如何使用Ctypes将此numpy数组传递给C? [英] How to pass this numpy array to C with Ctypes?

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本文介绍了如何使用Ctypes将此numpy数组传递给C?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

目前,我正在学习C类型.我的目标是产生一个numpypython中的数组A从500到4 * pi分500步.该数组传递给计算这些值的切线的C代码.C代码也将这些值传递回python中的numpy数组B.

Currently I'm learning about C types. My goal is to generate an numpy array A in python from 0 to 4*pi in 500 steps. That array is passed to C code which calculates the tangent of those values. The C code also passes those values back to an numpy array B in python.

昨天我只是尝试将一个值从python转换为C并(经过一些帮助)成功.今天,我尝试传递整个数组,而不是值.

Yesterday I tried simply to convert one value from python to C and (after some help) succeeded. Today I try to pass a whole array, not a value.

我认为向C库添加另一个函数是一个好主意,处理数组.新函数应该在循环中传递每个值将A转换为函数tan1()并将该值存储在数组B中.

I think it's an good idea to add another function to the C library to process the array. The new function should in a loop pass each value of A to the function tan1() and store that value in array B.

我有两个问题:

  • 编写处理numpy数组A的函数
  • 在python和C代码之间传递numpy数组.

我阅读了以下信息:

有帮助,但我仍然不知道如何解决我的问题.

Helpful, but I still don't know how to solve my problem.

C代码(仅显示相关的部分):

C code (Only the piece that seems relevant):

double tan1(f) double f;
{
    return sin1(f)/cos1(f); 
}


void loop(double A, int n);
{
    double *B;
    B = (double*) malloc(n * sizeof(double));
    for(i=0; i<= n, i++)
    {
        B[i] = tan1(A[i])
    }
}

Python代码:

import numpy as np
import ctypes


A = np.array(np.linspace(0,4*np.pi,500), dtype=np.float64)

testlib = ctypes.CDLL('./testlib.so')
testlib.loop.argtypes = ctypes.c_double,
testlib.loop.restype = ctypes.c_double


#print(testlib.tan1(3))
    

我知道在这种情况下ctypes.c_double是错误的,但这就是我在1值版本中所拥有的,并且还不知道要替代什么.

I'm aware that ctypes.c_double is wrong in this context, but that is what I had in the 1 value version and don't know yet for what to substitute.

请问关于如何实现这一目标的一些反馈意见?

Could I please get some feedback on how to achieve this goal?

推荐答案

您需要返回动态分配的内存,例如将您的C代码更改为:

You need to return the dynamically allocated memory, e.g. change your C code to something like:

#include <math.h>
#include <stdlib.h>
#include <stdio.h>

double tan1(double f) {
    return sin(f)/cos(f);
}

double *loop(double *arr, int n) {
    double *b = malloc(n * sizeof(double));
    for(int i = 0; i < n; i++) {
        b[i] = tan(arr[i]);
    }
    return b;
}

void freeArray(double *b) {
    free(b);
}

在Python方面,您必须声明参数并返回类型.正如其他人在评论中提到的那样,您还应该释放动态分配的内存.请注意,在C端,数组总是衰减为指针.因此,您需要一个附加参数来告诉您数组中元素的数量.

On the Python side you have to declare parameter and return types. As mentioned by others in comments, you should also free dynamically allocated memory. Note that on the C side, arrays always decay into pointers. Therefore, you need an additional parameter which tells you the number of elements in the array.

此外,如果您返回一个指向double的指针到Python页面,则必须指定数组的大小.使用 np.frombuffer ,您无需复制数据即可处理数据.

Also if you return a pointer to double to the Python page, you must specify the size of the array. With np.frombuffer you can work with the data without making a copy of it.

import numpy as np
from ctypes import *

testlib = ctypes.CDLL('./testlib.so')

n = 500
dtype = np.float64
input_array = np.array(np.linspace(0, 4 * np.pi, n), dtype=dtype)
input_ptr = input_array.ctypes.data_as(POINTER(c_double))

testlib.loop.argtypes = (POINTER(c_double), c_int)
testlib.loop.restype = POINTER(c_double * n)
testlib.freeArray.argtypes = POINTER(c_double * n),

result_ptr = testlib.loop(input_ptr, n)
result_array = np.frombuffer(result_ptr.contents)

# ...do some processing
for value in result_array:
    print(value)

# free buffer
testlib.freeArray(result_ptr)

这篇关于如何使用Ctypes将此numpy数组传递给C?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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