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我正在寻找关于我在 python 中创建的模型的一些统计数据。我想对其进行 t 检验，但想知道是否有一种简单的方法可以使用 numpy/scipy 来执行此操作。周围有什么好的解释吗？

例如，我有三个相关的数据集，如下所示：

[55.0, 55.0, 47.0, 47.0, 55.0, 55.0, 55.0, 63.0]

现在，我想对它们进行学生 t 检验。

4

3 回答 3

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在scipy.stats包中，函数很少ttest_...。请参阅此处的示例：

>>> print 't-statistic = %6.3f pvalue = %6.4f' %  stats.ttest_1samp(x, m)
t-statistic =  0.391 pvalue = 0.6955
于 2010-02-24T08:10:17.423 回答
8

van 使用 scipy 的回答是完全正确的，使用scipy.stats.ttest_*函数非常方便。

但是我来到这个页面寻找一个纯numpy的解决方案，如标题中所述，以避免对 scipy 的依赖。为此，让我指出这里给出的示例：https ://docs.scipy.org/doc/numpy/reference/generated/numpy.random.standard_t.html

主要问题是，numpy 没有累积分布函数，因此我的结论是你应该真正使用 scipy。无论如何，只使用 numpy 是可能的：

从最初的问题我猜你想比较你的数据集并用 t 检验判断是否存在显着偏差？此外，样本是配对的吗？（请参阅https://en.wikipedia.org/wiki/Student%27s_t-test#Unpaired_and_paired_two-sample_t-tests）在这种情况下，您可以像这样计算 t 值和 p 值：

import numpy as np
sample1 = np.array([55.0, 55.0, 47.0, 47.0, 55.0, 55.0, 55.0, 63.0])
sample2 = np.array([54.0, 56.0, 48.0, 46.0, 56.0, 56.0, 55.0, 62.0])
# paired sample -> the difference has mean 0
difference = sample1 - sample2
# the t-value is easily computed with numpy
t = (np.mean(difference))/(difference.std(ddof=1)/np.sqrt(len(difference)))
# unfortunately, numpy does not have a build in CDF
# here is a ridiculous work-around integrating by sampling
s = np.random.standard_t(len(difference), size=100000)
p = np.sum(s<t) / float(len(s))
# using a two-sided test
print("There is a {} % probability that the paired samples stem from distributions with the same means.".format(2 * min(p, 1 - p) * 100))

这将打印There is a 73.028 % probability that the paired samples stem from distributions with the same means.由于这远高于任何合理的置信区间（例如 5%），因此您不应针对具体案例得出任何结论。

于 2017-06-26T15:27:32.423 回答
-4

一旦你得到你的 t 值，你可能想知道如何将它解释为概率——我做到了。这是我写的一个函数来帮助解决这个问题。

它基于我从http://www.vassarstats.net/rsig.html和http://en.wikipedia.org/wiki/Student%27s_t_distribution收集的信息。

# Given (possibly random) variables, X and Y, and a correlation direction,
# returns:
#  (r, p),
# where r is the Pearson correlation coefficient, and p is the probability
# of getting the observed values if there is actually no correlation in the given
# direction.
#
# direction:
#  if positive, p is the probability of getting the observed result when there is no
#     positive correlation in the normally distributed full populations sampled by X
#     and Y
#  if negative, p is the probability of getting the observed result, when there is no
#     negative correlation
#  if 0, p is the probability of getting your result, if your hypothesis is true that
#    there is no correlation in either direction
def probabilityOfResult(X, Y, direction=0):
    x = len(X)
    if x != len(Y):
        raise ValueError("variables not same len: " + str(x) + ", and " + \
                         str(len(Y)))
    if x < 6:
        raise ValueError("must have at least 6 samples, but have " + str(x))
    (corr, prb_2_tail) = stats.pearsonr(X, Y)

    if not direction:
        return (corr, prb_2_tail)

    prb_1_tail = prb_2_tail / 2
    if corr * direction > 0:
        return (corr, prb_1_tail)

    return (corr, 1 - prb_1_tail)
于 2013-01-09T02:15:08.413 回答