Modern versions of gdb allow integration of python code to "pretty print" complex data structures. There are some great pretty printer implementations for C++'s STL classes as well as some of the more common boost.org types. In network programming, one commonly encounters select / poll calls. While poll() uses an array of data structures, select() uses fd_set . Has anyone run
gdb的现代版本允许将Python代码集成到“漂亮打印”复杂数据结构。 对于C ++的STL类以及一些更常见的boost.org类型，有一些非常漂亮的打印机实现。 在网络编程中，通常会遇到select / poll调用。 poll()使用一组数据结构， select()使用fd_set 。 有没有人跑过漂亮的打印机实现fd_set ，最好是便携式的，但即使是平台特定的也没关系。 理想情况下，它将是Linux / x86，但我会采取任何措施并希望能够适应。 Alrighty，这是
I'm trying to understand how yield works and after I had read this text I believed that it's quite understandable. However I still don't understand what's the connection between yield and __iter__ because I've just found out that this code works: class Vect(): def __init__(self, *args): self.__a = list(args) def print_(self): print self.__a def _
我试图理解产量是如何工作的，而且在阅读了这篇文章之后，我相信这是可以理解的。 然而，我仍然不明白yield和__iter__之间的联系是什么，因为我刚发现这个代码有效： class Vect(): def __init__(self, *args): self.__a = list(args) def print_(self): print self.__a def __iter__(self): yield self.__a asd = Vect(1,2,3,4,5) for foo in asd: print foo 我认为，当我有一个
meta_map = {} results = db.meta.find({'corpus_id':id, 'method':method}) #this Mongo query only takes 3ms print results.explain() #result is mongo queryset of 2000 documents count = 0 for r in results: count += 1 print count word = r.get('word') data = r.get('data',{}) if not meta_map.has_key(word): meta_map[word] = data
meta_map = {} results = db.meta.find({'corpus_id':id, 'method':method}) #this Mongo query only takes 3ms print results.explain() #result is mongo queryset of 2000 documents count = 0 for r in results: count += 1 print count word = r.get('word') data = r.get('data',{}) if not meta_map.has_key(word): meta_map[word] = data
This question already has an answer here: What does ** (double star/asterisk) and * (star/asterisk) do for parameters? 15 answers How to unzip a list of tuples into individual lists? [duplicate] 2 answers What does the Star operator mean? [duplicate] 5 answers Please explain a python zip and unpacking solution [duplicate] 1 answer How to iterate through two lists in parallel? 7 answ
这个问题在这里已经有了答案： **（双星/星号）和*（星号/星号）对参数做什么？ 15个答案 如何将元组列表解压缩到单个列表中？ [重复] 2个答案 Star运算符是什么意思？ [重复] 5个答案 请解释一个python zip和unpacking解决方案1个答案 如何并行迭代两个列表？ 7个答案
This question already has an answer here: What does the Star operator mean? [duplicate] 5 answers With this: zip(*zipped) you tell python the same as this: zip(zipped[0],zipped[1],zipped[2]) for this basic example. What does exactly that operator When used as argument of a function, it takes the elements of the argument and expand it before passing as argument. For instance: power
这个问题在这里已经有了答案： Star运算符是什么意思？ [重复] 5个答案 有了这个： zip(*zipped) 你告诉python与此相同： zip(zipped[0],zipped[1],zipped[2]) 对于这个基本的例子。 那个操作员究竟做了什么？ 当作为函数的参数使用时，它将获取参数的元素并在作为参数传递之前展开它。 例如： power = [2,3] math.pow(*power) 会给你2 3 = 8的价值。 http://ideone.com/D0R9FB
This question already has an answer here: What does the Star operator mean? [duplicate] 5 answers In python, * is the 'splat' operator. It is used for unpacking a list into arguments. For example: foo(*[1, 2, 3]) is the same as foo(1, 2, 3) . The zip() function takes n iterables, and returns y tuples, where y is the least of the length of all of the iterables provided. The y th
这个问题在这里已经有了答案： Star运算符是什么意思？ [重复] 5个答案 在python中，*是'splat'操作符。 它用于将列表解压缩为参数。 例如： foo(*[1, 2, 3])与foo(1, 2, 3) 。 zip()函数需要n迭代，并返回y元组，其中y是所提供的所有迭代的长度中最小的。 第y个元组将包含所有提供的迭代的第y个元素。 例如： zip(['a', 'b', 'c'], [1, 2, 3]) 会屈服 ('a', 1) ('b', 2) ('c', 3) 对于您提供的示例中的
Possible Duplicate: Python: Once and for all. What does the Star operator mean in Python? x = [1, 2, 3] y = [4, 5, 6] zipped = zip(x, y) list(zipped) x2, y2 = zip(*zip(x, y)) x == list(x2) and y == list(y2) What type of object does *zip(x, y) return? Why res = *zip(x, y) print(res) doesn't work? The asterisk "operator" in Python does not return an object; it's a synt
可能重复： Python：一劳永逸。 Star运算符在Python中意味着什么？ x = [1, 2, 3] y = [4, 5, 6] zipped = zip(x, y) list(zipped) x2, y2 = zip(*zip(x, y)) x == list(x2) and y == list(y2) *zip(x, y)返回什么类型的对象？ 为什么 res = *zip(x, y) print(res) 不起作用？ Python中的星号“运算符”不返回对象; 这是一个句法结构，意思是“用给出的列表作为参数来调用函数”。 所以： x = [1，2，3] F（* x）
This question already has an answer here: What does the Star operator mean? [duplicate] 5 answers zip wants a bunch of arguments to zip together, but what you have is a single argument (a list, whose elements are also lists). The * in a function call "unpacks" a list (or other iterable), making each of its elements a separate argument. So without the * , you're doing zip( [[1
这个问题在这里已经有了答案： Star运算符是什么意思？ [重复] 5个答案 zip想要一堆参数一起压缩，但你拥有的是一个参数（一个列表，其元素也是列表）。 *在函数调用中“解包”一个列表（或其他迭代），使其每个元素成为一个单独的参数。 所以如果没有* ，你在做zip( [[1,2,3],[4,5,6]] ) 。 用* ，你在做zip([1,2,3], [4,5,6]) 。 *运算符在函数调用语句中解压参数。 考虑这一点 def add(x, y): return x + y 如
I'm using itertools.chain to "flatten" a list of lists in this fashion: uniqueCrossTabs = list(itertools.chain(*uniqueCrossTabs)) how is this different than saying: uniqueCrossTabs = list(itertools.chain(uniqueCrossTabs)) * is the "splat" operator: It takes a list as input, and expands it into actual positional arguments in the function call. So if uniqueCrossTabs wa
我正在使用itertools.chain以这种方式“列出”列表清单： uniqueCrossTabs = list(itertools.chain(*uniqueCrossTabs)) 这与说不同有什么不同： uniqueCrossTabs = list(itertools.chain(uniqueCrossTabs)) *是“splat”操作符：它将一个列表作为输入，并将其展开为函数调用中的实际位置参数。 因此，如果uniqueCrossTabs [ [ 1, 2 ], [ 3, 4 ] ] ，则itertools.chain(*uniqueCrossTabs)与itertools.chain([ 1, 2 ], [ 3, 4 ])
Dealing with classes (nested etc) does not look easy in Python , surprisingly ! The following problem appeared to me recently and took several hours (try, search ...) without success. I read most of SO related links but none of them has pointed the issue presented here! #------------------------------------ class A: def __init__(self): self.a = 'a' print self.a class B(A)
处理类（嵌套等）在Python中看起来并不简单， 令人惊讶 ！ 以下问题最近出现在我身上，花了几个小时（尝试，搜索...），但没有成功。 我阅读了大部分与SO相关的链接，但他们都没有指出这里介绍的问题！ #------------------------------------ class A: def __init__(self): self.a = 'a' print self.a class B(A): def __init__(self): self.b = 'b' A.a = 'a_b' print se