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深入解析ATL - 文档下载 - 清泛网 - 专注C/C++及内核技术

...第1章 你好,ATL  1.1 什么是ATL  1.2 创建COM服务器  1.3 插入COM类  1.4 添加属性和方法  1.5 实现附加接口  1.6 脚本支持  1.7 添加永久性  1.8 添加和激发事件显示全部信息第2版序 第1版序 前言 致谢 关于作者 联系...
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Array slices in C#

... 13 Note using Array.Copy performs a lot faster than using LINQ's Take or Skip methods. – Michael Jun 23 ...
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How to do multiple arguments to map function where one remains the same in python?

... One option is a list comprehension: [add(x, 2) for x in [1, 2, 3]] More options: a = [1, 2, 3] import functools map(functools.partial(add, y=2), a) import itertools map(add, a, itertools.repeat(2, len(a))) ...
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Why isn't my Pandas 'apply' function referencing multiple columns working? [closed]

... 381 Seems you forgot the '' of your string. In [43]: df['Value'] = df.apply(lambda row: my_test(r...
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Convert a matrix to a 1 dimensional array

I have a matrix (32X48). 10 Answers 10 ...
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Should I use a data.frame or a matrix?

... | edited Aug 30 '17 at 0:12 answered Mar 1 '11 at 19:00 ...
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Weighted random numbers

...andom number that is 0 or greater and is less than the sum of the weights 3) go through the items one at a time, subtracting their weight from your random number, until you get the item where the random number is less than that item's weight Pseudo-code illustrating this: int sum_of_weight = 0; f...
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Finding the index of elements based on a condition using python list comprehension

... of a list. >>> import numpy >>> a = numpy.array([1, 2, 3, 1, 2, 3]) >>> a array([1, 2, 3, 1, 2, 3]) >>> numpy.where(a > 2) (array([2, 5]),) >>> a > 2 array([False, False, True, False, False, True], dtype=bool) >>> a[numpy.where(a > 2...
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Best way to find the intersection of multiple sets?

...tiple arguments to set.intersection(), like u = set.intersection(s1, s2, s3) If the sets are in a list, this translates to: u = set.intersection(*setlist) where *a_list is list expansion Note that set.intersection is not a static method, but this uses the functional notation to apply intersec...
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How do I compare two hashes?

... 163 You can compare hashes directly for equality: hash1 = {'a' => 1, 'b' => 2} hash2 = {'a' =...