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把numpy翻译为一个中文库
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QUESTION
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 11:09I'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.
QUESTION
I am looking for a tweak that I could apply to the code shown later or an alternate method that would lead to faster run time. In general, I want to find the values of two parameters that are governed by a set of multiple nonlinear equations. These equations use multiple variable arguments.
An Example of the problem at hand is presented for only a small array (4 values) as an example:
- I want to find the values of q and h that are governed by 2 non-linear equations
- I will need to apply this procedure to larger a and b arrays each of about 50k values
Equations are:
- q = k/2L * (h^2 - b^2)
- q = -2 * pi * k * (h-a) / ln((h-d)/r)
arguments/variables:
- k = 144.0
- L = 550.0
- d = 140.9
- r = 0.5
- a = [190.0, 185.0, 160.0, 150.0]
- b = [70.0, 70.0, 30.0, 10.0]
I followed suggestions in similar posts at stackoverflow, mostly This post, and modified my original code to avoid using a for loop. Though, after doing that, I tried applying the code to a big array of variables, but it took quite some time.
I should also mention that I am only interested in the solution when h is between a certain limit (when h is greater than d+r and less than 300), but I didn't know how to apply this in fsolve when I am solving for multiple parameters and using multiple non-linear functions.
The final code that I am currently using is shown in this post, but I am looking for something that is faster and more efficient.
...ANSWER
Answered 2021-Jun-06 at 19:53The simplest solution in your case is to use the Numba's JIT since many (short) calls to the same function are performed. Here is an example:
QUESTION
I have a numpy array and want to replace its values with some strings. I checked this solution but could not solve my issue. My strings are in a list:
...ANSWER
Answered 2021-Jun-02 at 08:25How about smth like this:
QUESTION
I have a (previously sorted) text file that, consisting of either a dash -
or a single alphabetical character. I'd greatly appreciate any help in better understanding the proper awk syntax to move through each column of the text file and retain only the first non-dash character in each row if a non-dash character exists, or else to retain that dash character if no alphabetical character exists. The result in either situation would be a single row of text. Files are always formatted in such a way that every row has the same number of columns, and the first non-dash character is always preferred, regardless if other alphabetical characters exist in 'lower' rows.
Two examples to clarify: given this text file:
...ANSWER
Answered 2021-Jun-02 at 02:29How about awk -f foo.awk input.txt
where
foo.awk:
QUESTION
How can I plot a 2D line from a chosen axis of a Numpy Array quickly?
An analogy: when sum
an arbitrary matrix sigma
with respect to axis = 0
, I would write:
ANSWER
Answered 2021-May-16 at 08:32If I understood your question, your first dimension is a time, for which you have a 2D array at each time point, and you want to see how a given index in that 2D array evolves.
One way to approach (so that you don't have to keep copying data, assuming you have a large dataset), is to flatten your original sigma
array and index the 2D array locations.
QUESTION
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 19:50This 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
QUESTION
I'm rendering a dynamically changing numpy bitmap array and trying to improve my framerate.
Currently I'm using openCV:
...ANSWER
Answered 2021-May-26 at 01:17You may render the video using external sub-process video renderer.
I tested the suggested solution by piping the video frames to FFplay.
The solution is not working so well - the last frames are not shown.
Treat the solution as conceptual solution.
The code opens FFplay as sub-process, and write the raw frames to stdin
pipe of FFplay.
Here is the code:
QUESTION
The code down below calculates the maximum number of times consecutive positive values Cons_Pos_results
, negative values Cons_Neg_results
, zero values Cons_Zero_results
. I am trying to implement a piece to the code to the already existing code where it shows the indexes of where the maximum number of consecutive values where between. So for the maximum number of consecutive values that are positive are between the Index 38-60. The issue for the used code: issue
Array:
...ANSWER
Answered 2021-May-23 at 21:19You can take advantage of the fact that the d array has non-zero values at the beginning and end of each found sequence. When the distance between two such non-zero values is equal to the count, you have found the desired indexes:
QUESTION
I want to reorder the nth axis of a ndarray of abitrary dimension d according to a list of indices order. If the axis was the last one, the solution from this question (Reordering last dimension of a numpy ndarray) would be enough. In my case, however, the axis is not, in general, the first or the last one, so that Ellipsis alone does not solve the issue.
This is the solution I have come up so far:
...ANSWER
Answered 2021-May-20 at 04:14Use the command np.take_along_axis and assign the output to a new variable. See the code below:
QUESTION
My question is similar to testing whether a Numpy array contains a given row but instead I need a non-trivial extension to the method offered in the linked question; the linked question is asking how to check if each row in an array is the same as a single other row. The point of this question is to do that for numerous rows, one does not obviously follow from the other.
Say I have an array:
...ANSWER
Answered 2021-May-19 at 11:28Broadcasting is an option:
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