Python 和 R 中卡方检验的不同 P 值

Question

我正在尝试对两类生物数据进行卡方检验。我有一个这样的数据框：

         Brain, Cerebelum, Heart, Kidney,  liver,  testis
expected 3        66       1        44       34       88
observed 6        57       4        45       35       69

structure(list(Brain = c(3L, 6L), Cerebelum = c(66L, 57L), heart = c(1L, 
4L), kidney = 44:45, liver = 34:35, testis = c(88L, 69L)), .Names = c("Brain", 
"Cerebelum", "heart", "kidney", "liver", "testis"), class = "data.frame", row.names = c("rand", 
"cns"))

我用Python做了测试：

from scipy.stats import chisquare
chisquare(obs,f_exp=exp)

结果为：

Power_divergenceResult(statistic=17.381684491978611, pvalue=0.0038300192430189722)

我尝试使用 R 复制结果，所以我制作了 csv 文件，作为数据框导入到 R，运行代码为：

d<-read.csv(file)
chisq.test(d)

Pearson's Chi-squared test

data:  d
X-squared = 4.9083, df = 5, p-value = 0.4272

为什么 python 和 R 的卡方值和 P 值不同？我用简单的 (O-E)^2/E 公式手工计算，卡方值等于 17.38由 python 计算，但我无法弄清楚 R 如何计算 4.90 的值。

Answer 1

我可以回答你的第一个问题。

chisq.test，当你给它一个具有 > 2 行和列的矩阵时，将其视为二维偶然性 table 并测试沿行和列的观察之间的独立性. Here's an example and another one.

另一方面，

scipy.stats.chisq 只是做 definition of the test stat 中熟悉的 X = sum( (O_i-E_i)^2 / E_i)。

那么如何对圆进行平方呢？首先，传递 R 观测值，然后在参数 p 中定义预期概率。其次，您还需要阻止 R 进行默认的连续性校正。

e <- d[1, ]
o <- d[2, ]
chisq.test(o, p = e / sum(e), correct = FALSE)

瞧

Chi-squared test for given probabilities

data:  o
X-squared = 17.139, df = 5, p-value = 0.004243

PS SO 的棘手问题，交叉验证可能更好？请注意，与 scipy 相比，R 的默认更正可能是一件好事。这是否属实肯定是为了交叉验证。

PPS ?chisq.test 中的帮助有点难以解析，但我认为这些都在某处；)

 If ‘x’ is a matrix with one row or column, or if ‘x’ is a vector
 and ‘y’ is not given, then a _goodness-of-fit test_ is performed
 (‘x’ is treated as a one-dimensional contingency table).  The
 entries of ‘x’ must be non-negative integers.  In this case, the
 hypothesis tested is whether the population probabilities equal
 those in ‘p’, or are all equal if ‘p’ is not given.

 If ‘x’ is a matrix with at least two rows and columns, it is taken
 as a two-dimensional contingency table: the entries of ‘x’ must be
 non-negative integers.  Otherwise, ‘x’ and ‘y’ must be vectors or
 factors of the same length; cases with missing values are removed,
 the objects are coerced to factors, and the contingency table is
 computed from these.  Then Pearson's chi-squared test is performed
 of the null hypothesis that the joint distribution of the cell
 counts in a 2-dimensional contingency table is the product of the
 row and column marginals.

和

 correct: a logical indicating whether to apply continuity correction
          when computing the test statistic for 2 by 2 tables: one half
          is subtracted from all |O - E| differences; however, the
          correction will not be bigger than the differences
          themselves.  No correction is done if ‘simulate.p.value =
          TRUE’.

Python 和 R 中卡方检验的不同 P 值

Different P values for chi square test in Python and R

r

chi-squared

python-3.x