python统计方法median、var、mean、diff

import numpy as np

'''
numpy.median 计算沿指定轴的中位数 返回数组元素的中位数
median(a, axis=None, out=None,overwrite_input=False, keepdims=False)
各参数为：
a：输入的数组；
axis：计算哪个 轴 上的中位数，比如输入是二维数组，那么axis=0对应行，axis=1对应列；
out：用于放置求取中位数后的数组。 它必须具有与预期输出相同的形状和缓冲区长度；
overwrite_input :一个bool型的参数，默认为Flase。
如果为True那么将直接在数组内存中计算，
这意味着计算之后，原数组没办法保存，
但是好处在于节省内存资源，Flase则相反；
keepdims：一个bool型的参数，默认为Flase。
如果为True那么求取中位数的那个轴将保留在结果中；
'''

a = np.array([[10,  7,  4],
[ 3,  2,  1]])
print( np.median(a) )   # 3.5
m = np.median(a, axis=0) # [ 6.5  4.5  2.5]
print( np.median(a, axis=1) )   # [ 7.  2.]
out = np.zeros_like(m)  # 返回与给定数组相同的形状和类型的0数组
#print(out)  # [ 0.  0.  0.]  都替换为 0
print( np.median(a, axis=0, out=m) ) # [ 6.5  4.5  2.5]

b = a.copy()  #复制 一个 a 对象
#print('b :',b)
c = np.median(b, axis=1, overwrite_input=True)
#print('c :',c)

#范例 - 统计分析
c=np.loadtxt('data.csv', delimiter=',', usecols=(6,), unpack=True)
#print( ' c : ', c )
print ("median =", np.median(c))  #取中位数
sorted = np.sort(c)  # 排序  也可用 np.msort
print ("sorted =", sorted)  # 一行数组

N = len(c)
print ("middle =", sorted[(N - 1)/2])
print ("average middle =", (sorted[N /2] + sorted[(N - 1) / 2]) / 2)

print ("variance =", np.var(c))  # var计算轴上的 方差
print ("variance from definition =", np.mean((c - c.mean())**2))
'''
方差的定义  ： c 减去 轴上 mean()算术平均值 ，再平方，再求均值
'''

#范例：股票收益率
returns = np.diff( c ) / c[ : -1]
print( 'np.diff( c )  ' , np.diff( c ) )   #
#print( 'c[ : -1]  ' , c[ : -1] )   # 截掉最后一个元素
print ("Standard deviation =", np.std(returns))   # numpy.std() 计算矩阵 标准差

logreturns = np.diff( np.log(c) )

posretindices = np.where(returns > 0)
print ("Indices with positive returns", posretindices)

annual_volatility = np.std(logreturns)/np.mean(logreturns)
annual_volatility = annual_volatility / np.sqrt(1./252.)
print ("Annual volatility", annual_volatility)
print ("Monthly volatility", annual_volatility * np.sqrt(1./12.))

'''
diff 沿着指定轴计算第N维的离散差值
参数：
a：输入矩阵
n：可选，代表要执行几次差值 this减去前一个元素
axis：默认是从最后一个开始
'''
# 从 2 到 14(不包括结束位置)各数值，reshape按照 3 行 4列 的矩阵排布
A = np.arange(2 , 14).reshape((3 , 4));
A[1 , 1] = 8 # A矩阵里  第一行 第一列 的 值 赋值 为 8
print('A:' , A)
'''
A: [[ 2  3  4  5]
[ 6  8  8  9]
[10 11 12 13]]
'''
# 从输出结果可以看出，其实 diff函数就是执行的是 this减去前一个元素 。
print(np.diff(A))
'''
[[1 1 1]
[2 0 1]
[1 1 1]]
'''
print(np.diff(A,n=2))  # n 可选，代表要执行几次差值 this减去前一个元素
'''
[[ 0  0]
[-2  1]
[ 0  0]]
'''

'''
data.csv
AAPL,28-01-2017, ,344.17,344.4,333.53,336.1,21144800
AAPL,31-01-2017, ,335.8,340.04,334.3,339.32,13473000
AAPL,01-02-2017, ,341.3,345.65,340.98,345.03,15236800
AAPL,02-02-2017, ,344.45,345.25,343.55,344.32,9242600
AAPL,03-02-2017, ,343.8,344.24,338.55,343.44,14064100
AAPL,04-02-2017, ,343.61,346.7,343.51,346.5,11494200
AAPL,07-02-2017, ,347.89,353.25,347.64,351.88,17322100
AAPL,08-02-2017, ,353.68,355.52,352.15,355.2,13608500
AAPL,09-02-2017, ,355.19,359,354.87,358.16,17240800
AAPL,10-02-2017, ,357.39,360,348,354.54,33162400
AAPL,11-02-2017, ,354.75,357.8,353.54,356.85,13127500
AAPL,14-02-2017, ,356.79,359.48,356.71,359.18,11086200
AAPL,15-02-2017, ,359.19,359.97,357.55,359.9,10149000
AAPL,16-02-2017, ,360.8,364.9,360.5,363.13,17184100
AAPL,17-02-2017, ,357.1,360.27,356.52,358.3,18949000
AAPL,18-02-2017, ,358.21,359.5,349.52,350.56,29144500
AAPL,22-02-2017, ,342.05,345.4,337.72,338.61,31162200
AAPL,23-02-2017, ,338.77,344.64,338.61,342.62,23994700
AAPL,24-02-2017, ,344.02,345.15,338.37,342.88,17853500
AAPL,25-02-2017, ,345.29,348.43,344.8,348.16,13572000
AAPL,28-02-2017, ,351.21,355.05,351.12,353.21,14395400
AAPL,01-03-2017, ,355.47,355.72,347.68,349.31,16290300
AAPL,02-03-2017, ,349.96,354.35,348.4,352.12,21521000
AAPL,03-03-2017, ,357.2,359.79,355.92,359.56,17885200
AAPL,04-03-2017, ,360.07,360.29,357.75,360,16188000
AAPL,07-03-2017, ,361.11,361.67,351.31,355.36,19504300
AAPL,08-03-2017, ,354.91,357.4,352.25,355.76,12718000
AAPL,09-03-2017, ,354.69,354.76,350.6,352.47,16192700
AAPL,10-03-2017, ,349.69,349.77,344.9,346.67,18138800
AAPL,11-03-2017, ,345.4,352.32,345,351.99,16824200

'''