Here is a function I wrote that will take a very long text file. Such as a text file containing an entire textbook. It will find any repeating substrings and output the largest string. Right now it doesn't work however, it just outputs the string I put in For example, if there was a typo in which an entire sentence was repeated. It would output that sentence; given it is the largest in
这是我写的一个函数，它将需要一个非常长的文本文件。 如包含整个教科书的文本文件。 它会查找任何重复的子字符串并输出最大的字符串。 现在它不起作用，它只是输出我放入的字符串 例如，如果有整个句子重复的错字。 它会输出这个句子; 因为它是整个文件中最大的。 如果在输入整个段落两次时出现错字，则会输出该段落。 该算法采用第一个字符，查找任何匹配项，如果匹配并且长度最大，则存储子字符串。 然后它需要前2
I found this example, which explains how to perform a 2-way ANOVA. I was wondering how to do the same for a repeated-measures design. I did see this question, but I cannot assume independence of my repeated measurements. I'm using statsmodels version 0.5.0.dev-Unknown . Ideally, I'd like to use statsmodels, but if there's a viable solution using another library, I'd be inte
我找到了这个例子，它解释了如何执行双向ANOVA。 我想知道如何为重复措施设计做同样的事情。 我确实看到了这个问题，但我不能假设我重复测量的独立性。 我正在使用statsmodels版本0.5.0.dev-Unknown 。 理想情况下，我想使用statsmodels，但如果有一个可行的解决方案使用另一个库，我也有兴趣听到它。 提前致谢！ 没有什么内置的重复措施设计ANOVA，但我怀疑它不会很难支持。 如果您可以提供或指向我一个示例并提交
I've been having trouble understanding what h(a)(b) means. I'd never seen one of those before yesterday, and I couldn't declare a function this way: def f (a)(b): return a(b) When I tried to do def f (a, b): , it didn't work either. What do these functions do? How can I declare them? And, finally, what's the difference between f(a, b) and f(a)(b) ? Functions with m
我一直无法理解h(a)(b)含义。 我以前从未见过其中之一，我也无法这样声明函数： def f (a)(b): return a(b) 当我试图做def f (a, b):它也不起作用。 这些功能是做什么的？ 我该如何申报？ 最后， f(a, b)和f(a)(b)之间有什么区别 ？ 正如您在尝试定义一个时所看到的那样，具有多个参数括号的函数不存在。 然而，有些函数返回（其他）函数： def func(a): def func2(b): return a + b return func2
I'm programmatically creating and setting up an s3 buckets with boto. I can create buckets, objects .. and write on objects ... I would like to configure a lifecyle for bucket but when I run the code below I get this exception: boto.exception.S3ResponseError: S3ResponseError: 403 Forbidden with code SignatureDoesNotMatch. lifecycle = Lifecycle() lifecycle.add_rule( 'rulename',
我正在以编程方式创建和设置一个与博托s3桶。 我可以创建桶，对象..并在对象上写... 我想为bucket配置一个lifecyle，但是当我运行下面的代码时，我得到这个异常： boto.exception.S3ResponseError：S3ResponseError：403禁止使用代码SignatureDoesNotMatch。 lifecycle = Lifecycle() lifecycle.add_rule( 'rulename', prefix='/', status='Enabled', expiration=Expiration(days=1) ) bucket = s3.g
What is the best pattern to handle Redis connections (both for interacting with Redis directly and indirectly through Python-RQ)? Generally, database connections need to be closed / returned to a pool when done, but I don't see how to do that with redis-py. That makes me wonder if I'm doing it the wrong way. Also, I have seen some performance dips when enqueuing jobs to RQ, which I&#
处理Redis连接的最佳模式是什么（通过Python-RQ直接或间接与Redis进行交互）？ 通常，数据库连接在完成时需要关闭/返回到池中，但是我不明白如何使用redis-py来完成此操作。 这让我怀疑我是否以错误的方式行事。 另外，在将工作排入RQ时，我看到一些性能下降，据我了解，这可能与连接使用/重用不良有关。 基本上，我有兴趣了解正确的模式，因此我可以验证或更正我们在应用程序中的内容。 非常感谢！ 如果有更多有用的
I doubt this is even possible, but here is the problem and proposed solution (the feasibility of the proposed solution is the object of this question): I have some "global data" that needs to be available for all requests. I'm persisting this data to Riak and using Redis as a caching layer for access speed (for now...). The data is split into about 30 logical chunks, each about
我怀疑这甚至是可能的，但这里是问题和提出的解决方案（提出的解决方案的可行性是这个问题的目标）： 我有一些“全局数据”需要适用于所有请求。 我坚持将这些数据保存到Riak中，并使用Redis作为访问速度的缓存层（现在...）。 数据分成大约30个逻辑块，每个大约8 KB。 每个请求都需要读取这些8KB块中的4块，从Redis或Riak中读取32KB数据。 这对于任何需要读取的请求特定数据都是附加的（这是相当多的）。 假设每秒甚至有
This question already has an answer here: Making a flat list out of list of lists in Python 31 answers If you're just looking to iterate over a flattened version of the data structure and don't need an indexable sequence, consider itertools.chain and company. >>> list_of_menuitems = [['image00', 'image01'], ['image10'], []] >>> import itertools >>> chain =
这个问题在这里已经有了答案： 在Python 31列表中列出一个扁平列表的答案 如果您只是在遍历数据结构的扁平版本并且不需要可索引序列，请考虑itertools.chain和company。 >>> list_of_menuitems = [['image00', 'image01'], ['image10'], []] >>> import itertools >>> chain = itertools.chain(*list_of_menuitems) >>> print(list(chain)) ['image00', 'image01', 'image10'] 它将处
I have a sample file that looks like this: @XXXXXXXXX VXVXVXVXVX + ZZZZZZZZZZZ @AAAAAA YBYBYBYBYBYBYB ZZZZZZZZZZZZ ... I wish to only read the lines that fall on the index 4i+2, where i starts at 0. So I should read the VXVXV (4*0+2 = 2)... line and the YBYB...(4*1 +2 = 6) line in the snippet above. I need to count the number of 'V's, 'X's,'Y
我有一个如下所示的示例文件： @XXXXXXXXX VXVXVXVXVX + ZZZZZZZZZZZ @AAAAAA YBYBYBYBYBYBYB ZZZZZZZZZZZZ ... 所以我应该读VXVXV (4*0+2 = 2)...行和YBYB...(4*1 +2 = 6) VXVXV (4*0+2 = 2)...行。我希望只读取索引4i + 2上的行， YBYB...(4*1 +2 = 6)在上面的代码段中。 我需要计算'V's, 'X's,'Y's and 'B's并存储在一个预先存在的字典中。 fp = open(fi
I have a problem with scipy.quad . In short, I have a really long and complicated set of nested functions and integrals which include an integral of a decreasing function which must be integrated over the specific range 10^2 < x < 10^20. To demonstrate this problem simply, consider the integral of y=x^(-2) between these values using numpy.quad : import numpy as np from scipy.int
我有一个scipy.quad的问题。 简而言之，我有一个非常长而复杂的嵌套函数和积分集合，其中包含积分递减函数，它必须在特定范围内集成10 ^ 2 <x <10 ^ 20。 为了简单地演示这个问题，使用numpy.quad这些值之间的y = x ^（ - 2）的numpy.quad ： import numpy as np from scipy.integrate import quad def func(x,z): "decreasing function to test" #print x,x**-2. return x**(-2
I would like to implement this: asset_hist = [] for key_host, val_hist_list in am_output.asset_history.items(): for index, hist_item in enumerate(val_hist_list): row = collections.OrderedDict([("computer_name", key_host), ("id", index), ("hist_item", hist_item)]) asset_hist.append(row) Using list comprehension. This is as close as I can get: asset_hist = [collections.Orde
我想实现这一点： asset_hist = [] for key_host, val_hist_list in am_output.asset_history.items(): for index, hist_item in enumerate(val_hist_list): row = collections.OrderedDict([("computer_name", key_host), ("id", index), ("hist_item", hist_item)]) asset_hist.append(row) 使用列表理解。 这是尽可能接近，我可以得到： asset_hist = [collections.OrderedDict([("computer_name", k