您可以将预先计算的距离矩阵作为链接传递给
clustermap()：
import pandas as pd, seaborn as snsimport scipy.spatial as sp, scipy.cluster.hierarchy as hcfrom sklearn.datasets import load_irissns.set(font="monospace")iris = load_iris()X, y = iris.data, iris.targetDF = pd.Dataframe(X, index = ["iris_%d" % (i) for i in range(X.shape[0])], columns = iris.feature_names)DF_corr = DF.T.corr()DF_dism = 1 - DF_corr # distance matrixlinkage = hc.linkage(sp.distance.squareform(DF_dism), method='average')sns.clustermap(DF_dism, row_linkage=linkage, col_linkage=linkage)
对于
clustermap(distance_matrix)（即，未通过链接），链接是基于距离矩阵中行和列的成对距离（在内部获得详细信息，在内部进行计算），而不是直接使用距离矩阵的元素（正确的解决方案）
。结果，输出与问题中的输出有些不同：
注意：如果没有
row_linkage将传递给
clustermap()，则通过将每行视为一个“点”（观察）并计算点之间的成对距离来内部确定行链接。因此行树状图反映了行相似性。与相似
col_linkage，其中每列均视为一个点。此说明可能应该添加到docs中。在此修改文档的第一个示例，以使内部链接计算明确：
import seaborn as sns; sns.set()import scipy.spatial as sp, scipy.cluster.hierarchy as hcflights = sns.load_dataset("flights")flights = flights.pivot("month", "year", "passengers")row_linkage, col_linkage = (hc.linkage(sp.distance.pdist(x), method='average') for x in (flights.values, flights.values.T))g = sns.clustermap(flights, row_linkage=row_linkage, col_linkage=col_linkage) # note: this produces the same plot as "sns.clustermap(flights)", where # clustermap() calculates the row and column linkages internally
