我编写了一个使用自制函数执行 kNN 分类的脚本。我已经将其性能与类似的脚本进行了比较，但使用了 sklearn 包。
结果：自制 ~ 20 秒SK学习 ~ 2 秒
所以现在我想知道性能差异主要是由于 sklearn 在较低级别执行(据我所知在 C 中(还是因为我的脚本效率不高。
如果你们中的一些人有参考资料，为编写高效的Python脚本和程序提供信息，我都知道
这是数据文件：数据文件
两个脚本中的文件名、os.environ['R_HOME']、os.environ['R_USER'] 必须根据您的目录结构设置为特定于用户
我的代码使用自制的kNN分类
#Start Timer
import time
tic = time.time()
# Begin Script
import os
os.environ['R_HOME'] = r'C:UsersMyUserDocumentsRR-3.4.1' #setting temporary PATH variables : R_HOME
#a permanent solution could be achieved but more complicated
os.environ['R_USER'] = r'C:UsersMyUserAppDataLocalProgramsPythonPython36Libsite-packagesrpy2'
#same story
import rpy2.robjects as robjects
import numpy as np
import matplotlib.pyplot as plt
## Read R data from ESLII book
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '../ESL.mixture.rda')
robjects.r['load'](filename) #load rda file in R workspace
rObject = robjects.r['ESL.mixture'] #read variable in R workspace and save it into python workspace
#Extract Blue and Orange classes data
classes = np.array(rObject[0]) #note that information about rObject are known by outputing the object into the console
#numpy is able to convert R data natively
BLUE = classes[0:100,:]
BLUE = np.concatenate((BLUE,np.zeros(np.size(BLUE,axis=0))[:,None]),axis=1)
#the [:,None] is necessary to make the 1D array 2D.
#Indeed concatenate requires identical dimensions
#other functions exist such as np.columns_stack but they take more time to execute than basic concatenate
ORANGE = classes[100:200]
ORANGE = np.concatenate((ORANGE,np.ones(np.size(ORANGE,axis=0))[:,None]),axis=1)
trainingSet = np.concatenate((BLUE,ORANGE),axis=0)
##create meshgrid
minBound = -3
maxBound = 4.5
xmesh = np.linspace(minBound, maxBound, 100)
ymesh = np.linspace(minBound, maxBound, 100)
xv, yv = np.meshgrid(xmesh, ymesh)
gridSet =np.stack((xv.ravel(),yv.ravel())).T
def predict(trainingSet, queryPoint, k):
# create list for distances and targets
distances = []
# compute euclidean distance
for i in range (np.size(trainingSet,0)):
distances.append(np.sqrt(np.sum(np.square(trainingSet[i,:-1]-queryPoint))))
#find k nearest neighbors to the query point and compute its outcome
distances=np.array(distances)
indices = np.argsort(distances) #provides indices, sorted from short to long distances
kindices = indices[0:k]
kNN = trainingSet[kindices,:]
queryOutput = np.average(kNN[:,2])
return queryOutput
k = 1
gridSet = np.concatenate((gridSet,np.zeros(np.size(gridSet,axis=0))[:,None]),axis=1)
i=0
for point in gridSet[:,:-1]:
gridSet[i,2] = predict(trainingSet, point, k)
i+=1
#k = 1
#test = predict(trainingSet, np.array([4.0, 1.2]), k)
col = np.where(gridSet[:,2]<0.5,'b','r').flatten() #flatten is necessary. 2D arrays are only accepted with RBA colors
plt.scatter(gridSet[:,0],gridSet[:,1],c=col,s=0.2)
col = np.where(trainingSet[:,2]<0.5,'b','r').flatten() #flatten is necessary. 2D arrays are only accepted with RBA colors
plt.scatter(trainingSet[:,0],trainingSet[:,1],c=col,s=1.0)
plt.contour(xv,yv,gridSet[:,2].reshape(xv.shape),0.5)
plt.savefig('kNN_homeMade.png', dpi=600)
plt.show()
#
#Stop timer
toc = time.time()
print(toc-tic, 'sec Elapsed')
我的代码使用 sklearn kNN
#Start Timer
import time
tic = time.time()
# Begin Script
import os
os.environ['R_HOME'] = r'C:UsersMyUserDocumentsRR-3.4.1' #setting temporary PATH variables : R_HOME
#a permanent solution could be achieved but more complicated
os.environ['R_USER'] = r'C:UsersMyUserAppDataLocalProgramsPythonPython36Libsite-packagesrpy2'
#same story
import rpy2.robjects as robjects
import numpy as np
import matplotlib.pyplot as plt
from sklearn import neighbors
## Read R data from ESLII book
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '../ESL.mixture.rda')
robjects.r['load'](filename) #load rda file in R workspace
rObject = robjects.r['ESL.mixture'] #read variable in R workspace and save it into python workspace
#Extract Blue and Orange classes data
classes = np.array(rObject[0]) #note that information about rObject are known by outputing the object into the console
#numpy is able to convert R data natively
BLUE = classes[0:100,:]
BLUE = np.concatenate((BLUE,np.zeros(np.size(BLUE,axis=0))[:,None]),axis=1)
#the [:,None] is necessary to make the 1D array 2D.
#Indeed concatenate requires identical dimensions
#other functions exist such as np.columns_stack but they take more time to execute than basic concatenate
ORANGE = classes[100:200]
ORANGE = np.concatenate((ORANGE,np.ones(np.size(ORANGE,axis=0))[:,None]),axis=1)
trainingSet = np.concatenate((BLUE,ORANGE),axis=0)
##create meshgrid
minBound = -3
maxBound = 4.5
xmesh = np.linspace(minBound, maxBound, 100)
ymesh = np.linspace(minBound, maxBound, 100)
xv, yv = np.meshgrid(xmesh, ymesh)
gridSet =np.stack((xv.ravel(),yv.ravel())).T
gridSet = np.concatenate((gridSet,np.zeros(np.size(gridSet,axis=0))[:,None]),axis=1)
##classify using kNN
k = 1
clf = neighbors.KNeighborsClassifier(k, weights='uniform',algorithm='brute')
clf.fit(trainingSet[:,:-1],trainingSet[:,-1:].ravel()) #learn, ravel necessary to obtain (n,) shape instead of a vector (n,1)
gridSet[:,2] = clf.predict(np.c_[xv.ravel(), yv.ravel()])
#Plot
col = np.where(gridSet[:,2]<0.5,'b','r').flatten() #flatten is necessary. 2D arrays are only accepted with RBA colors
plt.scatter(gridSet[:,0],gridSet[:,1],c=col,s=0.2)
col = np.where(trainingSet[:,2]<0.5,'b','r').flatten() #flatten is necessary. 2D arrays are only accepted with RBA colors
plt.scatter(trainingSet[:,0],trainingSet[:,1],c=col,s=1.0)
plt.contour(xv,yv,gridSet[:,2].reshape(xv.shape),0.5)
plt.savefig('kNN_sciKit.png', dpi=600)
plt.show()
#
#Stop timer
toc = time.time()
print(toc-tic, 'sec Elapsed')
按照andrew_reece建议并对我的代码进行一些分析，我将计算时间减少到大约 2 秒(而不是 20 秒(。
罪魁祸首是以下代码中的两个for循环：
def predict(trainingSet, queryPoint, k):
# create list for distances and targets
distances = []
#targets = []
# compute euclidean distance
for i in range (np.size(trainingSet,0)):
distances.append(np.sqrt(np.sum(np.square(trainingSet[i,:-1]-queryPoint))))
#find k nearest neighbors to the query point and compute its outcome
distances=np.array(distances)
indices = np.argsort(distances) #provides indices, sorted from short to long distances
kindices = indices[0:k]
kNN = trainingSet[kindices,:]
queryOutput = np.average(kNN[:,2])
return queryOutput
k = 1
gridSet = np.concatenate((gridSet,np.zeros(np.size(gridSet,axis=0))[:,None]),axis=1)
i=0
for point in gridSet[:,:-1]:
gridSet[i,2] = predict(trainingSet, point, k)
i+=1
从包中读取neighbors类sklearn我注意到他们以完全矢量化的方式执行欧几里得距离计算。所以我已经阅读了代码，现在了解了该功能...但我懒得完全重写它。相反，我只需导入函数euclidean_distances并将其直接用于我的数据，从而演示改进。修改后的部分如下：
def predict(trainingSet, queryPoints, k):
# create list for distances and targets
distances = euclidean_distances(trainingSet[:,:-1],queryPoints[:,:-1]) #provides distances between each training point and each query point
#line i is distance between training i and all query (in columns)
#so k neighbors of query j are k first lines for column j
#find k nearest neighbors to the query point and compute its outcome
indices = np.argsort(distances,axis=0) #provides indices, sorted from short to long distances for each query point
#kindices = indices[0:k]
kNN = trainingSet[indices[:k,:],2] #produce the kNN outputs
queryOutput = np.average(kNN,0)
return queryOutput
k = 1
gridSet = np.concatenate((gridSet,np.zeros(np.size(gridSet,axis=0))[:,None]),axis=1)
gridSet[:,2] = predict(trainingSet, gridSet, k)