处理 NumPy 矩阵和 ufunc

原文链接：https://github.com/wizardforcel/data-science-notebook/blob/master/numpy/%E5%A4%84%E7%90%86%20NumPy%20%E7%9F%A9%E9%98%B5%E5%92%8C%20ufunc.md

创建矩阵

import numpy as np# mat 函数创建矩阵# 空格分割行，分号分隔列A = np.mat('1 2 3; 4 5 6; 7 8 9')print "Creation from string", A'''Creation from string [[1 2 3] [4 5 6] [7 8 9]]'''# 矩阵转置print "transpose A", A.T'''transpose A [[1 4 7] [2 5 8] [3 6 9]]'''# 求逆（O(n^3)复杂度）print "Inverse A", A.I'''Inverse A [[ -4.50359963e+15   9.00719925e+15  -4.50359963e+15] [  9.00719925e+15  -1.80143985e+16   9.00719925e+15] [ -4.50359963e+15   9.00719925e+15  -4.50359963e+15]]'''print "Check Inverse", A * A.I # 从 ndarray 创建矩阵print "Creation from array", np.mat(np.arange(9).reshape(3, 3))'''Creation from array [[0 1 2] [3 4 5] [6 7 8]]'''

从其它矩阵创建矩阵

import numpy as np# eye 创建单位矩阵A = np.eye(2)print "A", A'''A [[ 1.  0.] [ 0.  1.]]'''# 数乘是乘每个元素B = 2 * Aprint "B", B'''B [[ 2.  0.] [ 0.  2.]]'''# 复合矩阵print "Compound matrix\n", np.bmat("A B; A B")'''Compound matrix[[ 1.  0.  2.  0.] [ 0.  1.  0.  2.] [ 1.  0.  2.  0.] [ 0.  1.  0.  2.]]'''

通用函数（ufunc）

import numpy as npdef ultimate_answer(a):   # 创建与 a 形状相同的零数组   result = np.zeros_like(a)   # 每个值都赋为 42   result.flat = 42   return result# 创建 ultimate_answer 的通用函数ufunc = np.frompyfunc(ultimate_answer, 1, 1) print "The answer", ufunc(np.arange(4))# The answer [42 42 42 42]print "The answer", ufunc(np.arange(4).reshape(2, 2))'''The answer [[42 42]           [[42 42]            [42 42]]'''

通用函数的方法

import numpy as npa = np.arange(9)# a[0] op a[1] op ... op a[n]print "Reduce", np.add.reduce(a)# Reduce 36# s[i] = a[0] + ... + a[i]print "Accumulate", np.add.accumulate(a)# Accumulate [ 0  1  3  6 10 15 21 28 36]print "Reduceat", np.add.reduceat(a, [0, 5, 2, 7])# Reduceat [10  5 20 15]print "Reduceat step I", np.add.reduce(a[0:5])# Reduceat step I 10print "Reduceat step II", a[5]# Reduceat step II 5print "Reduceat step III", np.add.reduce(a[2:7])# Reduceat step III 20print "Reduceat step IV", np.add.reduce(a[7:])# Reduceat step IV 15# 返回 a[i] op b[j] 的矩阵print "Outer", np.add.outer(np.arange(3), a)'''Outer [[ 0  1  2  3  4  5  6  7  8]       [ 1  2  3  4  5  6  7  8  9]       [ 2  3  4  5  6  7  8  9 10]]'''

数组除法

# 导入 Py3 的除法特性# 除法运算符变为真除法# 整数除以整数产生浮点from __future__ import divisionimport numpy as npa = np.array([2, 6, 5])b = np.array([1, 2, 3])# divide 经典除法# divide(1, 2): 0, divide(1.0, 2): 0.5print "Divide", np.divide(a, b), np.divide(b, a)# Divide [2 3 1] [0 0 0]# true_divide 真除法# true_divide(1, 2): 0.5, true_divide(1.0, 2): 0.5print "True Divide", np.true_divide(a, b), np.true_divide(b, a)# True Divide [ 2.          3.          1.66666667] [ 0.5        0.33333333  0.6       ]# floor_divide 地板除法# floor_divide(1, 2): 0, floor_divide(1.0, 2): 0.0print "Floor Divide", np.floor_divide(a, b), np.floor_divide(b, a)# Floor Divide [2 3 1] [0 0 0]c = 3.14 * bprint "Floor Divide 2", np.floor_divide(c, b), np.floor_divide(b, c)# Floor Divide 2 [ 3.  3.  3.] [ 0.  0.  0.]# 除法运算符已变成真除法print "/ operator", a/b, b/a# / operator [ 2.          3.          1.66666667] [ 0.5         0.33333333  0.6       ]# 地板除法print "// operator", a//b, b//a# // operator [2 3 1] [0 0 0]# 地板除法print "// operator 2", c//b, b//c# // operator 2 [ 3.  3.  3.] [ 0.  0.  0.]

求余数

import numpy as np# 余数的符号由除数决定# 3 除以 2 商 1 余 1# -3 除以 2 商 -2 余 1# 3 除以 -2 商 -2 余 -1# -3 除以 -2 商 1 余 -1a = np.arange(-4, 4)print "Remainder", np.remainder(a, 2)# Remainder [0 1 0 1 0 1 0 1]print "Mod", np.mod(a, 2)# Mod [0 1 0 1 0 1 0 1]print "% operator", a % 2# % operator [0 1 0 1 0 1 0 1]# fmod 会将被除数和除数的符号交换# 根据被除数来决定余数符号print "Fmod", np.fmod(a, 2)# Fmod [ 0 -1  0 -1  0  1  0  1]

斐波那契数

import numpy as npF = np.matrix([[1, 1], [1, 0]])print "F", F'''F [[1 1]  [1 0]]'''print "8th Fibonacci", (F ** 7)[0, 0]# 8th Fibonacci 21n = np.arange(1, 9)# 计算黄金比例数sqrt5 = np.sqrt(5)phi = (1 + sqrt5)/2# 斐波那契数的通项公式# rint 对浮点数取整，但不改变浮点类型fibonacci = np.rint((phi**n - (-1/phi)**n)/sqrt5)print "Fibonacci", fibonacci# Fibonacci [  1.   1.   2.   3.   5.   8.  13.  21.]

利萨茹曲线

# x 和 y 都是 t 的正弦函数# 但是频率和初相不同import numpy as npfrom matplotlib.pyplot import plotfrom matplotlib.pyplot import showimport sysa = float(sys.argv[1])b = float(sys.argv[2])t = np.linspace(-np.pi, np.pi, 201)x = np.sin(a * t + np.pi/2)y = np.sin(b * t)plot(x, y)show()

（a 为 9，b 为 8）

方波

import numpy as npfrom matplotlib.pyplot import plotfrom matplotlib.pyplot import showimport syst = np.linspace(-np.pi, np.pi, 201)'''array([-3.14159265, -3.11017673, -3.0787608 , ... ,        3.11017673,  3.14159265])'''k = np.arange(1, float(sys.argv[1]))k = 2 * k - 1# 假设传入 99：# array([1, 3, ..., 195])f = np.zeros_like(t)# array([0] * 201)for i in range(len(t)):   f[i] = np.sum(np.sin(k * t[i])/k)f = (4 / np.pi) * fplot(t, f)show()

三角波

import numpy as npfrom matplotlib.pyplot import plotfrom matplotlib.pyplot import showimport syst = np.linspace(-np.pi, np.pi, 201)k = np.arange(1, float(sys.argv[1]))f = np.zeros_like(t)for i in range(len(t)):   f[i] = np.sum(np.sin(2 * np.pi * k * t[i])/k)f = (-2 / np.pi) * fplot(t, f, lw=1.0)plot(t, np.abs(f), lw=2.0)show()

位操作

import numpy as np# 对于一个非零的数，取反之后符号位会发生变化# 异或之后符号位为 1，一定小于 0x = np.arange(-9, 9)y = -xprint "Sign different?", (x ^ y) < 0print "Sign different?", np.less(np.bitwise_xor(x, y), 0)'''Sign different? [ True  True  True  True  True  True  True  True  True False  True  True  True  True  True  True  True  True]Sign different? [ True  True  True  True  True  True  True  True  True False  True  True  True  True  True  True  True  True]'''# 检查是否是 2 的幂# 2 的幂只有一位是 1，其它都是 0# 减一之后，右边都是 1，其它都是 0# 位与操作之后得 0# 但是这里并没有排除 0print "Power of 2?\n", x, "\n", (x & (x - 1)) == 0print "Power of 2?\n", x, "\n", np.equal(np.bitwise_and(x,  (x - 1)), 0)'''Power of 2?[-9 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8] [False False False False False False False False False  True  True  True False  True False False False  True]Power of 2?[-9 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8] [False False False False False False False False False  True  True  True False  True False False False  True]'''# m % 4 相当于 m & 0b11，只保留后两位print "Modulus 4\n", x, "\n", x & ((1 << 2) - 1)print "Modulus 4\n", x, "\n", np.bitwise_and(x, np.left_shift(1, 2) - 1)'''Modulus 4[-9 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8] [3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0]Modulus 4[-9 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8] [3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0]'''