如何計(jì)算下面每個(gè)工人的均值？以下是我的示例NumPy ndarray。第0列是工作人員編號(hào)，第1列是緯度，第2列是經(jīng)度。我想計(jì)算每個(gè)workerid的平均緯度和經(jīng)度。我想保留所有使用NumPy（ndarray），而不轉(zhuǎn)換為熊貓。import numpyfrom scipy.spatial.distance import cdist, euclideanimport itertoolsfrom itertools import groupbyclass 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)上面的輸出是：[32.14303108 -94.36152893]
2 回答
30秒到達(dá)戰(zhàn)場(chǎng)
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超6個(gè)贊
您可以使用一些創(chuàng)造性的數(shù)組切片和where函數(shù)來解決此問題。
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]是從2d數(shù)組中提取列（在本例中為第一列）的便捷方法。為了獲得均值，我們從第一列中找到唯一的ID，然后針對(duì)每個(gè)ID，使用提取相應(yīng)的行where，然后合并。最終結(jié)果是元組的字典，其中鍵是ID，值是包含其他兩列平均值的元組。當(dāng)我運(yùn)行它時(shí)，它會(huì)產(chǎn)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)}
臨摹微笑
TA貢獻(xiàn)1982條經(jīng)驗(yàn) 獲得超2個(gè)贊
使用workerid和列表理解將是：
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]
但是考慮到似乎總是有3次連續(xù)測(cè)量，應(yīng)該有一個(gè)相對(duì)簡(jiǎn)單的方法對(duì)其進(jìn)行矢量化
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