大约有 46,100 项符合查询结果（耗时：0.0364秒） [XML]
游戏中的算法与计算思维 03：用洗牌理解随机与循环 · App Inventor 2 中文网
...生成 App 鸿蒙 · 一套积木，多端原生打包！ 下载 AI2伴侣 · 安卓/鸿蒙/iOS aia 源码商店 中文社区 中文教育版 各版本对比 关于 关于我们 发布日志 服务条款 ...
How to convert a unix timestamp (seconds since epoch) to Ruby DateTime?
...must be converted to a string:
require 'date'
DateTime.strptime("1318996912",'%s')
share
|
improve this answer
|
follow
|
...
Python, compute list difference
... matter, you can simply calculate the set difference:
>>> set([1,2,3,4]) - set([2,5])
set([1, 4, 3])
>>> set([2,5]) - set([1,2,3,4])
set([5])
share
|
improve this answer
...
Most popular screen sizes/resolutions on Android phones [closed]
...
You can see the resolutions for those categories in the Table 2, in this section: http://developer.android.com/guide/practices/screens_support.html#testing
share
|
improve this answer
...
How can I use numpy.correlate to do autocorrelation?
... result = numpy.correlate(x, x, mode='full')
return result[result.size/2:]
You will, of course, need error checking to make sure that x is actually a 1-d array. Also, this explanation probably isn't the most mathematically rigorous. I've been throwing around infinities because the definition o...
How do I use the includes method in lodash to check if an object is in the collection?
...
226
The includes (formerly called contains and include) method compares objects by reference (or m...
Install a module using pip for specific python version
On Ubuntu 10.04 by default Python 2.6 is installed, then I have installed Python 2.7. How can I use pip install to install packages for Python 2.7.
...
Regular expression to match URLs in Java
...
|
edited Sep 29 '13 at 20:09
Alan Moore
66.6k1111 gold badges8787 silver badges145145 bronze badges
...
Are Javascript arrays sparse?
...
|
edited May 23 '17 at 11:54
Community♦
111 silver badge
answered Oct 2 '09 at 17:21
...
Minimizing NExpectation for a custom distribution in Mathematica
...ur definition above into simple functions:
pdf[a_, b_, m_, s_, x_] := (1/(2*(a + b)))*a*b*
(E^(a*(m + (a*s^2)/2 - x))*Erfc[(m + a*s^2 - x)/(Sqrt[2]*s)] +
E^(b*(-m + (b*s^2)/2 + x))*Erfc[(-m + b*s^2 + x)/(Sqrt[2]*s)])
pdf2[a_, b_, m_, s_, x_] := pdf[a, b, m, s, Log[x]]/x;
If we plot pdf2 ...