def simulate2():
n = 10000
win1 = 0
win2 = 0
win3 = 0
win4 = 0
win5 = 0
for i in range(n): #simulate 10000 games
#simulate game whose total number is 10, and player1 and player3 won 1
result = Bookie2(10, 1, 0, 1, 0, 0)
if result == 1:
win1 += 1
elif result == 2:
win2 += 1
elif result == 3:
win3 += 1
elif result == 4:
win4 += 1
elif result == 5:
win5 += 1
log('player1 wins: ' + str(float(win1) / float(n)))
log('player2 wins: ' + str(float(win2) / float(n)))
log('player3 wins: ' + str(float(win3) / float(n)))
log('player4 wins: ' + str(float(win4) / float(n)))
log('player5 wins: ' + str(float(win5) / float(n)))
def solve(self): sum1=43+21+35+0.0 p1=1.0/3.0 p3=21.0/sum1 dis = p1-p3 dis = round(dis,2) result = [98,dis,0] log(result) return result
def solve(self): n=51 n=math.sqrt(n) s=4.9 z1=1.1/(s/n) z1=round(z1,2) z2=1.96 log([z1,z2]) return [0.95,z1,1]
def first_year():
X=sta.norm(loc=950, scale=20)#generate random data in normal distribution whose expectation is 950 and standard deviation is 20
wbread=[]
for i in range(365):
x=X.rvs(size=100)
wbread.append(x[0])#get the random data for one day
log(numpy.mean(wbread))#print mean value
log(sta.skew(wbread))#print skew value
plt.hist(wbread,color='grey')
plt.savefig('first_year.png')
def second_year():
X=sta.norm(loc=950, scale=20)#generate random data in normal distribution whose expectation is 950 and standard deviation is 20
wbread=[]
for i in range(365):
x=X.rvs(size=100)
wbread.append(max(x))#get the random data for one day
log(numpy.mean(wbread))#print mean value
log(sta.skew(wbread))#print skew value
plt.hist(wbread,color='grey')
plt.savefig('second_year.png')
def solve(self): u=5.0 x=4.6 s=2.2 n=20 n=math.sqrt(n) z=(x-u)/(s/n) z=round(z,2) result=[19.0,z,1] log(result) return result
def simulate1():
n = 10000
win1 = 0
win2 = 0
for i in range(n): #simulate 10000 games
#simulate game whose total number is 10, and player1 won 5, player2 won 2
result = Bookie1(10, 5, 2)
if result == 1:
win1 += 1
elif result == 2:
win2 += 1
log('player1 wins: ' + str(float(win1) / float(n)))
log('player2 wins: ' + str(float(win2) / float(n)))
def linregress1():
x = np.linspace(-5, 5, num=150)
y = x + np.random.normal(size=x.size)
y[12:14] += 10
slope, intercept, r_value, p_value, std_err = stats.linregress(x,y)
log(slope)
log(intercept)
log(r_value)
log(p_value)
log(std_err)
plt.plot(x, y, 'b.')
plt.plot(x, slope * x + intercept, 'r-')
plt.savefig('linregress1.png')
def nc_of_expon():
#1st non-center moment of expon distribution whose lambda is 0.5
E1 = lambda x: x*0.5*np.exp(-x/2)
#2nd non-center moment of expon distribution whose lambda is 0.5
E2 = lambda x: x**2*0.5*np.exp(-x/2)
log(integrate.quad(E1, 0, np.inf))
log(integrate.quad(E2, 0, np.inf))
log(expon(scale=2).moment(1))
log(expon(scale=2).moment(2))
def ttest():
x = stats.norm.rvs(loc=5, scale=10, size=50)
log(stats.ttest_1samp(x,5.0))
log(stats.ttest_1samp(x,1.0))
class Solution():
def solve(self, A):
result = {}
for str in A:
result[str] = result.get(str, 0) + 1
return result
'''
已知列表fruits中顺序保存了某商店每日出售的水果品名，例如fruits=['apple','banana','cherry','pineapple','banana','peach','pear','peach','cherry' ]，完成函数solve()计算每一种水果的出售次数，存入字典result中并将结果返回
Input / Output
Input Example
['apple', 'banana', 'cherry', 'pineapple', 'banana', 'peach', 'pear','peach', 'cherry' ]
Input Description
一个纪录了水果出售纪录的列表
Output Example
{'pear':1, 'banana':2, 'cherry':2, 'peach':2, 'pineapple':1, 'apple':1}
Output Description
水果出售次数统计结果
'''
log(Solution().solve(['apple', 'banana', 'cherry', 'pineapple', 'banana', 'peach', 'pear', 'peach', 'cherry']))
#-*- coding:utf-8 -*-
'''
log api example: log('output is: ' + str(output))
'''
import random
from log_api import log
'''
You can delete the MontyCarlo and write your Monty Carlo
'''
#this function may spend a little time to execute
def MontyCarlo():
n = 1000000
k = 0
for i in range(n):
x = random.uniform(-1, 1)
y = random.uniform(-1, 1)
if x**2 + y**2 < 1:
k = k + 1
return 4 * float(k) / float(n)
if __name__ == '__main__':
pi = MontyCarlo()
#print out the result
log(pi)
from log_api import log
class Solution():
def solve(self, A):
return np.poly1d(A) * np.poly1d(np.array([2.0, 0.0, -1.0, 1.0]))
'''
在Numpy中，多项式函数的系数可以用一维数组表示，例如对于f(x)=2x^3-x+1可表示为f=np.array([2.0,0.0,-1.0,1.0])，而np.poly1d()方法可以将多项式转换为poly1d(一元多项式)对象，返回多项式函数的值，请利用poly1d()方法计算多项式g(x)(例如g(x)=x^2+2x+1)和f(x)的乘积并将结果返回
Input / Output
Input Example
np.array([1.0, 2.0, 1.0])
Input Description
多项式g(x)的一维数组表示
Output Example
np.poly1d([2.0,4.0,1.0,-1.0,1.0,1.0])
Output Description
f(x)*g(x)的poly1d对象
'''
log(Solution().solve(np.array([1.0, 2.0, 1.0])))
# judge whether x is prime or not
def prime(self, x):
for i in range(2, int(x ** 0.5) + 1):
if x % i == 0:
return False
return True
'''
完成函数solve，判断传入的整数列表A中的数字是否是素数，并将所有的素数保存到另一个列表中并返回
Input / Output
Input Example
[23,45,76,67,17]
Input Description
一个含素数和非素数的多个整数列表
Output Example
[23,67,17]
Output Description
素数组成的列表
'''
log(Solution().solve([23, 45, 76, 67, 17]))
import random
from log_api import log
'''
You can delete the MontyCarlo and write your Monty Carlo
'''
#this function may spend a little time to execute
def MontyCarlo():
n = 1000000
k = 0
for i in range(n):
x = random.uniform(-1, 1)
y = random.uniform(-1, 1)
if x**2 + y**2 < 1:
k = k + 1
return 4 * float(k) / float(n)
if __name__ == '__main__':
pi = MontyCarlo()
#print out the result
log(pi)
#not hit at the first and change the mind
else:
return 1
if __name__ == '__main__':
#repeat 10000 times
n = 10000
#not sure whether to change the mind
win = 0
for i in range(n):
Dselect = random.randint(1, 3)
Dchange = random.randint(0, 1)
win = win + MontyHall(Dselect, Dchange)
log(float(win) / float(n))
#be sure not to change the mind
win = 0
for i in range(n):
Dselect = random.randint(1, 3)
Dchange = 0
win = win + MontyHall(Dselect, Dchange)
log(float(win) / float(n))
#be sure to change the mind
win = 0
for i in range(n):
Dselect = random.randint(1, 3)
Dchange = 1
win = win + MontyHall(Dselect, Dchange)
def nc_of_norm():
f1 = lambda x: x**4
f2 = lambda x: x**2-x+2
log(norm.expect(f1, loc=1, scale=2))
log(norm.expect(f2, loc=2, scale=5))
for str in A:
if self.isPalindrome(str):
result.append(str)
return result
def isPalindrome(self, x):
return x == x[::-1]
'''
对于一个包含一系列数字字符串的列表，寻找其中的回文串存入一个列表中并返回
Input / Output
Input Example
['123', '232', '4556554', '12123', '3443','1314131']
Input Description
一个包含一系列数字字符串的列表
Output Example
['232', '4556554', '3443','1314131']
Output Description
仅包含回文串的列表
'''
log(Solution().solve(['123', '232', '4556554', '12123', '3443', '1314131']))
def oneway_anova():
A1=[27.0, 26.2, 28.8, 33.5, 28.8]
A2=[22.8, 23.1, 27.7, 27.6, 24.0]
A3=[21.9, 23.4, 20.1, 27.8, 19.3]
A4=[23.5, 19.6, 23.7, 20.8, 23.9]
A=[A1, A2, A3, A4]
n=20
As=np.sum(A, axis=1)
QA=0.0
for i in range(4):
QA=QA+As[i]*As[i]
QA=QA/5
QT=0.0
for i in range(4):
for j in range(5):
QT=QT+A[i][j]*A[i][j]
C=np.sum(A)*np.sum(A)/n
ST=QT-C
SA=QA-C
Se=ST-SA
F=(SA/3)/(Se/16)
log('QA is ' + str(QA))
log('QT is ' + str(QT))
log('C is ' + str(C))
log('ST is ' + str(ST))
log('SA is ' + str(SA))
log('Se is ' + str(Se))
log('F is '+ str(F))
def chisquare():
A=[16, 18, 16, 14, 12, 12]
B=[16, 16, 16, 16, 16, 8]
log(stats.chisquare(A))
log(stats.chisquare(A, f_exp=B))
# -*- coding:utf-8 -*-
'''
log api example: log('output is: ' + str(output))
'''
import numpy as np
from log_api import log
class Solution():
def solve(self):
# 1 - P(X = 0) >= 0.98
# 1 - e^(-lambda) >= 0.98
# lambda >= -ln(0.02)
l = -np.log(0.02)
return np.ceil(l)
'''
假定一块蛋糕上的葡萄干粒数服从泊松分布，如果想让每块蛋糕上至少有一粒葡萄干的概率大于等于0.98，蛋糕上葡萄干的平均粒数应该是多少？
Output Example
1
Output Description
num_of_grape, int类型数字
'''
log(Solution().solve())
def nc_of_uniform():
rv = uniform(loc=2, scale=6)
log(rv.mean())
log(rv.var())
log(rv.moment(1))
log(rv.moment(2))
log(rv.moment(3))
log(rv.moment(4))
log(rv.stats(moments='mvsk'))
def nc_of_poisson():
rv = poisson(mu=5)
log(rv.mean())
log(rv.var())
log(rv.moment(1))
log(rv.moment(2))
log(rv.moment(3))
log(rv.moment(4))
log(rv.stats(moments='mvsk'))
def estimate():
number_experiments = 10000
upper_bound = 100
H1 = H2 = H3 = MSE1 = MSE2 = MSE3 = ME1 = ME2 = ME3 = 0.0
for j in range(number_experiments):
N = number_trans(upper_bound)
evidence = train_seen(N)
hypo1 = MLE_estimation(evidence)
hypo2 = MSE_estimation(evidence)
hypo3 = ME_estimation(evidence)
#calculating hits
H1 = H1 + 1 if hypo1==N else H1
H2 = H2 + 1 if hypo2==N else H2
H3 = H3 + 1 if hypo3==N else H3
#calculating mean squared error
MSE1 = MSE1 + (hypo1-N)**2
MSE2 = MSE2 + (hypo2-N)**2
MSE3 = MSE3 + (hypo3-N)**2
#calculating mean error
ME1 = ME1 + (hypo1-N)
ME2 = ME2 + (hypo2-N)
ME3 = ME3 + (hypo3-N)
log(H1/number_experiments)
log(H2/number_experiments)
log(H3/number_experiments)
log(MSE1/number_experiments)
log(MSE2/number_experiments)
log(MSE3/number_experiments)
log(ME1/number_experiments)
log(ME2/number_experiments)
log(ME3/number_experiments)
def estimate():
number_experiments = 10000
upper_bound = 100
H1 = H2 = H3 = MSE1 = MSE2 = MSE3 = ME1 = ME2 = ME3 = 0.0
for j in range(number_experiments):
N = number_trans(upper_bound)
evidence = train_seen(N)
hypo1 = MLE_estimation(evidence)
hypo2 = MSE_estimation(evidence)
hypo3 = ME_estimation(evidence)
#calculating hits
H1 = H1 + 1 if hypo1==N else H1
H2 = H2 + 1 if hypo2==N else H2
H3 = H3 + 1 if hypo3==N else H3
#calculating mean squared error
MSE1 = MSE1 + (hypo1-N)**2
MSE2 = MSE2 + (hypo2-N)**2
MSE3 = MSE3 + (hypo3-N)**2
#calculating mean error
ME1 = ME1 + (hypo1-N)
ME2 = ME2 + (hypo2-N)
ME3 = ME3 + (hypo3-N)
log(H1/number_experiments)
log(H2/number_experiments)
log(H3/number_experiments)
log(MSE1/number_experiments)
log(MSE2/number_experiments)
log(MSE3/number_experiments)
log(ME1/number_experiments)
log(ME2/number_experiments)
log(ME3/number_experiments)
x_delta = [xi - x_mean for xi in x]
y_delta = [yi - y_mean for yi in y]
sum_xy = np.sum([x_delta[i] * y_delta[i] for i in range(0, n)])
sum_x2 = np.sum([xd**2 for xd in x_delta])
sum_y2 = np.sum([yd**2 for yd in y_delta])
r_val = sum_xy / np.sqrt(sum_x2) / np.sqrt(sum_y2)
if abs(r_val) == 1.0:
p_val = 0.0
else:
t2 = r_val**2 * ((n - 2) / ((1.0 - r_val) * (1.0 + r_val)))
p_val = 1 - beta.sf((n - 2) / (n - 2 + t2), 0.5 * (n - 2), 0.5)
return [round(r_val, 6), round(p_val, 6)]
'''
利用python实现简化版皮尔森相关系数计算函数
注意事项：
1）scipy包只能使用scipy.x.ppf或scipy.x.sf函数
2）x,y为空时，返回值为[None,None]
3）结果保留6位小数点
Input Description
x : 一维数组；y : 一维数组，且x、y长度相同
Output Description
[r-val,p-value]分别代表皮尔森相关系数、检验结果P值
'''
log(Solution().solve([1, 2, 3], [2, 2, 3]))
def nc_of_binom():
rv = binom(10,0.2)
log(rv.mean())
log(rv.var())
log(rv.moment(1))
log(rv.moment(2))
log(rv.moment(3))
log(rv.moment(4))
log(rv.stats(moments='mvsk'))
result[i] = 0
for str in A:
for c in str:
i = int(c)
result[i] += 1
return result
'''
已知有一个由数字字符串构成的列表，统计列表中数字字符'0'-'9'各自出现的次数并返回统计结果
Input / Output
Input Example
['12','34','567', '36','809','120']
Input Description
一个由数字字符串构成的列表
Output Example
{0:2, 1:2, 2:2, 3:2, 4:1, 5:1, 6:2, 7:1, 8:1, 9:1}
Output Description
各数字字符的出现次数
'''
log(Solution().solve(['12', '34', '567', '36', '809', '120']))
def ttest():
x = stats.norm.rvs(loc=5, scale=1, size=50)
y = stats.norm.rvs(loc=2, scale=10, size=50)
log(stats.ttest_ind(x, y))
log(stats.ttest_ind(x, y, equal_var = False))
