Question

如何计算下面每个工人的均值？以下是我的示例NumPy ndarray。列0是工作人员编号，列1是纬度，列2是经度。
我想计算每个workerid的平均纬度和经度。我想使用NumPy（ndarray）保留所有这些，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

您可以使用一些创造性的数组切片和where函数来解决此问题。

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2d数组提取列（在本例中为第一个列）的便捷方法。为了获得均值，我们从第一列中找到唯一的ID，然后为每个ID用where提取适当的行，然后合并。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，它会产生以下命令：

{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)}

Answer 2

使用workerid和列表理解将是：

a=np.array(self.a)
ids=np.unique(a[:,0]) #array of unique ids
pos_mean=[np.mean(a[a[:,0]==i, 1:], axis=0) for i in ids]

但是考虑到似乎总是有3次连续测量，应该有一个相对简单的矢量化方法

Answer 3

给定这个数组，我们想按第一列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 且不使用循环

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

_ndx = np.argsort(groups)
_id, _pos, g_count  = np.unique(groups, 
                                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])}

通过numpy.mean分组

3 个答案: