', default=True)
bar.add_argument('-m', action='store_true', default=False)
#bar = parser.add_argument_group()
bar = parser.add_mutually_exclusive_group()
bar.add_argument('-y', metavar='year', type=int,
dest='iy', nargs='?', default=0)
baz = bar.add_argument_group()
g_13 = baz.add_mutually_exclusive_group()
g_13.add_argument('-1', dest='i1',
help='Display single month output.',
action='store_true', default=True)
g_13.add_argument('-3', dest='i3',
help='Display prev/current/next month output.',
action='store_true', default=False)
#aaa = bar.add_argument_group()
baz.add_argument(metavar='month', …在构造函数内部,有一个连接:
connect(&amskspace::on_board_computer_model::self(),
SIGNAL(camera_status_changed(const amskspace::camera_status_t&)),
this,
SLOT(set_camera_status(const amskspace::camera_status_t&)));
方法:
void camera_model::
set_camera_status(const amskspace::camera_status_t& status) {
disconnect(&amskspace::on_board_computer_model::self(),
SIGNAL(camera_status_changed(const amskspace::camera_status_t&)),
this,
SLOT(set_camera_status(const amskspace::camera_status_t&)));
// do the job
}
我想在第一次通话后断开此插槽.
问题是:有没有办法只调用一次插槽？没有明确的断开连接？像单枪法一样？可能吗？