我对编程和统计有点新,所以如果正式不正确,请帮助我改进这个问题。
我有很多参数和一些我在MonteCarlo模拟中生成的结果向量。现在我想测试每个参数对结果的影响。我已经有了一个与Kendall的Tau合作的剧本。现在我想与Spearman和Pearson rho进行比较。一个例子:
from scipy.stats import spearmanr, kendalltau, pearsonr
result = [106, 86, 100, 101, 99, 103, 97, 113, 112, 110]
parameter = ['A', 'B', 'A', 'B', 'A', 'B', 'A', 'B', 'A', 'B']
kendalltau(parameter, result)
>> (0.14907119849998596, 0.54850624613917143)
但是,如果我为spearmanr
或pearsonr
尝试相同操作,我会收到错误消息。显然,此功能未在Scipy中实现。您是否知道获取分类数据的相关系数的简单方法?
答案 0 :(得分:4)
实际上spearmanr有效,但是pearsonr不需要计算数组的平均值,dtype
对于字符串不正确。见下文:
from scipy.stats import spearmanr, kendalltau, pearsonr
result = [106, 86, 100, 101, 99, 103, 97, 113, 112, 110]
parameter = ['A', 'B', 'A', 'B', 'A', 'B', 'A', 'B', 'A', 'B']
spearmanr(result, parameter)
(0.17407765595569782,0.63053607555697644)
help(pearsonr)
Help on function pearsonr in module scipy.stats.stats:
pearsonr(x, y)
Calculates a Pearson correlation coefficient and the p-value for testing
non-correlation.
The Pearson correlation coefficient measures the linear relationship
between two datasets. Strictly speaking, Pearson's correlation requires
that each dataset be normally distributed. Like other correlation
coefficients, this one varies between -1 and +1 with 0 implying no
correlation. Correlations of -1 or +1 imply an exact linear
relationship. Positive correlations imply that as x increases, so does
y. Negative correlations imply that as x increases, y decreases.
The p-value roughly indicates the probability of an uncorrelated system
producing datasets that have a Pearson correlation at least as extreme
as the one computed from these datasets. The p-values are not entirely
reliable but are probably reasonable for datasets larger than 500 or so.
Parameters
----------
x : 1D array
y : 1D array the same length as x
Returns
-------
(Pearson's correlation coefficient,
2-tailed p-value)
References
----------
http://www.statsoft.com/textbook/glosp.html#Pearson%20Correlation
转换' A' 1,' B'到2,例如
params = [1 if el == 'A' else 2 for el in parameter]
print params
[1, 2, 1, 2, 1, 2, 1, 2, 1, 2]
pearsonr(params, result)
(-0.012995783552244984, 0.97157652425566488)
希望这会有所帮助。