zenith | Zenith installation consists of compute nodes | Storage library

 by   zenithdb Rust Version: Current License: Non-SPDX

kandi X-RAY | zenith Summary

kandi X-RAY | zenith Summary

zenith is a Rust library typically used in Storage, Nodejs, Docker, Prometheus, Amazon S3 applications. zenith has no bugs, it has no vulnerabilities and it has low support. However zenith has a Non-SPDX License. You can download it from GitHub.

A Zenith installation consists of compute nodes and Zenith storage engine. Compute nodes are stateless PostgreSQL nodes, backed by Zenith storage engine.
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              zenith has a low active ecosystem.
              It has 201 star(s) with 19 fork(s). There are 23 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 216 open issues and 369 have been closed. On average issues are closed in 79 days. There are 43 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of zenith is current.

            kandi-Quality Quality

              zenith has no bugs reported.

            kandi-Security Security

              zenith has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

            kandi-License License

              zenith has a Non-SPDX License.
              Non-SPDX licenses can be open source with a non SPDX compliant license, or non open source licenses, and you need to review them closely before use.

            kandi-Reuse Reuse

              zenith releases are not available. You will need to build from source code and install.
              Installation instructions, examples and code snippets are available.

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            zenith Key Features

            No Key Features are available at this moment for zenith.

            zenith Examples and Code Snippets

            No Code Snippets are available at this moment for zenith.

            Community Discussions

            QUESTION

            Issues with finding the clearness index and extraterrestrial irradiance using PVLIB functions
            Asked 2021-May-18 at 10:31

            I recently ran into an issue with calculating the clearness index and the extraterrestrial irradiance using the PVLIB functions. Basically, the numbers do not tally up.

            This is my raw data that I ran the function for (my Datetime is already timezone aware):

            I then ran the below code to get the clearness index:

            ...

            ANSWER

            Answered 2021-May-18 at 10:31

            The problem is in this line:

            Source https://stackoverflow.com/questions/67584069

            QUESTION

            Astral solar calculation
            Asked 2021-Apr-28 at 08:19

            I am fairly new to python and would like to calculate sun's position (azimuth and zenith angle) from a datetime column of a dataframe.

            I found Astral module for this task and also made some correct calculations from it using the individual timestamps. But when I try to process the whole datetime column, I get an error.

            I also tried to localize the dataframe (datetime column) to a specific timezone, but the error remains the same.

            Help appreciated.

            Here is the code:

            ...

            ANSWER

            Answered 2021-Apr-28 at 08:19

            Not sure how you got your code sample to work but basically, you can apply the solar_azimuth function to a pandas.Series like e.g.

            Source https://stackoverflow.com/questions/67295809

            QUESTION

            Solving Fick's second law of diffusion in 1D sphere using FiPy
            Asked 2021-Feb-18 at 14:29

            I'm trying to solve the second law of diffusion for spheres PDE using fipy. I've checked the documentation but there's no example for such a case, so I was wondering if it is actually possible to do it, as I was not successful reaching for the adequate equation definition structure. I consider azimuthal and zenith angular symmetries, so the equation I need to solve results in the following.

            Of course, boundary conditions are fixed at r=0 and r=R at fixed values and the initial concentration is also known. I also tried to follow the ideas presented in here but didn't get any clear result for it. Any ideas would be welcomed.

            The code I'm using at the moment is the following:

            ...

            ANSWER

            Answered 2021-Feb-18 at 14:29

            There are a few things going on:

            • Some solvers don't like the spherical mesh, probably because of the enormous range in cell volumes. The SciPy LinearLUSolver seems to work. Judicious preconditioning might help other solvers.
            • Eq. (45) in the paper you linked below defines the flux, but you are constraining the gradient. You need to divide by the diffusivity.
            • X_ca is in units of [stoichiometry], but BoundaryR1_ca is in units of mol/(m**2 * s) (or mol/m**4 after dividing by D_ca. You need to divide by C_ca_max to get [stoichiometry]/m, as you're solving something halfway between Eq. (43) and Eq. (52).
            • No-flux is the natural boundary condition for FiPy, so there's no need to constrain at mesh.facesLeft.
            • The gradient at mesh.facesRight should be constrained to a vector (even in 1D).

            With the changes below, I get behavior that looks consistent with Fig. 7 in Mayur et al..

            Source https://stackoverflow.com/questions/66164005

            QUESTION

            How to create a polar plot with azimuth, zenith and a averaged weight value?
            Asked 2020-Sep-07 at 17:22

            I want to use the code from Polar histogram in Python for given r, theta and z values, replaced with my dataframe columns df.azimuth, df.zenith, and df.ozone (average for each bin).

            I'm having issues converting dataframe .values format, to a polar plot format.

            I would like to produce a polar plot with my DataFrame columns df.azimuth, df.zenith, and df.o3 average for each bin as showed on the plot example that I attached here.

            I'm having issues converting DataFrame format, to a polar plot format. I don't know how to do this from a DataFrame format. Any help is welcome.

            That is my code so far, but it is not showing the plot correctly.

            Dataframe (o3Pan_wff):

            ...

            ANSWER

            Answered 2020-Sep-07 at 16:58
            • Do not change the scope of the question, after it has been answered.
            • There seems to be two main issues
              1. x, y, and z are the incorrect columns.
                • .iloc[:,0] is the time column
                • .iloc[:,1] is the zenith column
                • .iloc[:,2] is the aximuth column
                • UFuncTypeError occurs because radius is the 'time' column of the dataframe.
              2. azimuth should be in radians, according to the cited sample code.
            • zenith needs to be in radians for the density function to return the expected averges.
            • o3 is assumed to be the concentration and will be used as the weight
              • The concentration of the o3 column is normalized by using density=True inside np.histogram2d().
            • There is no reason to extract the values of each dataframe column, with .values. The operations will accept a dataframe column.

            Source https://stackoverflow.com/questions/63758336

            QUESTION

            Is there a Geopy Python function that converts dataframe column (Latitude/Longitude) from "degree minute second" to "degree decimal"
            Asked 2020-Aug-22 at 13:23

            I applied the following function for each row in my dataframe (df), and using a Lat0/Lon0, direction(azimuth) and distance(dist) I got the new Lat/Lon in "degree minute second":

            ...

            ANSWER

            Answered 2020-Aug-22 at 13:23

            If you remove .format_decimal(), your location function would return Point instances instead of strings (see https://geopy.readthedocs.io/en/stable/#geopy.point.Point), where decimal coordinates can easily be extracted as attributes:

            Source https://stackoverflow.com/questions/63525742

            QUESTION

            How do i insert array into database without getting error: Array to string conversion
            Asked 2020-Jun-28 at 16:46

            In m app I have a form with data meant to insert data into 2 tables :payers and Spouses and Payer has Many Spouse. In my PayerController i have

            ...

            ANSWER

            Answered 2020-Jun-28 at 14:20

            you can try something like this for payer and spouse model ....

            Source https://stackoverflow.com/questions/62623512

            QUESTION

            Why is this randomly generated spherical point cloud not uniformly distributed?
            Asked 2020-Jun-04 at 19:29

            I'm trying to simulate radiation emitting from a point source. To do this, given the coordinates of a source and the desired length of emitted rays, I randomly generate a direction vector in spherical coordinates, convert it to cartesian, and return the correct end point. However, when I run this, and visualize the resulting point cloud (consisting of all the randomly generated end points) in Blender, I see that it's more densely populated at the "poles" of the sphere. I'd like the points to be uniformly distributed along the sphere. How can I achieve this?

            The random generation function:

            ...

            ANSWER

            Answered 2020-Jun-04 at 19:29

            When you transform points by spherical coordinates and angle theta approaches pi, the circle which is an image of [0,2pi]x{theta} gets smaller and smaller. Since theta is uniformly distributed, there will be more points near poles. It could be seen on image of grid.

            If you want to generate uniformly distributed points on sphere, you can use the fact that if you cut a sphere with two parallel planes, the area of the strip of spherical surface between the planes depends only on the distance between the planes. Hence, you can get a uniform distribution on the sphere using two uniformly distributed random variables:

            • z coordinate between -r and r,
            • an angle theta between [0, 2pi) corresponding to a longitude.

            Then you can easily calculate x and y coordiantes.

            Example code:

            Source https://stackoverflow.com/questions/62199614

            QUESTION

            My absolute positioned particles.js is covering my relative div
            Asked 2020-May-04 at 16:14

            ANSWER

            Answered 2020-May-03 at 11:00
            #particles-js {
              position: relative;
            }
            

            Source https://stackoverflow.com/questions/61571655

            QUESTION

            How to show place with Latitude and Longitude?
            Asked 2020-Apr-04 at 16:44

            I am working with the Google Takeout. They give me data with json file.

            ...

            ANSWER

            Answered 2020-Apr-04 at 11:20

            Multiply each Latitude and Longitude with

            0.0000001

            Source https://stackoverflow.com/questions/61026453

            QUESTION

            Can't reconcile PVLIB output with NREL SAM
            Asked 2020-Feb-14 at 23:36

            Background

            Traditionally I've used NREL SAM tool to estimate solar output. I've been experimenting with PVLIB which is great due to the open nature and flexibility, however I can't seem to reconcile the solar production estimates between PVLIB and NREL SAM.

            What I've done

            I'm modeling a hypothetical solar farm near Gympie QLD. I've gone to the climate.onebuiling website, and downloaded the zip folder / epw file for "AUS_QLD_Gympie.AP.945660_TMYx.2003-2017". I've then used that weather file in NREL's SAM tool using PVwatts, with the following specs;

            • 200,000 KWdc
            • Module Type = Standard
            • 1.2 DC to AC Ratio
            • 96% inverter efficiency
            • 1 axis backtracking
            • tilt = 26 degrees
            • azimuth = 0 degrees
            • GCR = 0.4
            • losses, shading & curtailment = default

            In NREL SAM i get an annual Energy Yield (AC GWh) of 415.96 GWh p.a.

            I then took that same epw file and converted it to a csv, keeping just the columns for ghi, dni, dhi, temp_air & wind_speed (Google Drive link to CSV file). I've used this file as an import into PVLIB. I spec'd a PVLIB system the same specs above, with addition of albedo = 0.2 and max angle = 90 degrees (Code Below).

            The result I get in PVLIB is 395.61 GWh.

            Problem

            The results I got are pretty different. PVLIB = ~395 GWh p.a. vs SAM = ~415 GWH p.a. I expected around a 1-2% difference, but not 5%.

            Figures are even worse when I compare to a PVLIB system using clearsky.ineichen (adjusted with linke_turbidity) which yields ~475 GWh p.a.

            Help Requested

            Anyone know why my results are so different? is there anything I can do to narrow the gap?

            PVLIB Code

            ...

            ANSWER

            Answered 2020-Feb-14 at 23:36

            Hard to say exactly why the annual energy is different without a detailed comparison of intermediate results. One contributing factor appears to be the transposition model (GHI, DHI and DNI to plane-of-array): pvlib ModelChain defaults to the Hay/Davies model, and I believe SAM defaults to the Perez 1990 model. The two models will differ by a few percent in annual plane-of-array irradiance, which varies with the relative levels of diffuse and direct irradiance; see Lave et al. Figure 6.

            You can select the Perez 1990 model in pvlib by adding transposition_model = 'perez', to the mc instance. I expect that will narrow the difference between pvlib and SAM results, and am interested in what you find.

            Calculations using a TMY weather file won't give the same result as a calculation using irradiance from a clear sky model, since TMY is assembled from historical weather records and so includes cloudy periods.

            Source https://stackoverflow.com/questions/60218037

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install zenith

            On Ubuntu or Debian this set of packages should be sufficient to build the code:. [Rust] 1.56.1 or later is also required. To run the psql client, install the postgresql-client package or modify PATH and LD_LIBRARY_PATH to include tmp_install/bin and tmp_install/lib, respectively. To run the integration tests or Python scripts (not required to use the code), install Python (3.7 or higher), and install python3 packages using ./scripts/pysync (requires poetry) in the project directory.
            Install build dependencies and other useful packages
            Build zenith and patched postgres
            Start pageserver and postgres on top of it (should be called from repo root):
            Now it is possible to connect to postgres and run some queries:
            And create branches and run postgres on them:
            If you want to run tests afterwards (see below), you have to stop all the running the pageserver, safekeeper and postgres instances you have just started. You can stop them all with one command:

            Support

            Now we use README files to cover design ideas and overall architecture for each module and rustdoc style documentation comments. See also /docs/ a top-level overview of all available markdown documentation. To view your rustdoc documentation in a browser, try running cargo doc --no-deps --open.
            Find more information at:

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