我使用以下代码执行t-test:
def t_stat(na,abar,avar,nb,bbar,bvar):
logger.info("T-test to be performed")
logger.info("Set A count = %f mean = %f variance = %f" % (na,abar,avar))
logger.info("Set B count = %f mean = %f variance = %f" % (nb,bbar,bvar))
adof = na - 1
bdof = nb - 1
logger.info("Degrees of Freedom of a=%f" % adof)
logger.info("Degrees of Freedom of b=%f" % bdof)
tf = (abar - bbar) / np.sqrt(avar/na + bvar/nb)
dof = (avar/na + bvar/nb)**2 / (avar**2/(na**2*adof) + bvar**2/(nb**2*bdof))
logger.info("tf = %f, dof=%f"%(tf,dof))
pf = 2*stdtr(dof, -np.abs(tf))
我的输出如下:
Set A count = 3547465.000000 mean = 0.001123 variance = 0.000369
Set B count = 83759692.000000 mean = 0.001242 variance = 0.000424
Degrees of Freedom of a=3547464.000000
Degrees of Freedom of b=83759691.000000
tf = -11.374250, dof=-2176568.362223
formula: t = -11.3743 p = nan
当我传递与数组相同的数据并使用ttest_ind函数时,我得到t = -11.374250 p = 0.000000。
为什么我的函数将p作为nan? Afaik,我不能把nan视为0.我怎么能理解我的t_stat和ttest_ind之间的确切区别?任何帮助,将不胜感激。
答案 0 :(得分:1)
您传递给公式的自由度是负数。
In [6]:
import numpy as np
from scipy.special import stdtr
dof = -2176568
tf = -11.374250
2*stdtr(dof, -np.abs(tf))
Out[6]:
nan
如果是肯定的:
In [7]:
import numpy as np
from scipy.special import stdtr
dof = 2176568
tf = -11.374250
2*stdtr(dof, -np.abs(tf))
Out[7]:
5.6293517178917971e-30
我想知道你的情况是怎么发生的,我运行你的代码试图推断输入参数:
In [13]:
def t_stat(na,abar,avar,nb,bbar,bvar):
print("T-test to be performed")
print("Set A count = %f mean = %f variance = %f" % (na,abar,avar))
print("Set B count = %f mean = %f variance = %f" % (nb,bbar,bvar))
adof = na - 1
bdof = nb - 1
print("Degrees of Freedom of a=%f" % adof)
print("Degrees of Freedom of b=%f" % bdof)
tf = (abar - bbar) / np.sqrt(avar/na + bvar/nb)
dof = (avar/na + bvar/nb)**2 / (avar**2/(na**2*adof) + bvar**2/(nb**2*bdof))
print("tf = %f, dof=%f"%(tf,dof))
print(stdtr(dof, -np.abs(tf)))
In [14]:
t_stat(3547465, 0.001123, 0.000369, 83759692, 0.001242, 0.000424)
T-test to be performed
Set A count = 3547465.000000 mean = 0.001123 variance = 0.000369
Set B count = 83759692.000000 mean = 0.001242 variance = 0.000424
Degrees of Freedom of a=3547464.000000
Degrees of Freedom of b=83759691.000000
tf = -11.393950, dof=3900753.641275
2.2434573594e-30
希望无论如何都能帮助你找到问题。