获取分割NumPy数组的索引

Question

假设我有一个NumPy数组：

x = np.array([3, 9, 2, 1, 5, 4, 7, 7, 8, 6])

如果我对这个数组求和，得到52。我需要的是一种方法，从左到右将它分成大约n个块，其中n由用户选择。本质上，分裂是以贪婪的方式发生的。因此，对于一些块n，第一个n - 1块的总和必须至少达到52/n，并且它们必须是从左到右的连续索引

因此，如果n = 2，那么第一个块将由前7个元素组成：

chunk[0] = x[:7]  # [3, 9, 2, 1, 5, 4, 7], sum = 31
chunk[1] = x[7:]  # [7, 8, 6], sum = 21

请注意，第一个区块不会只包含前6个元素，因为总和将是小于52/2 = 26的24。另外，请注意，只要满足求和条件，每个块中的元素数量就可以改变。最后，最后一个块不接近52/2 = 26是很好的，因为其他块可能需要更多的时间

但是，我需要的输出是一个两列数组，第一列中包含开始索引，第二列中包含（独占）停止索引：

[[0, 7],
 [7, 10]]

如果n = 4，那么前3个块需要至少加起来52/4 = 13，如下所示：

chunk[0] = x[:3]  # [3, 9, 2], sum = 14
chunk[1] = x[3:7]  # [1, 5, 4], sum = 17
chunk[2] = x[7:9]  # [7, 8], sum = 15
chunk[3] = x[9:]  # [6], sum = 6

我需要的输出是：

[[0, 3],
 [3, 7],
 [7, 9],
 [9, 10]

因此，使用for循环的一种简单方法可能是：

ranges = np.zeros((n_chunks, 2), np.int64)
ranges_idx = 0
range_start_idx = start

sum = 0
for i in range(x.shape[0]):
    sum += x[i]
    if sum > x.sum() / n_chunks:
        ranges[ranges_idx, 0] = range_start_idx
        ranges[ranges_idx, 1] = min(
                i + 1, x.shape[0]
            )  # Exclusive stop index
        # Reset and Update
        range_start_idx = i + 1
        ranges_idx += 1
        sum = 0
# Handle final range outside of for loop
ranges[ranges_idx, 0] = range_start_idx
ranges[ranges_idx, 1] = x.shape[0]
if ranges_idx < n_chunks - 1:
    left[ranges_idx:] = x.shape[0]

return ranges

我正在寻找一个更好的矢量化解决方案

Answer 1

以下是一个不会迭代所有元素的解决方案：

def fun2(array, n):
    min_sum = np.sum(array) / n
    cumsum = np.cumsum(array)
    i = -1
    count = min_sum
    out = []
    while i < len(array)-1:
        j = np.searchsorted(cumsum, count) 
        out.append([i+1, j+1])
        i = j 
        if i < len(array):
            count = cumsum[i] + min_sum
    out[-1][1] -= 1
    return np.array(out)

对于这两个测试用例，它会产生您期望的结果。嗯

Answer 2

我从一个similar question that was answered中找到灵感：

def func(x, n):
    out = np.zeros((n, 2), np.int64)
    cum_arr = x.cumsum() / x.sum()
    idx = 1 + np.searchsorted(cum_arr, np.linspace(0, 1, n, endpoint=False)[1:])
    out[1:, 0] = idx  # Fill the first column with start indices
    out[:-1, 1] = idx  # Fill the second column with exclusive stop indices
    out[-1, 1] = x.shape[0]  # Handle the stop index for the final chunk
    return out

更新

为了涵盖病理病例，我们需要更精确一点，并采取如下措施：

def func(x, n, truncate=False):
    out = np.zeros((n_chunks, 2), np.int64)
    cum_arr = x.cumsum() / x.sum()
    idx = 1 + np.searchsorted(cum_arr, np.linspace(0, 1, n, endpoint=False)[1:])
    out[1:, 0] = idx  # Fill the first column with start indices
    out[:-1, 1] = idx  # Fill the second column with exclusive stop indices
    out[-1, 1] = x.shape[0]  # Handle the stop index for the final chunk

    # Handle pathological case
    diff_idx = np.diff(idx)
    if np.any(diff_idx == 0):
        row_truncation_idx = np.argmin(diff_idx) + 2
        out[row_truncation_idx:, 0] = x.shape[0]
        out[row_truncation_idx-1:, 1] = x.shape[0]
        if truncate:
            out = out[:row_truncation_idx]

    return out