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[英]How to Generate random code with 4 digit string and 4 digit number in Python?
[英]Code to Generate e one Digit at a Time
我正在尝试制作一个恒定的随机数生成器(我的意思是RNG输出一系列不会重复的数字,但是每次从头开始时都保持不变)。 我有一个pi。 我需要一种算法来逐位生成e位数字,最好以Python迭代器或生成器的形式输入到RNG中。 我也欢迎生成其他非理性数字的代码。 提前致谢。
是! 我做到了继续分数!
我从生成2的平方根的数字中找到了这些代码
def z(contfrac, a=1, b=0, c=0, d=1):
for x in contfrac:
while a > 0 and b > 0 and c > 0 and d > 0:
t = a // c
t2 = b // d
if not t == t2:
break
yield t
a = (10 * (a - c*t))
b = (10 * (b - d*t))
# continue with same fraction, don't pull new x
a, b = x*a+b, a
c, d = x*c+d, c
for digit in rdigits(a, c):
yield digit
def rdigits(p, q):
while p > 0:
if p > q:
d = p // q
p = p - q * d
else:
d = (10 * p) // q
p = 10 * p - q * d
yield d
我制作了连续分数生成器:
def e_cf_expansion():
yield 1
k = 0
while True:
yield k
k += 2
yield 1
yield 1
并将它们放在一起:
def e_dec():
return z(e_cf_expansion())
然后:
>>> gen = e_dec()
>>> e = [str(gen.next()) for i in xrange(1000)]
>>> e.insert(1, '.')
>>> print ''.join(e)
2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127736178215424999229576351482208269895193668033182528869398496465105820939239829488793320362509443117301238197068416140397019837679320683282376464804295311802328782509819455815301756717361332069811250996181881593041690351598888519345807273866738589422879228499892086805825749279610484198444363463244968487560233624827041978623209002160990235304369941849146314093431738143640546253152096183690888707016768396424378140592714563549061303107208510383750510115747704171898610687396965521267154688957035035
奖励:为sqrt(n)生成连续分数的代码,其中n是一个正整数,而sqrt(n)是不合理的:
def sqrt_cf_expansion(S):
"""Generalized generator to compute continued
fraction representation of sqrt(S)"""
m = 0
d = 1
a = int(math.sqrt(S))
a0 = a
while True:
yield a
m = d*a-m
d = (S-m**2)//d
a = (a0+m)//d
如果从带有已知n
的random
模块调用random.seed(n)
, 则每次的结果都相同:
>>> import random
>>> random.seed(4) # chosen by fair dice roll
>>> random.randint(0, 9)
2
>>> random.randint(0, 9)
1
>>> random.randint(0, 9)
3
>>> random.randint(0, 9)
1
>>> random.seed(4) # same seed as above
>>> random.randint(0, 9)
2
>>> random.randint(0, 9)
1
>>> random.randint(0, 9)
3
>>> random.randint(0, 9)
1
如果您需要传递状态,请使用Random
类(未充分记录):
>>> r = random.Random(4)
>>> r.randint(0, 9)
2
>>> r.randint(0, 9)
1
很容易从中产生一个生成器,使您可以生成多个不会踩到彼此脚趾的序列:
def random_digits(seed):
r = random.Random(seed)
while True:
yield r.randint(0, 9)
您是否正在寻找这样的东西:
>>> import math
>>> i = 1
>>> while i < 10:
... print('e = {0:.{1}f}'.format(math.e, i))
... i += 1
...
e = 2.7
e = 2.72
e = 2.718
e = 2.7183
e = 2.71828
e = 2.718282
e = 2.7182818
e = 2.71828183
e = 2.718281828
标准库可以为您提供math.e
: 数学常数e = 2.718281 ...,具有可用的精度 。
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