对于stackoverflow和编程来说都是新手,我来自以下统计背景,是KNN算法的实现。遇到错误
TypeError: unsupported operand type(s) for -: 'str' and 'str'
。
这些是我遇到的其他错误。预先感谢您的答复。
文件“ knn.py”,第78行,在 main()
文件“ knn.py”,第71行,位于主目录中 邻居= getNeighbors(trainingSet,testSet [x],k)
getNeighbors中的文件“ knn.py”,第33行 dist = euclideanDistance(testInstance,trainingSet [x],长度)
文件“ knn.py”,第26行,在euclideanDistance中 距离+ = pow((instance1 [x]-instance2 [x]),2)
import csv
import random
import math
import pandas
import numpy
def loadDataset(filename, split, trainingSet=[] , testSet=[]):
filename = 'data1.csv'
raw_data = open(filename, 'rt')
reader = csv.reader(raw_data, delimiter=',', quoting=csv.QUOTE_NONE)
dataset = list(reader)
for x in range(len(dataset)-1):
for y in range(4):
dataset[x][y] = float(dataset[x][y])
if random.random() < split:
trainingSet.append(dataset[x])
else:
testSet.append(dataset[x])
def euclideanDistance(instance1, instance2, length):
distance = 0
for x in range(length):
distance += pow((instance1[x] - instance2[x]), 2)
return math.sqrt(distance)
def getNeighbors(trainingSet, testInstance, k):
distances = []
length = len(testInstance)-1
for x in range(len(trainingSet)):
dist = euclideanDistance(testInstance, trainingSet[x], length)
distances.append((trainingSet[x], dist))
distances.sort(key=operator.itemgetter(1))
neighbors = []
for x in range(k):
neighbors.append(distances[x][0])
return neighbors
def getResponse(neighbors):
classVotes = {}
for x in range(len(neighbors)):
response = neighbors[x][-1]
if response in classVotes:
classVotes[response] += 1
else:
classVotes[response] = 1
sortedVotes = sorted(classVotes.iteritems(), key=operator.itemgetter(1), reverse=True)
return sortedVotes[0][0]
def getAccuracy(testSet, predictions):
correct = 0
for x in range(len(testSet)):
if testSet[x][-1] == predictions[x]:
correct += 1
return (correct/float(len(testSet))) * 100.0
def main():
# prepare data
trainingSet=[]
testSet=[]
split = 0.67
loadDataset('data1.csv', split, trainingSet, testSet)
print ('Train set: ' + repr(len(trainingSet)))
print ('Test set: ' + repr(len(testSet)))
# generate predictions
predictions=[]
k = 3
for x in range(len(testSet)):
neighbors = getNeighbors(trainingSet, testSet[x], k)
result = getResponse(neighbors)
predictions.append(result)
print('> predicted=' + repr(result) + ', actual=' + repr(testSet[x][-1]))
accuracy = getAccuracy(testSet, predictions)
print('Accuracy: ' + repr(accuracy) + '%')
main()
答案 0 :(得分:0)
Vamsi,
我注意到您正在使用numpy和pandas。在进行调试时,我确实想推荐另一个很棒的软件包,即sci-kit learning。
他们已经内置了KNN的实现:http://scikit-learn.org/stable/modules/neighbors.html
编辑: 我相信第26行存在强制转换问题。看来您正在尝试减去2个字符串。我认为,如果您使用整数数据,这可能会解决您的问题
def euclideanDistance(instance1, instance2, length):
distance = 0
for x in range(length):
distance += pow((int(instance1[x]) - int(instance2[x])), 2)
return math.sqrt(distance)
如果您使用浮点数据,那么以下方法将起作用:
distance += pow((float(instance1[x]) - float(instance2[x])), 2)
答案 1 :(得分:0)
df = pd.read_csv(filename)
默认返回字符串。您需要将适当的项目转换为浮点数。
此外,您可以直接使用
获取熊猫数据框data = df.values
并使用
获得一个Numpy数组(如果您愿意)euclideanDistance
然后对这些数据进行操作。
csv的第一行可能是标题,并且您没有跳过它。因此,您的第一个训练实例实际上是构成标头的字符串,并且您尝试在length = len(testInstance) - 1
for x in range(len(trainingSet)):
dist = euclideanDistance(testInstance, trainingSet[x], length)
distances.append((trainingSet[x], dist))
函数中减去字符串。
话虽如此,您的代码非常 unpythonic 。
例如,
length
您不需要传递len
,因为可以使用euclideanDistance
中的trainingSet
函数来查找。
您可以直接遍历for x in trainingSet:
dist = euclideanDistance(testInstance, x, length)
distances.append((x, dist))
中的实例
distances = [(x, euclideanDistance(testInstance, x) for x in trainingSet)]
或更佳
neighbors = []
for x in range(k):
neighbors.append(distances[x][0])
同样,
neighbors = [x[0] for x in distances[:k]]
可以是
def loadDataset(filename, split, trainingSet=[] , testSet=[]):
Python还允许您一次返回多个项目,并且像其他语言中常见的那样,通过引用返回是极其糟糕的做法。所以
def loadDataset(filename, split):
trainingSet = []
testSet = []
# ...
return trainingSet, testSet
trainingSet, testSet = loadDataset(filename, ',')
应该是
testInstance
对于这种应用程序,在这种情况下,应避免使用Python列表,而应使用Numpy数组存储数据。这样,许多操作都可以向量化,从而极大地提高了性能。
例如,要计算trainingSet
和# I'm deliberately converting them to numpy array but in general
# you should keep them in this form right from the start
testInstance = np.asarray(testInstance).reshape(1, -1)[:-1] # Your last item is label. Ideally remove them at the beginning
trainingSet = np.vstack(trainingSet)[:, :-1] # Same case as above.
# Here we use broadcasting to obtain difference
# between each row in trainingSet and testInstance
distances = np.linalg.norm(trainingSet - testInstance, axis=1)**2
之间的距离
https://start.spring.io/
如果允许/愿意使用Scipy,那么还有其他机会来减少代码行并加快操作速度。
用这种方式编写代码不仅可以提供更好的性能,而且还更简洁,更接近原始数学表达式。