AMBROSIA | Robust Distributed Programming Made Easy and Efficient | Architecture library

 by   microsoft C# Version: v2.0.0.0 License: Non-SPDX

kandi X-RAY | AMBROSIA Summary

kandi X-RAY | AMBROSIA Summary

AMBROSIA is a C# library typically used in Architecture applications. AMBROSIA has no bugs, it has no vulnerabilities and it has low support. However AMBROSIA has a Non-SPDX License. You can download it from GitHub.

Ambrosia: Robust Distributed Programming Made Easy and Efficient.
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              AMBROSIA has a low active ecosystem.
              It has 395 star(s) with 50 fork(s). There are 38 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 17 open issues and 19 have been closed. On average issues are closed in 24 days. There are 9 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of AMBROSIA is v2.0.0.0

            kandi-Quality Quality

              AMBROSIA has 0 bugs and 0 code smells.

            kandi-Security Security

              AMBROSIA has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              AMBROSIA code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              AMBROSIA 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

              AMBROSIA releases are available to install and integrate.
              Installation instructions are not available. Examples and code snippets are available.
              AMBROSIA saves you 79 person hours of effort in developing the same functionality from scratch.
              It has 9487 lines of code, 0 functions and 598 files.
              It has low code complexity. Code complexity directly impacts maintainability of the code.

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            AMBROSIA Key Features

            No Key Features are available at this moment for AMBROSIA.

            AMBROSIA Examples and Code Snippets

            No Code Snippets are available at this moment for AMBROSIA.

            Community Discussions

            QUESTION

            How to compare two consecutive rows grouped by another column?
            Asked 2022-Jan-11 at 10:47

            Using pandas, I have created a dataframe that contains the info about weed severity by field. Every field can contain various types of weed. Records are sorted by [field, weed, date] in order to observe the evolution of weed. Ideally, the severity of every weed should be supressed as the time goes by. In order to detect critical items, I would like to compare two consecutive weed severities, grouped by field and weed type and annotate the supression status as the difference between the two observed severities.

            Below is the example data

            ...

            ANSWER

            Answered 2022-Jan-11 at 10:47

            Subtract from right side by shifted values by Series.rsub with DataFrameGroupBy.shift:

            Source https://stackoverflow.com/questions/70665114

            QUESTION

            Identifying time signature with music21
            Asked 2021-Sep-24 at 19:40

            I'm trying to determine the time signature of a midi file in Python using music21. I can obtain the time signature of this sample midi for instance with:

            ...

            ANSWER

            Answered 2021-Sep-24 at 19:40

            music21 v.7.1 started making measures in parts parsed from MIDI files. For that reason, the .timeSignature property is not very useful when called on a Part (example from docs: "This property is not recursive, so a Part will not have the time signature of the measure within it"). The fact that this used to work for MIDI files was inconsistent with the rest of the system.

            If you just want the first time signature, perform a recursive search:

            Old syntax (still works): myPart.recurse().getElementsByClass(music21.meter.TimeSignature)[0]

            New shortcut syntax (v7): myPart[music21.meter.TimeSignature][0]

            Equivalently, you can use strings such as 'TimeSignature' in the getter instead of class names.

            You could also index into the first measure and access the .timeSignature property there: myPart['Measure'][0].timeSignature.

            Source https://stackoverflow.com/questions/69302934

            QUESTION

            How to change a value in nested array
            Asked 2021-Aug-18 at 23:13

            Code:

            ...

            ANSWER

            Answered 2021-Aug-18 at 23:13

            You can use a nested map() to return a modified attributes array in the object.

            Source https://stackoverflow.com/questions/68840191

            QUESTION

            Create in R a dataframe column that indicates how many times each row value has been sampled
            Asked 2020-Apr-03 at 10:16

            I am using R to draw a random sample of 24 elements (species) out of a vector containing 36 elements. As I have to repeat this process several times, I would like to create a data frame in which the first column is the name of the species and the second is a counter of the number of times in which the species has been sampled. So, at the end of the first sampling i would have something like:

            ...

            ANSWER

            Answered 2020-Apr-03 at 10:16

            Let's take a small example :

            Source https://stackoverflow.com/questions/61009165

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

            Vulnerabilities

            No vulnerabilities reported

            Install AMBROSIA

            You can download it from GitHub.

            Support

            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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