首先从 查询台开始查找你想要的置信区间。置信区间是 mean +/- z*sigma，其中 sigma是你的样本平均值的估计标准差，由 sigma = s / sqrt(n)给出，其中 s是从你的样本数据计算出来的标准差，而 n是你的样本量。

import numpy as np
import scipy.stats


def mean_confidence_interval(data, confidence=0.95):
a = 1.0 * np.array(data)
n = len(a)
m, se = np.mean(a), scipy.stats.sem(a)
h = se * scipy.stats.t.ppf((1 + confidence) / 2., n-1)
return m, m-h, m+h

你可以这样计算。

下面是沙山代码的简化版，计算了数组 abc 0平均值的95% 的置信区间:

import numpy as np, scipy.stats as st


st.t.interval(0.95, len(a)-1, loc=np.mean(a), scale=st.sem(a))

但使用 StatsModel 的 tconfint_mean可以说更好:

import statsmodels.stats.api as sms


sms.DescrStatsW(a).tconfint_mean()

两者的基本假设都是样本(数组 a)是独立于具有未知标准差的正态分布而绘制的(参见 数学世界或 维基百科)。

对于大样本量 n，样本平均值是正态分布的，我们可以使用 st.norm.interval()来计算它的置信区间(正如詹姆的评论所建议的)。但是上述解决方案对于小 n 也是正确的，其中 st.norm.interval()给出的置信区间太窄(即“假置信”)。有关更多细节，请参阅我的 回答以获得类似的问题(以及 Russ 的一条评论)。

这里有一个例子，正确的选项给出了(基本上)相同的置信区间:

In [9]: a = range(10,14)


In [10]: mean_confidence_interval(a)
Out[10]: (11.5, 9.4457397432391215, 13.554260256760879)


In [11]: st.t.interval(0.95, len(a)-1, loc=np.mean(a), scale=st.sem(a))
Out[11]: (9.4457397432391215, 13.554260256760879)


In [12]: sms.DescrStatsW(a).tconfint_mean()
Out[12]: (9.4457397432391197, 13.55426025676088)

最后，使用 st.norm.interval()得到的错误结果是:

In [13]: st.norm.interval(0.95, loc=np.mean(a), scale=st.sem(a))
Out[13]: (10.23484868811834, 12.76515131188166)

从 Python 3.8开始，标准库提供 NormalDist对象作为 statistics模块的一部分:

from statistics import NormalDist


def confidence_interval(data, confidence=0.95):
dist = NormalDist.from_samples(data)
z = NormalDist().inv_cdf((1 + confidence) / 2.)
h = dist.stdev * z / ((len(data) - 1) ** .5)
return dist.mean - h, dist.mean + h

这个:

• 从数据样本(NormalDist.from_samples(data))创建一个 NormalDist对象，这样我们就可以通过 NormalDist.mean和 NormalDist.stdev获得样本的平均值和标准差。

• 根据标准正态分布(由 NormalDist()表示) ，利用累积分布函数的倒数(inv_cdf)计算给定置信度的 Z-score。

• 根据样本的置信区间标准差和平均值来制作。

这里假设样本量足够大(假设大于100点) ，以便使用标准正态分布而不是学生的 t 分布来计算 z值。

关于 Ulrich 的回答，那就是使用 t 值。当真方差未知时，我们使用它。这时，您拥有的唯一数据就是样本数据。

对于 Bogatron 的回答，这涉及到 Z 表。当方差已知并提供时，将使用 z 表。然后你还有样本数据。西格玛并非样本平均值的估计标准差。大家都知道了。

假设方差已知，我们需要95% 的置信度:

from scipy.stats import norm
alpha = 0.95
# Define our z
ci = alpha + (1-alpha)/2
#Lower Interval, where n is sample siz
c_lb = sample_mean - norm.ppf(ci)*((sigma/(n**0.5)))
c_ub = sample_mean + norm.ppf(ci)*((sigma/(n**0.5)))

只有样本数据和一个未知的方差(这意味着方差将不得不计算只从样本数据) ，Ulrich 的答案工作完美。不过，你可能希望指定置信区间。如果你的数据是 a，你希望得到0.95的置信区间:

import statsmodels.stats.api as sms
conf = sms.DescrStatsW(a).tconfint_mean(alpha=0.05)
conf

基于原文，但附有一些具体例子:

import numpy as np


def mean_confidence_interval(data, confidence: float = 0.95) -> tuple[float, np.ndarray]:
"""
Returns (tuple of) the mean and confidence interval for given data.
Data is a np.arrayable iterable.


ref:
- https://stackoverflow.com/a/15034143/1601580
- https://github.com/WangYueFt/rfs/blob/f8c837ba93c62dd0ac68a2f4019c619aa86b8421/eval/meta_eval.py#L19
"""
import scipy.stats
import numpy as np


a: np.ndarray = 1.0 * np.array(data)
n: int = len(a)
if n == 1:
import logging
logging.warning('The first dimension of your data is 1, perhaps you meant to transpose your data? or remove the'
'singleton dimension?')
m, se = a.mean(), scipy.stats.sem(a)
tp = scipy.stats.t.ppf((1 + confidence) / 2., n - 1)
h = se * tp
return m, h


def ci_test_float():
import numpy as np
# - one WRONG data set of size 1 by N
data = np.random.randn(1, 30)  # gives an error becuase len sets n=1, so not this shape!
m, ci = mean_confidence_interval(data)
print('-- you should get a mean and a list of nan ci (since data is in wrong format, it thinks its 30 data sets of '
'length 1.')
print(m, ci)


# right data as N by 1
data = np.random.randn(30, 1)
m, ci = mean_confidence_interval(data)
print('-- gives a mean and a list of length 1 for a single CI (since it thinks you have a single dat aset)')
print(m, ci)


# multiple data sets (7) of size N (=30)
data = np.random.randn(30, 7)
print('-- gives 7 CIs for the 7 data sets of length 30. 30 is the number ud want large if you were using z(p)'
'due to the CLT.')
m, ci = mean_confidence_interval(data)
print(m, ci)


ci_test_float()

产出:

-- you should get a mean and a list of nan ci (since data is in wrong format, it thinks its 30 data sets of length 1.
0.1431623130952463 [nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan
nan nan nan nan nan nan nan nan nan nan nan nan]
-- gives a mean and a list of length 1 for a single CI (since it thinks you have a single dat aset)
0.04947206018132864 [0.40627264]
-- gives 7 CIs for the 7 data sets of length 30. 30 is the number ud want large if you were using z(p)due to the CLT.
-0.03585104402718902 [0.31867309 0.35619134 0.34860011 0.3812853  0.44334033 0.35841138
0.40739732]

我认为 Num _ data 集的 Num _ sample 是正确的，但是如果它不让我在注释部分知道。

它适用于什么类型的数据？

我认为它可以用于任何数据，原因如下:

我相信这是可行的，因为平均值和标准差是针对一般数值数据计算的，而 z _ p/t _ p 值只考虑置信区间和数据大小，因此它与数据分布的假设无关。

所以我相信它可以用于回归和分类。

作为额外收获，一个几乎只使用炬的炬实现:

def torch_compute_confidence_interval(data: Tensor,
confidence: float = 0.95
) -> Tensor:
"""
Computes the confidence interval for a given survey of a data set.
"""
n: int = len(data)
mean: Tensor = data.mean()
# se: Tensor = scipy.stats.sem(data)  # compute standard error
# se, mean: Tensor = torch.std_mean(data, unbiased=True)  # compute standard error
se: Tensor = data.std(unbiased=True) / (n ** 0.5)
t_p: float = float(scipy.stats.t.ppf((1 + confidence) / 2., n - 1))
ci = t_p * se
return mean, ci

对 CI 的一些评论(见 https://stats.stackexchange.com/questions/554332/confidence-interval-given-the-population-mean-and-standard-deviation?noredirect=1&lq=1) :

"""
Review for confidence intervals. Confidence intervals say that the true mean is inside the estimated confidence interval
(the r.v. the user generates). In particular it says:
Pr[mu^* \in [mu_n +- t.val(p) * std_n / sqrt(n) ] ] >= p
e.g. p = 0.95
This does not say that for a specific CI you compute the true mean is in that interval with prob 0.95. Instead it means
that if you surveyed/sampled 100 data sets D_n = {x_i}^n_{i=1} of size n (where n is ideally >=30) then for 95 of those
you'd expect to have the truee mean inside the CI compute for that current data set. Note you can never check for which
ones mu^* is in the CI since mu^* is unknown. If you knew mu^* you wouldn't need to estimate it. This analysis assumes
that the the estimator/value your estimating is the true mean using the sample mean (estimator). Since it usually uses
the t.val or z.val (second for the standardozed r.v. of a normal) then it means the approximation that mu_n ~ gaussian
must hold. This is most likely true if n >= 0. Note this is similar to statistical learning theory where we use
the MLE/ERM estimator to choose a function with delta, gamma etc reasoning. Note that if you do algebra you can also
say that the sample mean is in that interval but wrt mu^* but that is borning, no one cares since you do not know mu^*
so it's not helpful.


An example use could be for computing the CI of the loss (e.g. 0-1, CE loss, etc). The mu^* you want is the expected
risk. So x_i = loss(f(x_i), y_i) and you are computing the CI for what is the true expected risk for that specific loss
function you choose. So mu_n = emperical mean of the loss and std_n = (unbiased) estimate of the std and then you can
simply plug in the values.


Assumptions for p-CI:
- we are making a statement that mu^* is in mu+-pCI = mu+-t_p * sig_n / sqrt n, sig_n ~ Var[x] is inside the CI
p% of the time.
- we are estimating mu^, a mean
- since the quantity of interest is mu^, then the z_p value (or p-value, depending which one is the unknown), is
computed using the normal distribution.
- p(mu) ~ N(mu; mu_n, sig_n/ sqrt n), vial CTL which holds for sample means. Ideally n >= 30.
- x ~ p^*(x) are iid.


Std_n vs t_p*std_n/ sqrt(n)
- std_n = var(x) is more pessimistic but holds always. Never shrinks as n->infity
- but if n is small then pCI might be too small and your "lying to yourself". So if you have very small data
perhaps doing std_n for the CI is better. That holds with prob 99.9%. Hopefuly std is not too large for your
experiments to be invalidated.


ref:
- https://stats.stackexchange.com/questions/554332/confidence-interval-given-the-population-mean-and-standard-deviation?noredirect=1&lq=1
- https://stackoverflow.com/questions/70356922/what-is-the-proper-way-to-compute-95-confidence-intervals-with-pytorch-for-clas
- https://www.youtube.com/watch?v=MzvRQFYUEFU&list=PLUl4u3cNGP60hI9ATjSFgLZpbNJ7myAg6&index=205
"""