python使用

numpy.shape

>>> import numpy as np
>>> a = np.array([1, 2, 3, 4])
>>> a.shape
(4,)
>>> a.shape[0]
4
>>> c = np.array([[1, 2, 3, 4],[4, 5, 6, 7], [7, 8, 9, 10]])
>>> c.shape
(3, 4)
>>> c.shape[0]
3
>>> c.shape[1]
4

range

>>> range(1,5) #代表从1到5(不包含5)
[1, 2, 3, 4]
>>> range(1,5,2) #代表从1到5，间隔2(不包含5)
[1, 3]
>>> range(5) #代表从0到5(不包含5)
[0, 1, 2, 3, 4]

numpy.sum() 求和

>>> a
array([[6, 7, 1, 6],
[1, 0, 2, 3],
[7, 8, 2, 1]])
>>> np.sum(a)
44
>>> np.sum(a,axis=0)
array([14, 15,  5, 10])
>>> np.sum(a,axis=1)
array([20,  6, 18])
>>> np.sum(a,axis=-1)

numpy.mean() 均值

>>> a
array([[6, 7, 1, 6],
[1, 0, 2, 3],
[7, 8, 2, 1]])
>>> np.mean(a)
3.6666666666666665
>>> np.mean(a,axis=0)
array([ 4.66666667,  5.        ,  1.66666667,  3.33333333])

numpy.var() 方差

>>> np.var(a)
7.7222222222222223
>>> np.var(a,axis=0)
array([  6.88888889,  12.66666667,   0.22222222,   4.22222222])

numpy.std() 标准差

>>> np.std(a,axis=0)
array([ 2.62466929,  3.55902608,  0.47140452,  2.05480467])

参考 python 科学计算学习一：numpy快速处理数据（3）

numpy.max() 最大值

numpy.min() 最小值

numpy.argmax() 最大值的下标

numpy.argmin() 最小值的下标

numpy.sort() 排序

>>> a
array([[6, 7, 1, 6],
[1, 0, 2, 3],
[7, 8, 2, 1]])
>>> a.sort()
>>> a
array([[1, 6, 6, 7],
[0, 1, 2, 3],
[1, 2, 7, 8]])
>>> np.sort(a,axis=0)
array([[0, 1, 2, 3],
[1, 2, 6, 7],
[1, 6, 7, 8]])

numpy.argsort() 排序后的数据原来位置的下标

>>> np.argsort(a,axis=0)
array([[1, 1, 1, 1],
[0, 2, 0, 0],
[2, 0, 2, 2]])

pickle.load(file) Read a string from the open file object file and interpret it as a pickle data stream, reconstructing and returning the original object hierarchy

Welcome To PyCrust 0.7.2 - The Flakiest Python Shell
Sponsored by Orbtech - Your source for Python programming expertise.
Python 2.2.1 (#1, Aug 27 2002, 10:22:32)
[GCC 3.2 (Mandrake Linux 9.0 3.2-1mdk)] on linux-i386
Type "copyright", "credits" or "license" for more information.
>>> import cPickle as pickle
>>> t1 = ('this is a string', 42, [1, 2, 3], None)
>>> t1
('this is a string', 42, [1, 2, 3], None)
>>> p1 = pickle.dumps(t1)
>>> p1
"(S'this is a string'\nI42\n(lp1\nI1\naI2\naI3\naNtp2\n."
>>> print p1
(S'this is a string'
I42
(lp1
I1
aI2
aI3
aNtp2
.
>>> t2 = pickle.loads(p1)
>>> t2
('this is a string', 42, [1, 2, 3], None)
>>> p2 = pickle.dumps(t1, True)
>>> p2
'(U\x10this is a stringK*]q\x01(K\x01K\x02K\x03eNtq\x02.'
>>> t3 = pickle.loads(p2)
>>> t3
('this is a string', 42, [1, 2, 3], None)

14.