使用 numpy.mean 分组

Question

我如何计算以下每个 workerid 的平均值？下面是我的示例 NumPy ndarray。第 0 列是 workerid，第 1 列是纬度，第 2 列是经度。
我想计算每个工人的平均纬度和经度。我想使用 NumPy (ndarray) 保留这一切，而不转换为 Pandas。

import numpy
from scipy.spatial.distance import cdist, euclidean
import itertools
from itertools import groupby

class WorkerPatientScores:

    '''
    I read from the Patient and Worker tables in SchedulingOptimization.
    '''
    def __init__(self, dist_weight=1):
        self.a = []

        self.a = ([[25302, 32.133598100000000, -94.395845200000000],
                   [25302, 32.145095132560200, -94.358041585705600],
                   [25302, 32.160400000000000, -94.330700000000000],
                   [25305, 32.133598100000000, -94.395845200000000],
                   [25305, 32.115095132560200, -94.358041585705600],
                   [25305, 32.110400000000000, -94.330700000000000],
                   [25326, 32.123598100000000, -94.395845200000000],
                   [25326, 32.125095132560200, -94.358041585705600],
                   [25326, 32.120400000000000, -94.330700000000000],
                   [25341, 32.173598100000000, -94.395845200000000],
                   [25341, 32.175095132560200, -94.358041585705600],
                   [25341, 32.170400000000000, -94.330700000000000],
                   [25376, 32.153598100000000, -94.395845200000000],
                   [25376, 32.155095132560200, -94.358041585705600],
                   [25376, 32.150400000000000, -94.330700000000000]])

        ndarray = numpy.array(self.a)
        ndlist = ndarray.tolist()
        geo_tuple = [(p[1], p[2]) for p in ndlist]
        nd1 = numpy.array(geo_tuple)
        mean_tuple = numpy.mean(nd1, 0)
        print(mean_tuple)

上面的输出是：

[ 32.14303108 -94.36152893]

Answer 1

给定这个数组，我们希望按第一列进行分组并取其他两列的平均值

X = np.asarray([[25302, 32.133598100000000, -94.395845200000000],
                [25302, 32.145095132560200, -94.358041585705600],
                [25302, 32.160400000000000, -94.330700000000000],
                [25305, 32.133598100000000, -94.395845200000000],
                [25305, 32.115095132560200, -94.358041585705600],
                [25305, 32.110400000000000, -94.330700000000000],
                [25326, 32.123598100000000, -94.395845200000000],
                [25326, 32.125095132560200, -94.358041585705600],
                [25326, 32.120400000000000, -94.330700000000000],
                [25341, 32.173598100000000, -94.395845200000000],
                [25341, 32.175095132560200, -94.358041585705600],
                [25341, 32.170400000000000, -94.330700000000000],
                [25376, 32.153598100000000, -94.395845200000000],
                [25376, 32.155095132560200, -94.358041585705600],
                [25376, 32.150400000000000, -94.330700000000000]])

仅使用numpy循环和不使用循环

groups = X[:,0].copy()
X = np.delete(X, 0, axis=1)

_ndx = np.argsort(groups)
_id, _pos, g_count  = np.unique(groups[_ndx], 
                                return_index=True, 
                                return_counts=True)

g_sum = np.add.reduceat(X[_ndx], _pos, axis=0)
g_mean = g_sum / g_count[:,None]

将结果存储在字典中：

>>> dict(zip(_id, g_mean))
{25302.0: array([ 32.14636441, -94.36152893]),
 25305.0: array([ 32.11969774, -94.36152893]),
 25326.0: array([ 32.12303108, -94.36152893]),
 25341.0: array([ 32.17303108, -94.36152893]),
 25376.0: array([ 32.15303108, -94.36152893])}

Answer 2

你可以使用一些创造性的数组切片和where函数来解决这个问题。

means = {}
for i in numpy.unique(a[:,0]):
    tmp = a[numpy.where(a[:,0] == i)]
    means[i] = (numpy.mean(tmp[:,1]), numpy.mean(tmp[:,2]))

切片[:,0]是从二维数组中提取一列（在本例中为第一列）的便捷方式。为了得到平均值，我们从第一列中找到唯一的 ID，然后对于每一列，我们用 提取适当的行where，然后组合。最终结果是元组的字典，其中键是 ID，值是包含其他两列平均值的元组。当我运行它时，它会产生以下字典：

{25302.0: (32.1463644108534, -94.36152892856853),
 25305.0: (32.11969774418673, -94.36152892856853),
 25326.0: (32.12303107752007, -94.36152892856853),
 25341.0: (32.17303107752007, -94.36152892856853),
 25376.0: (32.15303107752007, -94.36152892856853)}