使用scipy.cluster.hierarchy monocrit标准仅展平特定树

Question

我正在使用scipy.cluster.hierarchy对一些时间序列进行聚合聚类

我在另一个答案{a1}中发现，使用fcluster上的monocrit标准，我可以手动选择我想要扩展的树。像这样：

import pandas

data = pandas.DataFrame({
'total_runs': {0: 2.489857755536053, 1: 1.2877651950650333, 2: 0.8898850111727028, 3: 0.77750321282732704, 4: 0.72593099987615461, 5: 0.70064977003207007, 6:0.68217502514600825,7: 0.67963194285399975, 8: 0.64238326692987524, 9:0.6102581538587678, 10: 0.52588765899448564, 11: 0.44813665774322564, 12: 0.30434031343774476, 13: 0.26151929543260161, 14: 0.18623657993534984, 15: 0.17494230269731209,16: 0.14023670906519603, 17: 0.096817318756050832, 18:0.085822227670014059, 19: 0.042178447746868117, 20: -0.073494398270518693,21: -0.13699665903273103, 22: -0.13733324345373216, 23: -0.31112299949731331, 24: -0.42369178918768974, 25: -0.54826542322710636,26: -0.56090603814914863, 27: -0.63252372328438811, 28: -0.68787316140457322,29: -1.1981351436422796, 30: -1.944118415387774,31: -2.1899746357945964, 32: -2.9077222144449961}, 
'total_salaries': {0: 3.5998991340231234,1: 1.6158435140488829, 2: 0.87501176080187315, 3: 0.57584734201367749, 4: 0.54559862861592978, 5: 0.85178295446270169,6: 0.18345463930386757, 7: 0.81380836410678736, 8: 0.43412670908952178, 9: 0.29560433676606418, 10: 1.0636736398252848, 11: 0.08930130612600648, 12: -0.20839133305170349, 13: 0.33676911316165403, 14: -0.12404710480916628, 15: 0.82454221267393346,16: -0.34510456295395986, 17: -0.17162157282367937, 18: -0.064803261585569982, 19: -0.22807757277294818, 20: -0.61709008778669083,21: -0.42506873158089231, 22: -0.42637946918743924, 23: -0.53516500398181921, 24: -0.68219830809296633, 25: -1.0051418692474947,26: -1.0900316082184143, 27: -0.82421065378673986, 28: 0.095758053930450004, 29: -0.91540963929213015, 30: -1.3296449323844519,31: -1.5512503530547552, 32: -1.6573856443389405}
})


from scipy.spatial.distance import pdist
from scipy.cluster.hierarchy import linkage, dendrogram

distanceMatrix = pdist(data)
Z = linkage(distanceMatrix, method='complete')
monocrit = np.zeros((Z.shape[0], ))
monocrit[[-1, -3]] = 1
fc = fcluster(Z, 0, criterion='monocrit', monocrit=monocrit)

此代码将在顶部fork和第三个to fork上展开。 然而，过了一段时间，很难在视觉上确定叉子的“索引”是什么，我不知道如何用其他方法来做

例如，在这段代码中，如果我想在蓝色圆圈的树上展开，我需要找到它的索引和它前面的叉。有没有办法把这些指数画在密度图上？或者至少知道它们是什么