此问题参考的是《面向数据科学家的O'Relly实用统计学第二版》第3章,卡方检验。
这本书提供了一个卡方测试用例的示例,其中假设一个网站的标题由三个不同的标题组成,并由1000个访问者管理。结果显示每个标题的点击次数。
观察到的数据如下:
Headline A B C
Click 14 8 12
No-click 986 992 988
期望值的计算如下:
Headline A B C
Click 11.13 11.13 11.13
No-click 988.67 988.67 988.67
表格现在在哪里:
Headline A B C
Click 0.792 -0.990 0.198
No-click -0.085 0.106 -0.021
到目前为止一切顺利。 现在是重采样部分:
1. Assuming a box of 34 ones and 2966 zeros
2. Shuffle, and take three samples of 1000 and count how many ones(Clicks)
3. Find the squared differences between the shuffled counts and expected counts then sum them.
4. Repeat steps 2 to 3, a few thousand times.
5. The P-value is how often does the resampled sum of squared deviations exceed the observed.
本书提供了重采样python测试代码,如下所示: (可以从https://github.com/gedeck/practical-statistics-for-data-scientists/tree/master/python/code下载)
## Practical Statistics for Data Scientists (Python)
## Chapter 3. Statistial Experiments and Significance Testing
# > (c) 2019 Peter C. Bruce, Andrew Bruce, Peter Gedeck
# Import required Python packages.
from pathlib import Path
import random
import pandas as pd
import numpy as np
from scipy import stats
import statsmodels.api as sm
import statsmodels.formula.api as smf
from statsmodels.stats import power
import matplotlib.pylab as plt
DATA = Path('.').resolve().parents[1] / 'data'
# Define paths to data sets. If you don't keep your data in the same directory as the code, adapt the path names.
CLICK_RATE_CSV = DATA / 'click_rates.csv'
...
## Chi-Square Test
### Chi-Square Test: A Resampling Approach
# Table 3-4
click_rate = pd.read_csv(CLICK_RATE_CSV)
clicks = click_rate.pivot(index='Click', columns='Headline', values='Rate')
print(clicks)
# Table 3-5
row_average = clicks.mean(axis=1)
pd.DataFrame({
'Headline A': row_average,
'Headline B': row_average,
'Headline C': row_average,
})
# Resampling approach
box = [1] * 34
box.extend([0] * 2966)
random.shuffle(box)
def chi2(observed, expected):
pearson_residuals = []
for row, expect in zip(observed, expected):
pearson_residuals.append([(observe - expect) ** 2 / expect
for observe in row])
# return sum of squares
return np.sum(pearson_residuals)
expected_clicks = 34 / 3
expected_noclicks = 1000 - expected_clicks
expected = [34 / 3, 1000 - 34 / 3]
chi2observed = chi2(clicks.values, expected)
def perm_fun(box):
sample_clicks = [sum(random.sample(box, 1000)),
sum(random.sample(box, 1000)),
sum(random.sample(box, 1000))]
sample_noclicks = [1000 - n for n in sample_clicks]
return chi2([sample_clicks, sample_noclicks], expected)
perm_chi2 = [perm_fun(box) for _ in range(2000)]
resampled_p_value = sum(perm_chi2 > chi2observed) / len(perm_chi2)
print(f'Observed chi2: {chi2observed:.4f}')
print(f'Resampled p-value: {resampled_p_value:.4f}')
chisq, pvalue, df, expected = stats.chi2_contingency(clicks)
print(f'Observed chi2: {chi2observed:.4f}')
print(f'p-value: {pvalue:.4f}')
现在,我运行perm_fun(box)2,000次,并获得0.4775的重采样P值。 但是,如果我运行perm_fun(box)10,000次和100,000次,则两次都可以获得重采样的P值0.84。在我看来,P值应在0.84左右。 为什么stats.chi2_contigency显示的数字这么小?
我运行2000次得到的结果是:
Observed chi2: 1.6659
Resampled p-value: 0.8300
Observed chi2: 1.6659
p-value: 0.4348
如果我要运行10,000次,结果是:
Observed chi2: 1.6659
Resampled p-value: 0.8386
Observed chi2: 1.6659
p-value: 0.4348
软件版本:
pandas.__version__: 0.25.1
numpy.__version__: 1.16.5
scipy.__version__: 1.3.1
statsmodels.__version__: 0.10.1
sys.version_info: 3.7.4
答案 0 :(得分:0)
我运行了2000、10000和100000个循环的代码,这三遍我都接近.47。但是,我确实在此行遇到了必须修复的错误:
resampled_p_value = sum(perm_chi2 > chi2observed) / len(perm_chi2)
这里perm_chi2
是一个列表,而chi2observed
是一个浮点数,所以我想知道这段代码是如何为您运行的(也许您为修复它所做的一切都是错误的根源)。无论如何,将其更改为预期的
resampled_p_value = sum([1*(x > chi2observed) for x in perm_chi2]) / len(perm_chi2)
允许我运行它并接近.47。
请确保在更改迭代次数时,仅更改2000,而不更改其他数字。