python 3.1 - 创建正态分布
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python 3.1 - Creating normal distribution
提问by jimy
I have scipy and numpy, Python v3.1
我有 scipy 和 numpy，Python v3.1
I need to create a 1D array of length 3million, using random numbers between (and including) 100-60,000. It has to fit a normal distribution.
我需要创建一个长度为 300 万的一维数组，使用介于（包括）100-60,000 之间的随机数。它必须符合正态分布。
Using 'a = numpy.random.standard_normal(3000000)', I get a normal distribution for that required length; not sure how to achieve the required range.
使用'a = numpy.random.standard_normal(3000000)'，我得到了所需长度的正态分布；不确定如何达到所需的范围。
采纳答案by Apalala
A standard normal distribution has mean 0 and standard deviation 1. What I understand from your requirements is that you need a ((60000-100)/2, (60000-100)/2) one. Take each value from standard_normal()result, multiply it by the new variance, and add the new mean.
标准正态分布的平均值为 0，标准差为 1。我从您的要求中了解到，您需要一个 ((60000-100)/2, (60000-100)/2) 一个。从standard_normal()结果中取出每个值，乘以新的方差，然后加上新的均值。
I haven't used NumPy, but a quick search of the docs says that you can achieve what you want directly bu using numpy.random.normal()
我没有使用过 NumPy，但是快速搜索文档说你可以直接实现你想要的 bu 使用 numpy.random.normal()
One last tidbit: normal distributions are not bounded. That means there isn't a value with probability zero. Your requirements should be in terms of means and variances (or standard deviations), and not of limits.
最后一点：正态分布是无界的。这意味着不存在概率为零的值。您的要求应该是均值和方差（或标准差），而不是限制。
回答by Eric Pauley
try this nice little method:
试试这个不错的小方法：
You'll want a method that just makes one random number.
您将需要一种仅生成一个随机数的方法。
import random
list = [random.randint(min,max) for i in range(numitems)]
This will give you a list with numitems random numbers between min and max.
这将为您提供一个列表，其中包含最小和最大之间的 numitems 随机数。
Of course, 3000000 is a lot of items to have in memory. Consider making the random numbers as they are needed by the program.
当然，3000000 在内存中是很多项目。考虑制作程序需要的随机数。
回答by fmark
If you want a truly random normal distribution, you can't guarentee how far the numbers will spread. You can reduce the probability of outliers, however, by specifying the standard deviation
如果你想要一个真正随机的正态分布，你不能保证数字会传播多远。但是，您可以通过指定标准偏差来降低异常值的概率
>>> n = 3000000
>>> sigma5 = 1.0 / 1744278
>>> n * sigma5
1.7199093263803131 # Expect one values in 3 mil outside range at 5 stdev.
>>> sigma6 = 1.0 / 1 / 506800000
>>> sigma6 = 1.0 / 506800000
>>> n * sigma6
0.0059194948697711127 # Expect 0.005 values in 3 mil outside range at 6 stdev.
>>> sigma7 = 1.0 / 390600000000
>>> n * sigma7
7.6804915514592934e-06
Therefore, in this case, ensuring that the standard deviation is only 1/6 or 1/7 of half the range will give you reasonable confidence that your data will not exceed the range.
因此，在这种情况下，确保标准偏差仅为范围一半的 1/6 或 1/7 将使您有理由相信您的数据不会超出范围。
>>> range = 60000 - 100
>>> spread = (range / 2) / 6 # Anything outside of the range will be six std. dev. from the mean
>>> mean = (60000 + 100) / 2
>>> a = numpy.random.normal(loc = mean, scale = spread, size = n)
>>> min(a)
6320.0238199673404
>>> max(a)
55044.015566089176
Of course, you can still can values that fall outside the range here
当然，您仍然可以在这里设置超出范围的值
