我希望能够用Python生成一些有关模型的统计数据。我想要生成t检验,但不确定是否可以使用numpy/scipy轻松实现。是否有相关的好的解释文档?
例如,我有三个相关的数据集,如下所示:
现在,我想对它们进行学生t检验。
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。
[55.0, 55.0, 47.0, 47.0, 55.0, 55.0, 55.0, 63.0]
在 scipy.stats 包中有一些 ttest_...
函数。请参见来自此处的示例:
>>> print 't-statistic = %6.3f pvalue = %6.4f' % stats.ttest_1samp(x, m)
t-statistic = 0.391 pvalue = 0.6955
Van 的回答使用了 scipy,scipy.stats.ttest_*
函数非常方便。这很正确。
但我来到这个页面是想寻找纯 numpy 的解决方案,就像标题所述,避免使用 scipy。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
主要问题在于 numpy 没有累积分布函数,因此我的结论是你真的应该使用 scipy。无论如何,仅使用 numpy 是可能的:
从原始问题中,我猜测您想比较数据集,并通过 t-test 判断是否存在显着偏差?此外,样本成对吗? (见https://en.wikipedia.org/wiki/Student%27s_t-test#Unpaired_and_paired_two-sample_t-tests )。 在这种情况下,可以这样计算 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))
有73.028%的概率,成对样本来自具有相同平均值的分布。
由于这远高于任何合理的置信区间(例如5%),因此您不应该针对具体情况得出任何结论。当你获得了你的t值,你可能会想知道如何将其解释为概率——我也是这样。这里是我编写的一段函数来帮助你。
它基于我从http://www.vassarstats.net/rsig.html和http://en.wikipedia.org/wiki/Student%27s_t_distribution中收集到的信息。
# 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)