Imagine I have a function that looks like this : myFunction(arg, arg, kwarg, arg, arg, kwarg, etc...): Where arg is an *arg and kwarg is a *kwarg. Before now, my function looked like myFunction(*args): and I was using just a long list of *args and I would just pass in a big list like this myFunction(*bigList): The bigList looked like = [[1,2,3],[4,5,6],'hello',[1,3,5],[2,4,6],'wo
想象一下，我有一个看起来像这样的函数： myFunction(arg, arg, kwarg, arg, arg, kwarg, etc...): arg是* arg，kwarg是* kwarg。 在此之前，我的函数看起来像myFunction(*args):并且我只使用了一个很长的* args列表，并且我会传入一个像这样的大列表 myFunction(*bigList): bigList看起来像= [[1,2,3],[4,5,6],'hello',[1,3,5],[2,4,6],'world',etc...] 但是，现在我需要每三分钟就有一次克瓦格。 所以
This is something that's bugged me for awhile now: def test (*args, **kwargs): print target test(foo='bar', target='baz') I would presume that target='test' in the aFunc call at the bottom would end up in kwargs (and it does), and I would also presume that **would unpack kwargs in the function call, so target would exist as a keyword argument inside of aFunc. It doesn't.
这是现在有一段时间给我听的： def test (*args, **kwargs): print target test(foo='bar', target='baz') 我会假设在底部的aFunc调用中的target='test'最终会出现在kwargs中（我们也会这样做），并且我还会认为**会在函数调用中解压kwargs，所以target会作为关键字存在aFunc内部的参数。 它没有。 我知道它是以字典的形式出现的，但我需要在参数列表中解开这个字典。 这可能吗？ 总之，有没有办法让* args和
I'm working with xlwt which has a 4k limit on how many styles can be defined in an excel doc. Normally, one creates styles like so: style = xlwt.easyxf("font: bold 1") Which I simply replaced with def cached_easyxf(self, format): return self._cache.setdefault(format, xlwt.easyxf(format)) Which works perfectly. Now, I've found out that I need to pass in keyword arguments sometim
我正在使用xlwt ，它可以在excel文档中定义多少种样式，具有4k的限制。 通常，创建如下所示的样式： style = xlwt.easyxf("font: bold 1") 我简单地取而代之 def cached_easyxf(self, format): return self._cache.setdefault(format, xlwt.easyxf(format)) 这完美地工作。 现在，我发现我需要传递关键字参数，这让我想到：我应该如何散列args / kwargs签名？ 我应该创建一个基于str（value）的缓存键吗？ 泡菜？
I want to build a query for sunburnt(solr interface) using class inheritance and therefore adding key - value pairs together. The sunburnt interface takes keyword arguments. How can I transform a dict ({'type':'Event'}) into keyword arguments (type='Event') ? Use the double-star (aka double-splat?) operator: func(**{'type':'Event'}) is equivalent to func(type='Event
我想使用类继承来构建对sunburnt（solr接口）的查询，并因此将键 - 值对添加到一起。 sunburnt接口采用关键字参数。 如何将字典({'type':'Event'})转换为关键字参数(type='Event') ？ 使用双星（又名双啪？）运算符： func(**{'type':'Event'}) 相当于 func(type='Event')
What is the proper way to use **kwargs in Python when it comes to default values? kwargs returns a dictionary, but what is the best way to set default values, or is there one? Should I just access it as a dictionary? Use get function? class ExampleClass: def __init__(self, **kwargs): self.val = kwargs['val'] self.val2 = kwargs.get('val2') A simple question, but one that
当谈到默认值时，在Python中使用**kwargs的正确方法是什么？ kwargs返回一个字典，但是设置默认值的最佳方式是什么，或者是否有？ 我应该只是作为字典来访问它？ 使用get函数？ class ExampleClass: def __init__(self, **kwargs): self.val = kwargs['val'] self.val2 = kwargs.get('val2') 一个简单的问题，但我无法找到很好的资源。 人们在我见过的代码中做了不同的方式，很难知道要使用什么。
This question already has an answer here: How do I sort a dictionary by value? 38 answers Keep in mind that dictionaries are unordered. So you can imagine what will happen if the dictionary got sorted and a dictionary was returned. To sort the dictionary keys and values you should use: sorted(phoneBook.items()) Calling an iterator on a dictionary will naturally return only its list of k
这个问题在这里已经有了答案： 如何按价值对字典进行排序？ 38个答案 请记住，词典是无序的。 所以你可以想象如果字典被排序并返回字典会发生什么。 要对字典键和值进行排序，您应该使用： sorted(phoneBook.items()) 在字典上调用iterator自然会返回唯一的键列表。 .items()确保返回键和值。 为了在排序后保持顺序，将得到的元组列表（通过sorted返回）放入OrderedDict ： from collections import OrderedDict p
This question already has an answer here: How do I sort a dictionary by value? 38 answers If you want to keep the sorted result with a key/value structure I recommend collections.OrderedDict : from collections import OrderedDict from operator import itemgetter dct = {'Rick Porcello, Bos SP': 579.0, 'Chris Sale, CWS SP': 575.0, 'Justin Verlander, Det SP': 601.0, 'Madison Bumgarner, SF SP':
这个问题在这里已经有了答案： 如何按价值对字典进行排序？ 38个答案 如果你想保留一个键/值结构的排序结果，我推荐collections.OrderedDict ： from collections import OrderedDict from operator import itemgetter dct = {'Rick Porcello, Bos SP': 579.0, 'Chris Sale, CWS SP': 575.0, 'Justin Verlander, Det SP': 601.0, 'Madison Bumgarner, SF SP': 617.0, 'Max Scherzer, Wsh SP': 668.0, 'Johnny Cueto, SF SP
This question already has an answer here: How do I sort a dictionary by value? 38 answers Simply use Counter. It will both shorten your code and get you the ordering that you want. Quoting from the documentation: A Counter is a dict subclass for counting hashable objects. It is an unordered collection where elements are stored as dictionary keys and their counts are stored as dictionar
这个问题在这里已经有了答案： 如何按价值对字典进行排序？ 38个答案 只需使用计数器。 它将缩短您的代码并获得您想要的订购。 从文档引用： Counter是用于计算可哈希对象的字典子类。 它是一个无序的集合，其元素以字典键的形式存储，并将其计数存储为字典值。 计数允许为包括零或负计数的任何整数值。 Counter类与其他语言的bag或multisets类似。 >>> c = Counter(['eggs', 'ham']) >>> c['bac
Possible Duplicate: Python: Sort a dictionary by value I need to sort by values a original dictionary on a descending order. As keys I have numbers and as values I have some date and time (string). This means I have: {1: '2011-09-25 16:28:18', 2: '2011-09-25 16:28:19', 3: '2011-09-25 16:28:13', 4: '2011-09-25 16:28:25'} And I want to have: {4: '2011-09-25 16:28:25', 2: '2011-09-25 16:2
可能重复： Python：按值排序字典 我需要按降序对原始字典进行排序。 作为键我有数字和值我有一些日期和时间（字符串）。 这意味着我有： {1: '2011-09-25 16:28:18', 2: '2011-09-25 16:28:19', 3: '2011-09-25 16:28:13', 4: '2011-09-25 16:28:25'} 我想要： {4: '2011-09-25 16:28:25', 2: '2011-09-25 16:28:19', 1: '2011-09-25 16:28:18', 3: '2011-09-25 16:28:13'} 请看，时间（价值）。 我想按降序排列时
This question already has an answer here: Python reverse / invert a mapping 27 answers How do I sort a dictionary by value? 38 answers Yes. You can call items on the dictionary to get a list of pairs of (key, value). Then you can reverse the tuples and pass the new list into dict : transposed = dict((value, key) for (key, value) in my_dict.items()) Python 2.7 and 3.x also have dictiona
这个问题在这里已经有了答案： Python反转/映射27个答案 如何按价值对字典进行排序？ 38个答案 是。 您可以调用字典上的items以获取（键，值）对的列表。 然后，您可以反转元组并将新列表传递给dict ： transposed = dict((value, key) for (key, value) in my_dict.items()) Python 2.7和3.x也有字典解析，这使得它更好： transposed = {value: key for (key, value) in my_dict.items()} transposed = dict(zip(d.v