random.sample python中的“样本大于总体”
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"sample larger than population" in random.sample python
提问by Giladiald
creating a simple pass generator for myself, i noticed that if i want my population to be digits only (0-9) which is overall 10 options, if i want my length over 10, it wont use any of the digits more then once and return the "sample larger then population" error.
为自己创建一个简单的通行证生成器，我注意到如果我希望我的人口只有数字（0-9），这是总共 10 个选项，如果我希望我的长度超过 10，它不会使用任何数字超过一次和返回“样本大于总体”错误。
is it possible to maintain the code, but add/reduce code lines so it works? or do i HAVE To use a use random choice?
是否可以维护代码，但添加/减少代码行使其有效？还是我必须使用使用随机选择？
import string
import random
z=int(raw_input("for: \n numbers only choose 1, \n letters only choose 2, \n letters and numbers choose 3, \n for everything choose 4:"))
if z==1:
x=string.digits
elif z==2:
x=string.letters
elif z==3:
x=string.letters+string.digits
elif z==4:
x=string.letters+string.digits+string.punctuation
else:
print "die in a fire"
y=int(raw_input("How many passwords would you like?:"))
v=int(raw_input("How long would you like the password to be?:"))
for i in range(y):
string=""
for n in random.sample(x,v):
string+=n
print string
ty
泰
采纳答案by Martijn Pieters
The purposeof random.sample()is to pick a subsetof the input sequence, randomly, without picking any one element more than once. If your input sequence has no repetitions, neither will your output.
该目的的random.sample()是选择一个子集的输入序列的，随机，无需拾取任何一种元素多于一次。如果您的输入序列没有重复，您的输出也不会。
You are notlooking for a subset; you want single random choices from the input sequence, repeated a number of times. Elements can be used more than once. Use random.choice()in a loop for this:
你不是在寻找一个子集；你想从输入序列中随机选择一个，重复多次。元素可以多次使用。random.choice()为此在循环中使用：
for i in range(y):
string = ''.join([random.choice(x) for _ in range(v)])
print string
This creates a string of length v, where characters from xcan be used more than once.
这将创建一个长度为的字符串v，其中的字符x可以多次使用。
Quick demo:
快速演示：
>>> import string
>>> import random
>>> x = string.letters + string.digits + string.punctuation
>>> v = 20
>>> ''.join([random.choice(x) for _ in range(v)])
'Ms>V\0Mf|W@R,#/.P~Rv'
>>> ''.join([random.choice(x) for _ in range(v)])
'TsPnvN&qlm#mBj-!~}3W'
>>> ''.join([random.choice(x) for _ in range(v)])
'{:dfE;VhR:=_~O*,QG<f'
回答by lucas0x7B
@Martijn Pieters is right. But since they state at https://docs.python.org/3.4/library/random.html:
@Martijn Pieters 是对的。但由于他们在https://docs.python.org/3.4/library/random.html 声明：
Warning: The pseudo-random generators of this module should not be used for security purposes. Use os.urandom() or SystemRandom if you require a cryptographically secure pseudo-random number generator.
警告：该模块的伪随机生成器不应用于安全目的。如果您需要加密安全的伪随机数生成器，请使用 os.urandom() 或 SystemRandom。
and the purpose of this is for generating passwords, I suggest this approach:
这样做的目的是为了生成密码，我建议使用这种方法：
import string
import random
set = string.letters + string.digits + string.punctuation
length = 20
password = ''.join( [ random.SystemRandom().choice( set) for _ in range( length) ] )
print( password)
Could anybody please confirm that this is more secure?
有人可以确认这更安全吗？
回答by Александр Морковин
Since the python_3.6 you can use random.choises(x, k=v)for your purpose. It returns a k sized list of elements chosen from the population with replacement. If the population is empty, raises IndexError.
从 python_3.6 开始，您可以random.choises(x, k=v)用于您的目的。它返回从带有替换的总体中选择的 ak 大小的元素列表。如果人口为空，则引发 IndexError。
