denominator | Portably control DNS clouds using java or bash | DNS library
kandi X-RAY | denominator Summary
kandi X-RAY | denominator Summary
Portably control DNS clouds using java or bash
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Top functions reviewed by kandi - BETA
- Adds a resource record to the cache
- Updates records of the record
- Creates a resource record
- Add rdatas to a pool
- Replace zone information with the request
- Creates a query from the given path
- Converts an iterator to JSON response
- Handles a recorded request
- Returns an iterator to iterate over the records for a given query
- Adds a resource record set to the registry
- Puts a GeoRecordSet into the cache
- Creates the denominator module
- Apply credentials to the request
- Runs the CLI command
- Returns a string representation of this ResourceRecordSet
- Returns the next resource record set
- Decodes and returns an exception thrown by the client
- Decodes the response
- Adds a zone
- Adds additional regions to a resource set
- Updates the specified resource record
- Decode the JSON document to retrieve an access token
- Provides a map of countries for the country
- Deserializes an access token from JSON
- Adds a new resource record to the Zone
- Adds a resource record to the domain
denominator Key Features
denominator Examples and Code Snippets
Lines of Code : 203License : Non-SPDX (Apache License 2.0)
Copydef reduce(self, reduce_op, value, axis):
"""Reduce `value` across replicas and return result on current device.
>>> strategy = tf.distribute.MirroredStrategy(["GPU:0", "GPU:1"])
>>> def step_fn():
... i = tf.dist def audio_microfrontend(audio,
sample_rate=16000,
window_size=25,
window_step=10,
num_channels=32,
upper_band_limit=7500.0,
def watch_gradients_by_tensor_names(self, graph, tensor_name_regex):
"""Watch gradient tensors by name(s) of the x-tensor(s).
The side effect of this method is that when gradient tensor(s) are created
with respect to the x-tensors, the g Community Discussions
Trending Discussions on denominator
QUESTION
The header lets you uses template meta-programming to work with and manipulate rational values.
However - it was introduced in C++11, when we already had constexpr. Why is it not good enough to have a fully-constexpr'ifed library type for rationals, i.e. basically:
ANSWER
Answered 2022-Apr-01 at 13:40Is there some concrete benefit to using std::ratio that C++11 constexpr functionality would not be well-suited enough for?
You can pass ratio as a template type argument, which is what std::chrono::duration does. To do that with a value-based ratio, you need C++20 or newer.
In C++20 and newer I don't see any benefits of the current design.
QUESTION
This comes from near the end of the codelab found here:
Intro to debugging - Debugging example: accessing a value that doesn't exist
This is all inside the MainActivity.kt file
Here's my onCreate
...ANSWER
Answered 2022-Jan-10 at 16:51I honestly have no idea what that Codelab is doing, based off the code they provide. The app isn't going to render anything (not the layout, not any changes to the layout) before onCreate finishes, and onCreate won't finish until it's run all its code, including that repeat block in the division function it calls.
division isn't starting any worker threads, so all Thread.sleep is doing is blocking the main thread - it's hanging the app. And you're right, sleep does take a millis value, not seconds - I get the feeling they didn't actually run this code, it's full of other mistakes and inconsistencies that honestly made it hard to work out what you were meant to be doing. Change which Log.d call? The ones in onCreate? (They actually mean the Log.v call in division, I assume)
Here's how you'd use a thread in Kotlin - you need to create a new one (so you're off the main thread, so it can actually finish creating the activity and run the UI):
QUESTION
I would like to calculate the sum of reciprocals of a list of integers (and see if it is larger or equal to 1):
I want to work with integers to avoid floating-point rounding issues. To do so, I want to work it out like this:
I have done this:
...ANSWER
Answered 2022-Mar-21 at 20:43The Fraction type can do this easily and exactly:
QUESTION
I am new to Grafana and Prometheus. I have read a lot of documentation and now I"m trying to work backwards by reviewing some existing queries and making sure I understand them
I have downloaded the Node Exporter Full dashboard (https://grafana.com/grafana/dashboards/1860). I have been reviewing the CPU Busy query and I"m a bit confused. I am quoting it below, spaced out so we can see the nested sections better:
In this query, job is node-exporter while instance is the IP and port of the server. This is my base understanding of the query:
node_cpu_seconds_total is a counter of the number of seconds the CPU took at a given sample.
- Line 5: Get cpu seconds at a given instant, broken down by the individual CPU cores
- Line 4: Add up all CPU seconds across all cores
- Line 3: Why is there an additional count()? Does it do anything?
- Line 12: Rate vector - get cpu seconds of when the cpu was idle over the given rate period
- Line 11: Take a rate to transfer that into the rate of change of cpu seconds (and return an instant vector)
- Line 10: Sum up all rates, broken down by CPU modes
- Line 9: Take the single average rate across all CPU mode rates
- Line 8: Subtract the average rate of change (Line 9) from total CPU seconds (Line 3)
- Line 16: Multiple by 100 to convert minutes to seconds 10: Line 18-20: Divide Line 19 by the count of the count of all CPU seconds across all CPUs
My questions are as follows:
- I would have thought that CPU usage would simply be (all non idle cpu usage) / (total cpu usage). I therefore don't understand why take into account rate at all (#6 and #8)
- The numerator here seems to be trying to get all non-idle usage and does so by getting the full sum and subtracting the idle time. But why does one use count and the other sum?
- If we grab cpu seconds by filtering by
mode=idle, then does adding theby (mode)add anything? There is only one mode anyways? My understanding ofby (something)is more relevant when there are multiple values and we group the values by that category (as we do bycpuin this query) - Lastly, as mentioned in bold above, what is with the double count(), in the numerator and denominator?
ANSWER
Answered 2022-Mar-19 at 12:37Both of these count functions return the amount of CPU cores. If you take them out of this long query and execute, it'll immediately make sense:
QUESTION
I'm trying to calculate interdaily stability as a feature for machine learning classification in Python. My data is for multiple days - I'm using this dataset (sample CSV). This data is sampled with minute frequency, i.e. we have 60 measurements per hour. The formula is:
So my approach is:
...ANSWER
Answered 2022-Mar-14 at 13:31Looks like your implementation is correct, as i am also trying to quantify the rest-activity rhythm and using calculated score, how feasible is Modelling that time series or not.
After, exploration, could simplify Interdaily stability quantifies how consistent the activity patterns are, given over a period of time.
QUESTION
I am trying to convert a calculation in matlab to python. This is code in matlab:
...ANSWER
Answered 2022-Mar-08 at 02:03Often when translating MATLAB it's important to get shapes/sizes correct. But when I run your code in Octave I see all variables are (1,1), "scalar". So dimensions shouldn't be an issue.
Let's check function values:
QUESTION
I have an array of positive integers. For example:
...ANSWER
Answered 2022-Feb-27 at 22:44This problem has a fun O(n) solution.
If you draw a graph of cumulative sum vs index, then:
The average value in the subarray between any two indexes is the slope of the line between those points on the graph.
The first highest-average-prefix will end at the point that makes the highest angle from 0. The next highest-average-prefix must then have a smaller average, and it will end at the point that makes the highest angle from the first ending. Continuing to the end of the array, we find that...
These segments of highest average are exactly the segments in the upper convex hull of the cumulative sum graph.
Find these segments using the monotone chain algorithm. Since the points are already sorted, it takes O(n) time.
QUESTION
I implemented an algorithm which, for a given non-negative rational number r and a positive interger b, computes all of the digits of the expansion of r in base b. The algorithm itself actually outputs a function f(i: int) satisfying the equation n = ... + f(-2)*b**-2 + f(-1)*b**-1 + f(0)*b**0 + f(1)*b**1 + f(2)*b**2 + ..., and it computes the digits of the whole and fractional parts of r separately through two other auxiliary functions.
Below is my code in Python (3.10.2). It looks weird for Python code because I'm actually implementing the algorithm in MIT/GNU Scheme (15.3) and "sketching" it on Python. I'm showing the Python implementation instead of the Scheme one mostly because I believe it's easier to formulate this question (and actually have it answered) in Python.
...ANSWER
Answered 2022-Feb-20 at 22:23Steps:
- Split into whole and fractional part
- Extract the whole digits by repeatedly dividing by
buntil there's nothing left. - Extract the fractional digits by repeatedly multiplying with
buntil we reach a fraction we've seen before. - Split the fractional digits at the place where we first saw the final fractional value.
QUESTION
I want to calculate the %'s of items within groups. For example, there are 2 groups and each contain 3 fruits. I want to know within each group, what are the proportions of fruit (i.e. each group should add up to 100%). I can achieve this using the below code but it feels too verbose. Can anyone suggests any improvements or a function that already exists to simplify it?
...ANSWER
Answered 2022-Feb-19 at 14:01Use the base R code.
QUESTION
As the title suggests I've been trying to find most efficient way to reduce a fraction (i.e. 10/20 -> 1/2) and I came up with this.
ANSWER
Answered 2022-Feb-12 at 03:42For the 'most efficient' solution, you're just looking for the GCD of n, d, so the Euclidean algorithm solves this in O(log(max(n,d)) multiplications. Unless you're a theoretician or dealing with massive numbers, it's probably not worth trying to optimize much beyond that. See, for example, the wiki on greatest common divisors for more information on GCD calculation, but to summarize, you'd have to use fast multiplication algorithms with inputs in the 'thousands of digits' range to start outperforming normal multiplication.
For the surprising timing results, it's likely due to while loop overhead compared to for-loops from Python internals. Try disassembling both functions-- the for-loop iteration is done in C, and the while loop iteration is done in Python bytecode, which is slower.
On the short time-scales you're measuring, performance can be highly unpredictable. As a result, timing many short loops might tell you more about the efficiency of the language's loop implementation than the algorithmic efficiency of your code.
The fact that you only saw results compatible with the asymptotic predictions when your inputs got large isn't all too surprising-- it's the core idea of asymptotic analysis.
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Install denominator
Download denominator
Place it on your $PATH. (ex. ~/bin)
Set it to be executable. (chmod 755 ~/bin/denominator)
Advanced usage, including ec2 hooks are covered in the readme. Here's a quick start for the impatient. If you just want to fool around, you can use the mock provider.
The current version of denominator is 4.6.0. Denominator can be resolved as maven dependencies, or equivalent in lein, gradle, etc. Here are the coordinates, noting you only need to list the providers you use.
Creating a connection to a provider requires that you have access to two things: the name of the provider, and as necessary, credentials for it.
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