我想自动将数字分组到不同的地区。根据图（样本数据集），我们可以看到它们是数字所在的三个区域，即 [0,100]，[650,750]，[1220,1300] 。我想只指出那些地区。可以有任何数量的此类区域。我们需要自动找到号码。这些地区和这些地区的范围。两个区域之间的距离将非常高。 有什么方法可以用Python做到这一点吗？
样本数据= [69,8,30,45,89,61,80,45,9,18,19,11,1255,1299,1296,1293,1287,1250,1265,1291,1281,1250 ，1286,1286,1251,1287,1266,1288,1254,1260,1260,1254,1267,1299,1273,1250,1300,1250,1279,1255,1293,1292,1278,1277,1252,1299,1278 ，1258,1268,1274,1285,1258,1279,1270,1278,1286,1278,1253,1267,1300,1295,1298,1285,1288,1274,1272,1252,1256,1283,1289,1251,1258 ，1253,1257,1297,1269,1292,1253,1273,1281,1251,1280,1253,1274,1275,1287,1296,1298,1296,1291,1284,1261,1267,1290,1273,1281,1263 ，1270,1264,1269,1278,1284,67,8,50,59,97,64,45,72,45,90,94,7,33,58,97,97,1252,1297,1265,1278 ，1272,1252,1258,1261,1287,1260,1260,1258,1280,1263,1256,1296,1269,1270,1296， 1282,696,678,665,700,700,691,689,688,650,663,662,698,655,660,662,684,690,657,653,663,670,691,687,675， 694,670,676,659,661,664,664,689,683,675,687,691,676,659,689,657,659,656,654,679,669,687,666,662,691， 1260,1276,1252,1295,1257,1277,1281,1257,1295,1269,1265,1290,1266,1269,1286,1254,1260,1265,1290,1294,1286,1279,1254,1256,1276， 1285,1282,1251,1282,1261,1253,56,74,85,94,18,83,38,80,8,4,78,43,7,79,68,78,1275,1250,1268， 1297,1284,1255,1294,1262,1250,1252,680,693,677,676,670,653,670,661,658,695,665,671,656,686,662,691,675,658， 671,650,667,653,652,686,667,682， 694,654,689,682,667,658,651,652,692,652,655,651,650,698,655,650,679,672,697,696,696,683,1277,1264,1274， 1260,1285,1285,1283,1259,1260,1288,1281,1284,1281,1257,1285,1295,1273,1264,1283,1284,1300,1299,1257,1297,1254,1257,1270,1257， 1295,34,5,73,42,27,36,91,85,19,50,34,21,73,38,18,73]
答案 0 :(得分：0)
一种可能性是使用sklearns高斯混合模型来找到簇。也许这有点过分，因为只有一个维度，但它无论如何都应该有效。
以下是如何完成的示例：
import numpy as np
from sklearn import mixture
# Generate some example data
centers = [50, 700, 1250] # Approximate
y = []
for i in range(20):
y.append(np.random.randn(100) * 50 + centers[np.random.randint(len(centers))])
y = np.c_[np.concatenate(y)]
N = 3 # Clusters
gmix = mixture.GaussianMixture(n_components=3, covariance_type='full')
gmix.fit(y) # Now it thinks it is trained
for i in range(N):
center = gmix.means_[i][0]
std = np.sqrt(gmix.covariances_[i][0])
print "Range %d: %.0f to %.0f" % (i + 1, center - std, center + std)
它应输出类似（使用IPython 2.7）：
Range 1: 1199 to 1298
Range 2: 649 to 752
Range 3: 4 to 102
它当然取决于它找到的集群，这可能并不总是一个很好的解决方案。
答案 1 :(得分：0)
我按照@schwobaseggl的建议引用Unsupervised clustering with unknown number of clusters ，并根据我的需要稍微更改了代码。 这是新代码：
import numpy
import scipy.cluster.hierarchy as hcluster
temp_data = [31,68,74,46,47,83,29,11,9,52,1272,1288,1297,1285,1294,1251,1265,1257,1280,1265,1292,1297,1271,1273,1253,1273,1291,1251,1295,1298,1264,1281,1294,1280,1250,1279,1298,1290,1294,1299,1266,1260,1298,1292,1280,1259,1266,1276,1253,1252,1271,1280,1284,1266,1254,1259,1291,1268,1253,1298,1288,1271,1298,1300,1274,1294,1263,1298,1270,1254,1266,1269,1283,1285,1286,1276,1257,1266,1272,1298,1261,1251,1272,1260,1291,1269,1260,1294,1287,1256,1253,1284,1269,1287,1292,1269,1272,1275,1250,1289,56,35,19,80,47,22,92,8,10,24,87,76,60,63,64,0,1295,1268,1280,1281,1277,1300,1278,1273,1250,1296,1266,1269,1282,1281,1272,1260,1292,1272,1253,1255,1299,1269,1268,1294,1250,1299,1292,1254,1281,1289,1259,1290,1271,1280,1272,1300,1258,1290,1289,1300,1299,1261,1300,1276,1290,1299,1280,1267,1283,1282,1269,1260,1285,1252,1250,1263,1297,1300,1292,1266,1260,1263,1292,1296,1289,1297,1251,1261,1250,1294,1278,1284,1291,1281,1269,1261,1257,1267,1265,1288,1291,1257,1296,1251,1260,1272,1294,1285,1269,1283,1297,1287,1253,1292,1299,1295,1286,1288,1283,1290,20,73,81,6,49,88,96,61,49,94,57,16,61,16,17,19,1280,1257,1259,1277,1257,1262,1263,1280,1292,1250,1287,1272,1258,1253,1285,1285,1257,1291,1273,1260,1267,1250,1280,1281,1263,1269,1292,1250,1282,1263,1274,1288,1296,1266,1291,1271,1273,1281,1261,1289,1269,1287,1296,1283,1280,1298,1259,1270,1259,1289,1269,1284,1295,1297,1256,1300,1281,1296,1284,1288,1285,1296,1277,1251,1279,1295,1281,1264,1280,1263,69,8,30,45,89,61,80,45,9,18,19,11,1255,1299,1296,1293,1287,1250,1265,1291,1281,1250,1286,1286,1251,1287,1266,1288,1254,1260,1260,1254,1267,1299,1273,1250,1300,1250,1279,1255,1293,1292,1278,1277,1252,1299,1278,1258,1268,1274,1285,1258,1279,1270,1278,1286,1278,1253,1267,1300,1295,1298,1285,1288,1274,1272,1252,1256,1283,1289,1251,1258,1253,1257,1297,1269,1292,1253,1273,1281,1251,1280,1253,1274,1275,1287,1296,1298,1296,1291,1284,1261,1267,1290,1273,1281,1263,1270,1264,1269,1278,1284,67,8,40,59,97,64,45,72,45,90,94,7,33,58,97,97,1252,1297,1265,1278,1272,1252,1258,1261,1287,1260,1260,1258,1280,1263,1256,1296,1269,1270,1296,1282,696,678,665,700,700,691,689,688,650,663,662,698,655,660,662,684,690,657,653,663,670,691,687,675,694,670,676,659,661,664,664,689,683,675,687,691,676,659,689,657,659,656,654,679,669,687,666,662,691,1260,1276,1252,1295,1257,1277,1281,1257,1295,1269,1265,1290,1266,1269,1286,1254,1260,1265,1290,1294,1286,1279,1254,1256,1276,1285,1282,1251,1282,1261,1253,56,74,85,94,18,83,38,80,8,4,78,43,7,79,68,78,1275,1250,1268,1297,1284,1255,1294,1262,1250,1252,680,693,677,676,670,653,670,661,658,695,665,671,656,686,662,691,675,658,671,650,667,653,652,686,667,682,694,654,689,682,667,658,651,652,692,652,655,651,650,698,655,650,679,672,697,696,696,683,1277,1264,1274,1260,1285,1285,1283,1259,1260,1288,1281,1284,1281,1257,1285,1295,1273,1264,1283,1284,1300,1299,1257,1297,1254,1257,1270,1257,1295,34,5,73,42,27,36,91,85,19,50,34,21,73,38,18,73]
ndata = [[td, td] for td in temp_data]
data = numpy.array(ndata)
# clustering
thresh = (11.0/100.0) * (max(temp_data) - min(temp_data)) #Threshold 11% of the total range of data
clusters = hcluster.fclusterdata(data, thresh, criterion="distance")
total_clusters = max(clusters)
clustered_index = []
for i in range(total_clusters):
clustered_index.append([])
for i in range(len(clusters)):
clustered_index[clusters[i] - 1].append(i)
clustered_range = []
for x in clustered_index:
clustered_index_x = [temp_data[y] for y in x]
clustered_range.append((min(clustered_index_x) , max(clustered_index_x)))
print clustered_range
我选择了阈值（thres）作为总数据范围的11％
因此该数据集的输出为：
[(0, 97), (1250, 1300), (650, 700)]