Return an array of arrays of size *returnSize.The sizes of the arrays are returned as *returnColumnSizes array.Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().

返回一个数组的数组，大小为*returnSize。数组的大小通过*returnColumnSizes数组返回。注意：返回的数组和*columnSizes数组必须是malloc分配的，假设调用者调用free（）。

相关问题

ValueError: Unexpected detail coefficient type: <class 'tuple'>. Detail coefficients must be a 3-tuple of arrays as returned by wavedec2. If you are using pywt.array_to_coeffs or pywt.unravel_coeffs, please specify output_format='wavedec2'

非常抱歉，看来我在修改代码时出现了新的错误。出现 `ValueError: Unexpected detail coefficient type: <class 'tuple'>...` 的错误是因为在量化子带时，遇到了一个非预期的类型，这可能是因为前面的DWT分解没有使用正确的输出格式。请在DWT分解时，将输出格式设置为 'wavedec2'，即：

coeffs = pywt.wavedec2(img, wavelet, level=level, mode='symmetric', axis=(-2,-1), output_format='wavedec2')

这样就可以避免这个错误了。下面是修改后的完整代码：

import numpy as np
import cv2
import pywt

def dwt_compress(img, wavelet='haar', level=1):
    # 转换为浮点数类型
    img = np.float32(img)
    # DWT分解
    coeffs = pywt.wavedec2(img, wavelet, level=level, mode='symmetric', axis=(-2,-1), output_format='wavedec2')
    # 将每个子带进行量化
    coeffs_list = list(coeffs)
    for i in range(1, len(coeffs_list)):
        coeffs_list[i] = (np.round(coeffs_list[i][0] / (2 ** (level - i + 1))) * (2 ** (level - i + 1)), coeffs_list[i][1], coeffs_list[i][2])
    # DWT重构
    img_compress = pywt.waverec2(coeffs_list, wavelet)
    # 转换为整数类型
    img_compress = np.uint8(img_compress)
    return img_compress

# 读取原始图像
img = cv2.imread('test.jpg', cv2.IMREAD_COLOR)
# 对原始图像进行DWT压缩
img_compress = dwt_compress(img, wavelet='haar', level=1)
# 计算PSNR
psnr = cv2.PSNR(img, img_compress)
print('PSNR:', psnr)

希望这次能够解决您的问题，再次感谢您的指正！

Design a recursive function to solve the following problem: sum up a given array a[0]..a[m-1] & return the sum to the caller

Here's an example of a recursive function that sums up an array:

int array_sum(int a[], int m) {
    // Base case: if the array has only one element, return it
    if (m == 1) {
        return a[0];
    }
    // Recursive case: split the array in half and recursively sum up each half
    int mid = m / 2;
    int left_sum = array_sum(a, mid);
    int right_sum = array_sum(a + mid, m - mid);
    // Return the sum of the left and right halves
    return left_sum + right_sum;
}

This function takes an array `a` and the number of elements `m` as parameters. It uses recursion to divide the array in half until each sub-array has only one element. At that point, it returns the value of the single element. The function then adds up the values returned by the left and right sub-arrays and returns the sum to the caller.