本總結僅為防止遺忘而作

                               常見的分佈有

                        關於具體分佈的理論部分在此不做過多闡述，可自行查閱資料。其中負二項分佈

                        下面給出生活中具體的例子

對於上面的例子，若11分制，劉詩雯若獲勝那麼最後一球肯定是劉詩雯贏的，則對於公式

x=11,r=6,最後一球為劉詩雯贏的，那麼前十球中劉詩雯贏5球，即有上面的公式。

下面附上python執行的程式碼：

import numpy as np
from scipy import special
if 
 
__name__ =='__main__':
method='strict'
#暴力模擬
if method =='simulation':
p = 0.6
a,b,c = 0,0,0
t,T = 0,1000000
while t < T:
a = b = 0
while (a <= 11) and (b <= 11):
if np.random.uniform()<p:
a += 1
else:
b += 1
if a>b:
c += 1
t += 1
print(float(c)/float(T))

    #直接計算
elif method == 'simple':
answer = 
 
0
p = 0.6
N = 11
for x in np.arange(N):   #x為對手得分
answer += special.comb(N+x-1,x)*((1-p)**x)*(p**N)  #scipy.special.comb(N, k)   二項分佈
print(answer)                                         #The number of combinations of N things taken k at a time.
    #嚴格計算
else:
answer = 0
p = 0.6
N = 11
for x in np.arange(N-
 
1):  #x為對手得分 11：9  11:8 11:7 ...
answer += special.comb(N+x-1,x)*((1-p)**x)*(p**N)
        print(answer)
        p10 = special.comb(2*(N-1),N-1)*((1-p)*p)**(N-1)   # 10:10的概率
t = 0
for n in np.arange(100):
t += (2*p*(1-p))**n*p*p
        answer += p10*t
        print(answer)

import numpy as np
from scipy import stats
import matplotlib as mpl
import matplotlib.pyplot as plt
import warnings

mpl.rcParams['axes.unicode_minus'] = False
mpl.rcParams['font.sans-serif'] = 'SimHei'
def calc_pearson(x, y):
std1 = np.std(x)                               #標準差
# np.sqrt(np.mean(x**2) - np.mean(x)**2)
std2 = np.std(y)
    cov = np.cov(x, y, bias=True)[0,1]
    print(np.cov(x,y))
    return cov / (std1 * std2)         # ρ
def intro():
N = 10
x = np.random.rand(N)
    y = 2 * x + np.random.randn(N) * 0.1
print(x)
    print(y)
    print('系統計算：', stats.pearsonr(x, y)[0])         #r是相關係數，取值[-1,1] 表示線性相關程度
# print(0.0005[0])
print('手動計算：', calc_pearson(x, y))


def rotate(x, y, theta=45):
data = np.vstack((x, y))   #vstack   vertical stack ，hstack horizon stack
print (data)
    mu = np.mean(data, axis=1)
    mu = mu.reshape((-1, 1))
    print('mu=',mu)
    data -= mu
    print ('data-mu=',data)
    theta *= (np.pi / 180)
    c = np.cos(theta)
    s = np.sin(theta)
    m = np.array((c, -s), (s, c))
    print('m=',m)
    return np.dot(m,data) + mu


def pearson(x, y, tip):
clrs = list('rgbmycrgbmycrgbmycrgbmyc')
    plt.figure(figsize=(10, 8), facecolor='w')
    for i, theta in enumerate(np.linspace(0, 90, 6)):
xr, yr = rotate(x, y, theta)
        p = stats.pearsonr(xr, yr)[0]
        print (calc_pearson(xr, yr))
        print('旋轉角度：', theta, 'Pearson相關係數：', p)
        str = '相關係數：%.3f' % p
        plt.scatter(xr, yr, s=40, alpha=0.9, linewidths=0.5, c=clrs[i], marker='o', label=str)
    plt.legend(loc='upper left', shadow=True)
    plt.xlabel('X')
    plt.ylabel('Y')
    plt.title('Pearson相關係數與資料分佈：%s' % tip, fontsize=18)
    plt.grid(b=True)
    plt.show()


if __name__ == '__main__':
# warnings.filterwarnings(action='ignore', category=RuntimeWarning)
np.random.seed(0)

    intro()

    N = 1000
# tip = '一次函式關係'
    # x = np.random.rand(N)
    # y = np.zeros(N) + np.random.randn(N)*0.001
    #
tip = u'二次函式關係'
x = np.random.rand(N)
    y = x ** 2 + np.random.randn(N)*0.002
#
    # tip = u'正切關係'
    # x = np.random.rand(N) * 1.4
    # y = np.tan(x)
    #
    # tip = u'二次函式關係'
    # x = np.linspace(-1, 1, 101)
    # y = x ** 2
    #
    # tip = u'橢圓'
    # x, y = np.random.rand(2, N) * 60 - 30
    # y /= 5
    # idx = (x**2 / 900 + y**2 / 36 < 1)
    # x = x[idx]
    # y = y[idx]
pearson(x, y, tip)