如何关联R中的多个子集

Question

如何将 8 个子集分别与两个不同的因变量相关联？对于两个不同的子集，我一直得到相同的相关系数（下面的示例）。这是输入：

with(subset(mydata2, PARTYID_Strength = 1), cor.test(PARTYID_Strength,
                                                     mean.legit))

with(subset(mydata2, PARTYID_Strength = 1), cor.test(PARTYID_Strength,
                                                     mean.leegauthor))

with(subset(mydata2, PARTYID_Strength = 2), cor.test(PARTYID_Strength,
                                                     mean.legit))

with(subset(mydata2, PARTYID_Strength = 2), cor.test(PARTYID_Strength,
                                                     mean.leegauthor))

输出（我得到了 PARTY_Strength = 1 和 2）：

皮尔逊积矩相关性

数据：PARTYID_Strength 和 mean.legit t = 3.1005，df = 607，p 值 = 0.002022 备择假设：真实相关性不等于 0 95% 置信区间：
0.0458644 0.2023031 样本估计：
相关
0.1248597

皮尔逊积矩相关性

数据：PARTYID_Strength 和 mean.leegauthor t = 2.8474, df = 607, p 值 = 0.004557 备择假设：真正的相关性不是 等于 0 95% 置信区间：
0.03568431 0.19250344 样本估计：
相关
0.1148091

样本数据：

> dput(head(mydata2, 10))
``structure(list(PARTYID = c(1, 3, 1, 1, 1, 4, 3, 1, 1, 1), PARTYID_Other = 
c("NA", 
"NA", "NA", "NA", "NA", "Green", "NA", "NA", "NA", "NA"), PARTYID_Strength = 
c(1, 
7, 1, 2, 1, 8, 1, 6, 1, 1), PARTYID_Strength_Other = c("NA", 
"NA", "NA", "NA", "NA", "Green", "NA", "NA", "NA", "NA"), THERM_Dem = c(80, 
65, 85, 30, 76, 15, 55, 62, 90, 95), THERM_Rep = c(1, 45, 10, 
5, 14, 14, 0, 4, 10, 3), Gender = c("Female", "Male", "Male", 
"Female", "Female", "Male", "Male", "Female", "Female", "Male"
), `MEAN Age` = c(29.5, 49.5, 29.5, 39.5, 29.5, 21, 39.5, 39.5, 
29.5, 65), Age = c("25 - 34", "45 - 54", "25 - 34", "35 - 44", 
"25 - 34", "18 - 24", "35 - 44", "35 - 44", "25 - 34", "65+"), 
Ethnicity = c("White or Caucasian", "Asian or Asian American", 
"White or Caucasian", "White or Caucasian", "Hispanic or Latino", 
"White or Caucasian", "White or Caucasian", "White or Caucasian", 
"White or Caucasian", "White or Caucasian"), Ethnicity_Other = c("NA", 
"NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA"), States = c("Texas", 
"Texas", "Ohio", "Texas", "Puerto Rico", "New Hampshire", 
"South Carolina", "Texas", "Texas", "Texas"), Education = c("Master's 
degree", 
"Bachelor's degree in college (4-year)", "Bachelor's degree in college (4- 
 year)", 
"Master's degree", "Master's degree", "Less than high school degree", 
"Some college but no degree", "Master's degree", "Master's degree", 
"Some college but no degree"), `MEAN Income` = c(30000, 140000, 
150000, 60000, 80000, 30000, 30000, 120000, 150000, 60000
), Income = c("Less than $30,000", "$130,001 to $150,000", 
"More than $150,000", "$50,001 to $70,000", "$70,001 to $90,000", 
"Less than $30,000", "Less than $30,000", "$110,001 to $130,000", 
"More than $150,000", "$50,001 to $70,000"), mean.partystrength = c(3.875, 
2.875, 2.375, 3.5, 2.625, 3.125, 3.375, 3.125, 3.25, 3.625
), mean.traitrep = c(2.5, 2.625, 2.25, 2.625, 2.75, 1.875, 
2.75, 2.875, 2.75, 3), mean.traitdem = c(2.25, 2.625, 2.375, 
2.75, 2.625, 2.125, 1.875, 3, 2, 2.5), mean.leegauthor = c(1, 
2, 2, 4, 1, 4, 1, 1, 1, 1), mean.legit = c(1.71428571428571, 
3.28571428571429, 2.42857142857143, 2.42857142857143, 2.14285714285714, 
1.28571428571429, 1.42857142857143, 1.14285714285714, 2.14285714285714, 
1.28571428571429)), row.names = c(NA, -10L), class = c("tbl_df", 
"tbl", "data.frame"))``

谢谢！

Answer 1

要运行测试，请为感兴趣的列创建一个向量，然后为每个列创建一个匿名函数 sapply。

fixed <- "PARTYID_Strength"
cols <- c("mean.leegauthor", "mean.legit")

cor_test_result <- sapply(cols, function(x){
  fmla <- paste(fixed, x, sep = "+")
  fmla <- as.formula(paste("~", fmla))
  cor.test(fmla, mydata2)
}, simplify = FALSE)

cor_test_result$mean.leegauthor
#
#        Pearson's product-moment correlation
#
#data:  PARTYID_Strength and mean.leegauthor
#t = 1.4804, df = 8, p-value = 0.177
#alternative hypothesis: true correlation is not equal to 0
#95 percent confidence interval:
# -0.2343269  0.8462610
#sample estimates:
#      cor 
#0.4637152