python - 通过合并邻近的值来减少排序的数字列表的更好方法？

Question

我有一个包含以下值的数字列表作为示例。

2, 99, 101, 150, 198, 201, 300

我们假设上面的数组总是按升序排序。

我的目标是对由变量x定义的彼此接近的值进行平均。假设 x = 5 预期结果应如下所示

2, 100, 150, 200, 300

以下是我编写的代码的工作副本

import statistics
def mean_reduce_by_proximity(data, distance):
    new_data = []
    temp = []
    for val in data:
        # if temp list is empty
        if (len(temp) == 0):
            temp.append(val)
        # if temp list if not empty
        else:
            # Last item of temp list is in proximity to current value
            if (val - temp[-1]) <= distance:
                temp.append(val)
            # Not in proximity
            else:
                temp_mean = round(statistics.mean(temp))
                new_data.extend([temp_mean])
                temp = []
                temp.append(val)

    # Handle last set of item(s)
    if (len(temp) > 0):
        temp_mean = round(statistics.mean(temp))
        new_data.extend([temp_mean])
        temp = []

    return new_data

In[1]: mean_reduce_by_proximity([2, 99, 101, 150, 198, 201, 300], 5)
Out[1]: [2, 100, 150, 200, 300]

我的查询

• 是否有针对此类减少的已知技术术语？
• 流行的Python库中是否有内置函数可以做到这一点这个？

Answer 1

这是一种矢量化方法 np.add.reduceat -

def mean_reduce_by_proximity_vectorized(a, thresh):
    # Get an array with indices off input array, such that each index
    # represent start of a group of close proximity elements
    i = np.flatnonzero(np.r_[True,np.diff(a)>thresh,True])

    # Sum based on those indices. Hence, each group is summed.
    sums = np.add.reduceat(a,i[:-1])

    # Get counts of each group
    counts = np.diff(i)

    # Get average of each group and round those for final o/p
    return np.round(sums/counts.astype(float))

示例运行 -

In [90]: a
Out[90]: array([  2,  99, 101, 150, 198, 201, 300])

In [91]: mean_reduce_by_proximity_vectorized(a, thresh=5)
Out[91]: array([  2., 100., 150., 200., 300.])