numpy.ndarray语法理解确认

时间:2016-08-23 06:04:28

标签: python python-2.7 numpy machine-learning logistic-regression

我在这里引用代码示例(http://scikit-learn.org/stable/auto_examples/linear_model/plot_iris_logistic.html),并且特别被此行iris.data[:, :2]搞糊涂,因为iris.data是150(行)* 4(列)维度我认为这意味着,选择所有行和前两列。我在这里要求确认我的理解是否正确,因为我需要时间但却找不到这样的语法定义官方文档。

另一个问题是,我使用以下代码获取行数和#列,不确定是否更优雅的方式?我的代码更多是Python原生样式,并且不确定numpy是否有更好的样式来获取相关值。

print len(iris.data) # for number of rows
print len(iris.data[0]) # for number of columns

将Python 2.7与miniconda解释器一起使用。

print(__doc__)


# Code source: Gaël Varoquaux
# Modified for documentation by Jaques Grobler
# License: BSD 3 clause

import numpy as np
import matplotlib.pyplot as plt
from sklearn import linear_model, datasets

# import some data to play with
iris = datasets.load_iris()
X = iris.data[:, :2]  # we only take the first two features.
Y = iris.target

h = .02  # step size in the mesh

logreg = linear_model.LogisticRegression(C=1e5)

# we create an instance of Neighbours Classifier and fit the data.
logreg.fit(X, Y)

# Plot the decision boundary. For that, we will assign a color to each
# point in the mesh [x_min, m_max]x[y_min, y_max].
x_min, x_max = X[:, 0].min() - .5, X[:, 0].max() + .5
y_min, y_max = X[:, 1].min() - .5, X[:, 1].max() + .5
xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h))
Z = logreg.predict(np.c_[xx.ravel(), yy.ravel()])

# Put the result into a color plot
Z = Z.reshape(xx.shape)
plt.figure(1, figsize=(4, 3))
plt.pcolormesh(xx, yy, Z, cmap=plt.cm.Paired)

# Plot also the training points
plt.scatter(X[:, 0], X[:, 1], c=Y, edgecolors='k', cmap=plt.cm.Paired)
plt.xlabel('Sepal length')
plt.ylabel('Sepal width')

plt.xlim(xx.min(), xx.max())
plt.ylim(yy.min(), yy.max())
plt.xticks(())
plt.yticks(())

plt.show()

的问候, 林

1 个答案:

答案 0 :(得分:1)

你是对的。第一种语法选择前2列/功能。查询维度的另一种方法是查看iris.data.shape。这将返回具有长度的n维元组。您可以在此处找到一些文档:http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html

import numpy as np
x = np.random.rand(100, 200)
# Select the first 2 columns
y = x[:, :2]
# Get the row length
print (y.shape[0])
# Get the column length
print (y.shape[1])
# Number of dimensions
print (len(y.shape))