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kandi X-RAY | numpy.org Summary

numpy.org is a HTML library typically used in Web Site, Blog, Numpy applications. numpy.org has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

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              numpy.org has a low active ecosystem.
              It has 85 star(s) with 97 fork(s). There are 32 watchers for this library.
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              It had no major release in the last 6 months.
              There are 39 open issues and 179 have been closed. On average issues are closed in 199 days. There are 7 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of numpy.org is current.

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            Community Discussions

            QUESTION

            `np.linalg.solve` get solution matrix?
            Asked 2021-Jun-13 at 06:48

            np.linalg.solve solves for x in a problem of the form Ax = b.

            For my application, this is done to avoid calculating the inverse explicitly (i.e inverse(A)b = x)

            I'd like to access what the effective inverse is that was used to solve this problem but looking at the documentation it doesn't appear to be an option... Is there a reasonable alternative approach I can follow to recover the inverse of A?

            (np.linalg.inv(A) is not accurate enough for my use case)

            ...

            ANSWER

            Answered 2021-Jun-13 at 06:48

            Following the docs and source code, it seems NumPy is calling LAPACK's _gesv to compute the solution, the documentation of which reads:

            The routine solves for X the system of linear equations A*X = B, where A is an n-by-n matrix, the columns of matrix B are individual right-hand sides, and the columns of X are the corresponding solutions.

            The LU decomposition with partial pivoting and row interchanges is used to factor A as A = P * L * U, where P is a permutation matrix, L is unit lower triangular, and U is upper triangular. The factored form of A is then used to solve the system of equations A * X = B.

            The NumPy implementation for solve doesn't return the inverted matrix back to the caller, and just frees the memory for the inverted matrix, so there's no hope there. SciPy provides low-level access to LAPACK so you should be able to access the result from there. You can follow the actual implementation in LAPACK's Fortran source code dgesv.f, dgetrf.f and dgetrs.f. Alternatively, you could note that NumPy's inv still calls the same underlying code, so it might be enough for your use case... You didn't specify why is it that you need the approximate inverse matrix.

            Source https://stackoverflow.com/questions/67955351

            QUESTION

            Python: pandas.replace doesn't work with np.nan
            Asked 2021-Jun-11 at 09:31

            I wanted to replace np.nan with None values, but weird behaviour occurred:

            ...

            ANSWER

            Answered 2021-Jun-11 at 08:41

            the second parameter is none, so it equalss.replace(np.nan)
            you can try replase all of the 'nan'using : df = df.fillna(value=''),it will replace all nan with ''

            Source https://stackoverflow.com/questions/67932935

            QUESTION

            Numpy iteration over all dimensions but the last one with unknown number of dimensions
            Asked 2021-Jun-07 at 11:09

            Physical Background

            I'm working on a function that calculates some metrics for each vertical profile in an up to four dimensional temperature field (time, longitude, latitude, pressure as height measure). I have a working function that takes the pressure and temperature at a single location and returns the metrics (tropopause information). I want to wrap it with a function that applies it to every vertical profile in the data passed.

            Technical Description of the Problem

            I want my function to apply another function to every 1D array corresponding to the last dimension in my N-dimensional array, where N <= 4. So I need an efficient loop over all dimensions but the last one without knowing the number of dimensions beforehand.

            Why I Open a New Question

            I am aware of several questions (e.g., iterating over some dimensions of a ndarray, Iterating over the last dimensions of a numpy array, Iterating over 3D numpy using one dimension as iterator remaining dimensions in the loop, Iterating over a numpy matrix with unknown dimension) asking how to iterate over a specific dimension or how to iterate over an array with unknown dimensions. The combination of these two problems is new as far as I know. Using numpy.nditer for example I haven't found out how to exclude just the last dimension regardless of the number of dimensions left.

            EDIT

            I tried to do a minimal, reproducible example:

            ...

            ANSWER

            Answered 2021-Jun-07 at 11:09

            I've used @hpaulj 's reshape approach several times. It means the loop can iterate the whole array by 1d slices.

            Simplified the function and data to have something to test.

            Source https://stackoverflow.com/questions/67857864

            QUESTION

            Write numpy's polyval() function
            Asked 2021-Jun-04 at 13:09

            I am trying to write a function that can replicate the numpy function, np.polyval()

            here is my current code:

            ...

            ANSWER

            Answered 2021-Jun-04 at 12:59

            Have you tried math.pow function ?

            Source https://stackoverflow.com/questions/67837543

            QUESTION

            Numpy: Calculate Covariance of large array
            Asked 2021-May-28 at 13:09

            I have a large numpy array of shape (32,2048,2048), where every [i,:,:] is a 2D set of pixels which are samples from a spatially correlated statistical distribution. With i=32 samples for each pixel.

            I now need to calculate the covariance matrix for everey 2x2 ROI (overlapping) on the 2D image, resulting in a a set of 4x4 covariance matrices with total shape of (4,4,2047,2047).

            Looping over all ROIs in a loop is of cause possible and takes about 4 minutes on my maschine:

            ...

            ANSWER

            Answered 2021-May-28 at 13:09

            From numpy 1.20.0+, there is sliding_window_view that allows you to extract the sliding windows. Then you can perform the covariant calculation:

            Source https://stackoverflow.com/questions/67079993

            QUESTION

            How can I specify start and step size in numpy.nditer?
            Asked 2021-May-26 at 18:36

            I have the following Python code which computes the first n Euler Totient function values:

            ...

            ANSWER

            Answered 2021-May-26 at 18:36

            Plain python version is faster:

            Source https://stackoverflow.com/questions/67706205

            QUESTION

            Why do I get "is not a constant expression" error in this case ? (templates)
            Asked 2021-May-26 at 15:45

            I am trying to implement some kind of numpy.where() for my ITK images in C++. ITK's way seems to be with Functors. I am not very experienced with templates, so my whole approach might be flawed, but here is my go:

            ...

            ANSWER

            Answered 2021-May-26 at 13:01

            You have needlessly many template parameters. You could follow the way it is done in a corresponding test. Define your function, set it via filter->SetFunctor() and call Update().

            Source https://stackoverflow.com/questions/67701013

            QUESTION

            Numpy Multi-stage Container Build -- Alpine
            Asked 2021-May-26 at 14:24

            I'm attempting a multi-stage container build to try and keep my image smaller. The offending package is numpy which apparently doesn't play nicely with Alpine.

            My error from numpy:

            ...

            ANSWER

            Answered 2021-May-25 at 19:50

            This might be related to linking problem, which appears as 'not found' many times in Alpine Linux when something is wrong with symlinks. When you build your numpy dependencies in the Debian based distro, it is linked to glibc in specific path. Usually, similar paths are also in Alpine. I'm not sure how this works with venv.

            I would suggest that, as you managed (I think) to install numpy directly in Alpine based container, to use Alpine distro as builder as well. Otherwise you might need to change some linked paths manually.

            If you look for example folder /lib64 in Alpine, the ld-linux-x86-64.so.2 should be in there, so it is not totally missing. (after installing libc6-compat package).

            But in general, Alpine is not the best choice for Python as the programming language is based on C, and musl is not perfect. Alpine could be also much slower for Python

            Source https://stackoverflow.com/questions/67694006

            QUESTION

            How to create nested weighted distributions in Python/numpy?
            Asked 2021-May-25 at 18:35

            I am trying to optimize (vectorize?) the creation of a monte-carlo style simulation and I have not been able to figure out how to create nested-weighted random values using numpy or similar libraries. Consider the scenario, inspired by an interviewqs question: "Students in three classrooms have to vote for one of two class president candidates. Classroom A has 40% of the students and are split 50/50 on candidate X and Y. B has 25% of the students and is split 60/40. C has 35% of the students and is split 35/65"

            Creating that data with vanilla Python might look something like this,

            ...

            ANSWER

            Answered 2021-May-25 at 15:29

            You can significantly reduce the number of python looping, making the code more vectorized:

            Source https://stackoverflow.com/questions/67688837

            QUESTION

            numpy roots negative exponential function
            Asked 2021-May-21 at 13:12

            I have read numpy.roots, which works out common algebraic function's y axis intersections.

            which

            ...

            ANSWER

            Answered 2021-May-21 at 13:12

            If you got a "polynomial" with negative exponents, it is not really a polynomial but a fractional function, so to find the roots you can simply find the roots of the numerator polynomial.

            In a function

            you can factor out the last term

            which is equal to

            So we can say

            The function g(x) is a polynomial of n+m degree and finding the roots of g(x) you'll get the roots of f(x) because the denominator cannot be zero (x=0 is outside the domain of f(x), when it's zero you get a singularity, in this case a division by zero which is impossible).

            EDIT

            We can graphically verify. For example, let's take a function with these coefficients

            Source https://stackoverflow.com/questions/67605082

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