I want to clear this up once and for all. Can someone please explain the exact meaning of having leading underscores before an object's name in Python? Also explain the difference between a single and a double leading underscore. Also, does that meaning stay the same whether the object in question is a variable, a function, a method, etc? Single Underscore Names, in a class, with a lea
我想彻底澄清这一点。 有人可以解释在Python中的对象名前面加下划线的确切含义吗？ 同时解释单个和双重前导下划线之间的区别。 另外，无论所讨论的对象是变量，函数还是方法等，这个意义是否保持不变？ 单下划线 带有前导下划线的类中的名称仅用于向其他程序员指示该属性或方法是私有的。 然而，这个名字本身并没有什么特别之处。 引用PEP-8： _single_leading_underscore：弱的“内部使用”指标。 例如， from M imp
Is it possible to create a dictionary comprehension in Python (for the keys)? Without list comprehensions, you can use something like this: l = [] for n in range(1, 11): l.append(n) We can shorten this to a list comprehension: l = [n for n in range(1, 11)] . However, say I want to set a dictionary's keys to the same value. I can do: d = {} for n in range(1, 11): d[n] = True #
是否有可能在Python中创建字典理解（用于键）？ 没有列表解析，你可以使用这样的东西： l = [] for n in range(1, 11): l.append(n) 我们可以将其缩短为列表理解： l = [n for n in range(1, 11)] 。 但是，假设我想将字典的键设置为相同的值。 我可以： d = {} for n in range(1, 11): d[n] = True # same value for each 我试过这个： d = {} d[i for i in range(1, 11)] = True 不过，我得到一个SyntaxEr
什么是解析Python命令行参数的最简单 ， 最 先进 ，最灵活的方法或库？ This answer suggests optparse which is appropriate for older Python versions. For Python 2.7 and above, argparse replaces optparse . See this answer for more information. As other people pointed out, you are better off going with optparse over getopt. getopt is pretty much a one-to-one mapping of the standard getopt(3) C libr
什么是解析Python命令行参数的最简单 ， 最 先进 ，最灵活的方法或库？ 这个答案建议optparse适用于较老的Python版本。 对于Python 2.7及更高版本， argparse替换optparse 。 请参阅此答案以获取更多信息。 正如其他人所指出的那样，您最好通过optopt进行optparse。 getopt几乎是标准getopt（3）C库函数的一对一映射，并且不是很容易使用。 optparse虽然稍微冗长些，但稍后可以更好地进行结构化和简化。 下面是一个典型
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers T is the key, so you should iterate with for T in instruments : import math instruments = {} def add_instrument(par, T, coup, price, compounding_freq = 2): instruments[T] = (par, coup, price, compounding_freq) add_instrument(100, 0.25, 0., 97.5) add_instrument(100, 0.5, 0., 94.9)
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 T是关键，所以你应该for T in instruments迭代for T in instruments ： import math instruments = {} def add_instrument(par, T, coup, price, compounding_freq = 2): instruments[T] = (par, coup, price, compounding_freq) add_instrument(100, 0.25, 0., 97.5) add_instrument(100, 0.5, 0., 94.9) add_instrument(100, 1.0, 3., 90.) add
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers You can use the keys method in the dictionary object to fetch the keys and then iterate over to get the required value. Example: d = { "available_projects": { "model001": { "available_models": [ "model_20171004-090552" ],
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 您可以使用字典对象中的keys方法来获取密钥，然后迭代以获取所需的值。 例： d = { "available_projects": { "model001": { "available_models": [ "model_20171004-090552" ], "status": "ready" }, "model002": { "available_models": [
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers Like this - first: for obj in objArray: for key in obj: value = obj[key] print(key) print(value) Second (python 3): for obj in objArray: for key, value in obj.items(): print(key) print(value) For python 2 you can use for key, valu
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 像这样 - 第一： for obj in objArray: for key in obj: value = obj[key] print(key) print(value) 第二（python 3）： for obj in objArray: for key, value in obj.items(): print(key) print(value) 对于python 2，你可以使用for key, value in d.iteritems()
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers You are getting this error because kwargs is a dict . In your case, this dict looks like this: {'key': lambda x: x*x} So when you iterate over the dict , you are really iterating over the keys, which in the case of kwargs , are strings (and strings are not callable). If you really w
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 你得到这个错误是因为kwargs是一个dict 。 在你的情况下，这个字典看起来像这样： {'key': lambda x: x*x} 所以当你迭代dict ，你真的在迭代键，在kwargs的情况下，它是字符串（并且字符串不可调用）。 如果你真的想要得到lambda，那么你应该在那个关键字中访问kwargs中的值： for fart in kwargs: print(fart) print(kwargs[fart[](4))
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers 将您的代码更改为： for data in digit_count: contents.append(digit_count[data]) You want to get the values from your dictionary; just use the .values() method. vals = digit_count.values() Instead of this: for data in digit_count: contents.append(data) You can use: c
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 将您的代码更改为： for data in digit_count: contents.append(digit_count[data]) 你想从你的字典中获得值; 只需使用.values()方法即可。 vals = digit_count.values() 取而代之的是： for data in digit_count: contents.append(data) 您可以使用： contents = digit_count.values()
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers It seems you want to iterate over keys and values . You can accomplish this by iterating over the items() : for channel_num, channel_det in self.dictionary.items(): In your case for something in self.dictionary: iterates only over the keys. And the keys are integers. But you try to
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 看来你想迭代keys和values 。 您可以通过迭代items()来完成此操作： for channel_num, channel_det in self.dictionary.items(): 在你的情况下for something in self.dictionary:迭代只能通过键。 钥匙是整数。 但是你尝试将整数解包为两个值： channel_num, channel_det ，这是它失败的原因。 补充评论： 你只需要for -loop中的值，这样你也
This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers theBoard = {'top-L': ' ', 'top-M': ' ', 'top-R': ' ', 'mid-L': ' ', 'mid-M': ' ', 'mid-R': ' ', 'low-L': ' ', 'low-M': ' ', 'low-R': ' ' } # <--- Close your dictionary
这个问题在这里已经有了答案： 使用'for'循环遍历字典12个答案 theBoard = {'top-L': ' ', 'top-M': ' ', 'top-R': ' ', 'mid-L': ' ', 'mid-M': ' ', 'mid-R': ' ', 'low-L': ' ', 'low-M': ' ', 'low-R': ' ' } # <--- Close your dictionary # <--- remove random string 'Che