我正在使用sklearn.cluster KMeans包。当我完成聚类后,如果我需要知道哪些值被分组在一起,我该如何做?
比如说我有100个数据点,KMeans给了我5个簇。现在我想知道哪些数据点在第5个簇中,我该怎么办。
是否有一个函数可以给出簇的id并列出该簇中的所有数据点?
我正在使用sklearn.cluster KMeans包。当我完成聚类后,如果我需要知道哪些值被分组在一起,我该如何做?
比如说我有100个数据点,KMeans给了我5个簇。现在我想知道哪些数据点在第5个簇中,我该怎么办。
是否有一个函数可以给出簇的id并列出该簇中的所有数据点?
我有类似的需求,正在使用pandas创建一个新数据框,以数据集的索引为行,标签为列。
data = pd.read_csv('filename')
km = KMeans(n_clusters=5).fit(data)
cluster_map = pd.DataFrame()
cluster_map['data_index'] = data.index.values
cluster_map['cluster'] = km.labels_
一旦DataFrame可用,过滤起来非常容易, 例如,要过滤聚类3中的所有数据点
cluster_map[cluster_map.cluster == 3]
km.labels_
重新构建数据框时,你的解决方案是否保留了行的顺序,就像聚类之前那样? - PeterB如果您有一个大型数据集,并且需要按需提取聚类,则使用numpy.where
会有一定的加速效果。以下是在鸢尾花数据集上的示例:
from sklearn.cluster import KMeans
from sklearn import datasets
import numpy as np
centers = [[1, 1], [-1, -1], [1, -1]]
iris = datasets.load_iris()
X = iris.data
y = iris.target
km = KMeans(n_clusters=3)
km.fit(X)
定义一个函数来提取您提供的 cluster_id 的索引。以下是两个函数,用于基准测试,它们都返回相同的值:
定义一个函数来提取您提供的cluster_id的索引。(这里有两个函数,用于基准测试,它们都返回相同的值):
def ClusterIndicesNumpy(clustNum, labels_array): #numpy
return np.where(labels_array == clustNum)[0]
def ClusterIndicesComp(clustNum, labels_array): #list comprehension
return np.array([i for i, x in enumerate(labels_array) if x == clustNum])
假设你想要所有属于聚类2
的样本:
ClusterIndicesNumpy(2, km.labels_)
array([ 52, 77, 100, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112,
115, 116, 117, 118, 120, 122, 124, 125, 128, 129, 130, 131, 132,
134, 135, 136, 137, 139, 140, 141, 143, 144, 145, 147, 148])
Numpy胜出基准测试:
%timeit ClusterIndicesNumpy(2,km.labels_)
100000 loops, best of 3: 4 µs per loop
%timeit ClusterIndicesComp(2,km.labels_)
1000 loops, best of 3: 479 µs per loop
现在您可以这样提取所有聚类2的数据点:X[ClusterIndicesNumpy(2,km.labels_)]
array([[ 6.9, 3.1, 4.9, 1.5],
[ 6.7, 3. , 5. , 1.7],
[ 6.3, 3.3, 6. , 2.5],
... #truncated
请仔细检查上述截断数组的前三个索引:
print X[52], km.labels_[52]
print X[77], km.labels_[77]
print X[100], km.labels_[100]
[ 6.9 3.1 4.9 1.5] 2
[ 6.7 3. 5. 1.7] 2
[ 6.3 3.3 6. 2.5] 2
实际上做这件事的方法非常简单:
clusters=KMeans(n_clusters=5)
df[clusters.labels_==0]
df
元素。同样,您可以找到其他聚类元素。df
中的索引。 - galactica要获取每个聚类内包含的点/样本/观测值的ID,请按照以下步骤进行:
使用鸢尾花数据和一种优美的Python方式的示例:
import numpy as np
from sklearn.cluster import KMeans
from sklearn import datasets
np.random.seed(0)
# Use Iris data
iris = datasets.load_iris()
X = iris.data
y = iris.target
# KMeans with 3 clusters
clf = KMeans(n_clusters=3)
clf.fit(X,y)
#Coordinates of cluster centers with shape [n_clusters, n_features]
clf.cluster_centers_
#Labels of each point
clf.labels_
# Nice Pythonic way to get the indices of the points for each corresponding cluster
mydict = {i: np.where(clf.labels_ == i)[0] for i in range(clf.n_clusters)}
# Transform this dictionary into list (if you need a list as result)
dictlist = []
for key, value in mydict.iteritems():
temp = [key,value]
dictlist.append(temp)
结果
#dict format
{0: array([ 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63,
64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76,
78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90,
91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 106, 113, 114,
119, 121, 123, 126, 127, 133, 138, 142, 146, 149]),
1: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33,
34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49]),
2: array([ 52, 77, 100, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112,
115, 116, 117, 118, 120, 122, 124, 125, 128, 129, 130, 131, 132,
134, 135, 136, 137, 139, 140, 141, 143, 144, 145, 147, 148])}
# list format
[[0, array([ 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63,
64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76,
78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90,
91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 106, 113, 114,
119, 121, 123, 126, 127, 133, 138, 142, 146, 149])],
[1, array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33,
34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49])],
[2, array([ 52, 77, 100, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112,
115, 116, 117, 118, 120, 122, 124, 125, 128, 129, 130, 131, 132,
134, 135, 136, 137, 139, 140, 141, 143, 144, 145, 147, 148])]]
只需更改
for key, value in mydict.iteritems():
to
for key, value in mydict.items():
labels_
。km = KMeans(2)
km.fit([[1,2,3],[2,3,4],[5,6,7]])
print km.labels_
output: array([1, 1, 0], dtype=int32)
你可以看到第一个和第二个点是属于集群 1
的,而最后一个点则属于集群 0
。