小编hal*_*_me的帖子

我通过scikit-learn图书馆学习机器学习.我使用以下代码将决策树分类器和随机森林分类器应用于我的数据:

def decision_tree(train_X, train_Y, test_X, test_Y):

    clf = tree.DecisionTreeClassifier()
    clf.fit(train_X, train_Y)

    return clf.score(test_X, test_Y)


def random_forest(train_X, train_Y, test_X, test_Y):
    clf = RandomForestClassifier(n_estimators=1)
    clf = clf.fit(X, Y)

    return clf.score(test_X, test_Y)

为什么随机森林分类器的结果更好(100次运行,随机抽样2/3的数据用于训练,1/3用于测试)？

100%|???????????????????????????????????????| 100/100 [00:01<00:00, 73.59it/s]
Algorithm: Decision Tree
  Min     : 0.3883495145631068
  Max     : 0.6476190476190476
  Mean    : 0.4861783113770316
  Median  : 0.48868030937802126
  Stdev   : 0.047158171852401135
  Variance: 0.0022238931724605985
100%|???????????????????????????????????????| 100/100 [00:01<00:00, 85.38it/s]
Algorithm: Random Forest
  Min     : 0.6846846846846847
  Max     : 0.8653846153846154
  Mean    : 0.7894823428836184
  Median  : 0.7906101571063208
  Stdev   : 0.03231671150915106
  Variance: 0.0010443698427656967 …

我正在尝试分析我的代码以检测瓶颈。我搜索了一些分析器，但从未找到我要找的东西。

我过去使用过很多Python，有一个软件： line_profiler，它给出了这样的回报：

0         Line     Hits  Time  Per Hit   % Time  Line Contents

11                                           @profile
12                                           def compute_prior(folder):
13                                               """
14                                               Given a folder, we compute the prior of neg and pos
15                                               folder = "./movie-reviews-en/train/"
16                                               """
17                                               # we compute the number of positive reviews
18         3         1719    573.0     52.9      number_positive = len([f for f in listdir(folder + "pos/")])
19                                               # then the negative
20         3         1512    504.0     46.6      number_negative = len([f for f in listdir(folder + …

我正在尝试解析这个forme的文件:

12
0,1,2,3,1,2,3,4,2,3,4,5
1,0,1,2,2,1,2,3,3,2,3,4
2,1,0,1,3,2,1,2,4,3,2,3
3,2,1,0,4,3,2,1,5,4,3,2
1,2,3,4,0,1,2,3,1,2,3,4
2,1,2,3,1,0,1,2,2,1,2,3
3,2,1,2,2,1,0,1,3,2,1,2
4,3,2,1,3,2,1,0,4,3,2,1
2,3,4,5,1,2,3,4,0,1,2,3
3,2,3,4,2,1,2,3,1,0,1,2
4,3,2,3,3,2,1,2,2,1,0,1
5,4,3,2,4,3,2,1,3,2,1,0
0,5,2,4,1,0,0,6,2,1,1,1
5,0,3,0,2,2,2,0,4,5,0,0
2,3,0,0,0,0,0,5,5,2,2,2
4,0,0,0,5,2,2,10,0,0,5,5
1,2,0,5,0,10,0,0,0,5,1,1
0,2,0,2,10,0,5,1,1,5,4,0
0,2,0,2,0,5,0,10,5,2,3,3
6,0,5,10,0,1,10,0,0,0,5,0
2,4,5,0,0,1,5,0,0,0,10,10
1,5,2,0,5,5,2,0,0,0,5,0
1,0,2,5,1,4,3,5,10,5,0,2
1,0,2,5,1,0,3,0,10,0,2,0

第一行给出矩阵的大小:n x n n跟随线是矩阵D.然后n跟随线是矩阵W.所以2n + 1有线.

这是解析它并将其放入变量的代码.

func readFile(path string) (int64, Matrix, Matrix) {
    // open the file
    f, _ := os.Open(path)
    defer f.Close()

    // init the new reader on the opened file
    r := bufio.NewReader(f)

    // we get the n value
    line, _ := r.ReadString('\n')
    splitedLine := strings.Fields(line)
    tmp, _ …