Question

使用cor(mtcars, method='pearson')生成一个矩阵，显示mtcars中所有变量与mtcars中所有其他变量的皮尔逊相关性。例如：

head(cor(mtcars, method='pearson'))
            mpg        cyl       disp         hp       drat         wt        qsec         vs         am       gear
mpg   1.0000000 -0.8521620 -0.8475514 -0.7761684  0.6811719 -0.8676594  0.41868403  0.6640389  0.5998324  0.4802848
cyl  -0.8521620  1.0000000  0.9020329  0.8324475 -0.6999381  0.7824958 -0.59124207 -0.8108118 -0.5226070 -0.4926866
disp -0.8475514  0.9020329  1.0000000  0.7909486 -0.7102139  0.8879799 -0.43369788 -0.7104159 -0.5912270 -0.5555692
hp   -0.7761684  0.8324475  0.7909486  1.0000000 -0.4487591  0.6587479 -0.70822339 -0.7230967 -0.2432043 -0.1257043
drat  0.6811719 -0.6999381 -0.7102139 -0.4487591  1.0000000 -0.7124406  0.09120476  0.4402785  0.7127111  0.6996101
wt   -0.8676594  0.7824958  0.8879799  0.6587479 -0.7124406  1.0000000 -0.17471588 -0.5549157 -0.6924953 -0.5832870
           carb
mpg  -0.5509251
cyl   0.5269883
disp  0.3949769
hp    0.7498125
drat -0.0907898
wt    0.4276059

如何获得上面相同的矩阵，除了每个值都是皮尔逊在每个变量之间的相关性之外，它是线性模型的r.squared值？例如，第一列，第二行与summary(lm(mtcars$mpg~ mtcars$cyl))$r.squared相同。谢谢

Answer 1

library(tidyverse)

# kepp names of dataset
names = names(mtcars)

expand.grid(names, names, stringsAsFactors = F) %>%  # create pairs of names
  filter(Var1 != Var2) %>%                           # exclude same variables (creates warnings)
  rowwise() %>%                                      # for each row
  mutate(r = summary(lm(paste(Var1, "~" ,Var2), data = mtcars))$r.squared) %>%  # get the r squared
  spread(Var2, r)                                    # reshape

# # A tibble: 11 x 12
# Var1        am     carb    cyl   disp     drat    gear      hp    mpg
# <chr>    <dbl>    <dbl>  <dbl>  <dbl>    <dbl>   <dbl>   <dbl>  <dbl>
# 1 am    NA        0.00331  0.273  0.350  0.508    0.631   0.0591  0.360
# 2 carb   0.00331 NA        0.278  0.156  0.00824  0.0751  0.562   0.304
# 3 cyl    0.273    0.278   NA      0.814  0.490    0.243   0.693   0.726
# 4 disp   0.350    0.156    0.814 NA      0.504    0.309   0.626   0.718
# 5 drat   0.508    0.00824  0.490  0.504 NA        0.489   0.201   0.464
# 6 gear   0.631    0.0751   0.243  0.309  0.489   NA       0.0158  0.231
# 7 hp     0.0591   0.562    0.693  0.626  0.201    0.0158 NA       0.602
# 8 mpg    0.360    0.304    0.726  0.718  0.464    0.231   0.602  NA    
# 9 qsec   0.0528   0.431    0.350  0.188  0.00832  0.0452  0.502   0.175
# 10 vs     0.0283   0.324    0.657  0.505  0.194    0.0424  0.523   0.441
# 11 wt     0.480    0.183    0.612  0.789  0.508    0.340   0.434   0.753
# # ... with 3 more variables: qsec <dbl>, vs <dbl>, wt <dbl>

如果您想要行名而不是第一列（Var1），可以在上面的管道末尾添加

... %>%
  data.frame() %>%
  column_to_rownames("Var1")

这将更接近cor(mtcars, method='pearson')

的输出

Answer 2

我创建了一个corlm函数，用for循环填充条目

corlm <- function(df){
mat <- matrix(NA, ncol(df), ncol(df), dimnames = list(colnames(df),colnames(df)))
suppressWarnings(for(i in 1:ncol(df)){
    for(j in 1:ncol(df)){
      mat[i,j] = summary(lm(df[,j]  ~ df[,i]))$r.squared}})
diag(mat) = NA; return(mat)
}

round(corlm(mtcars),3)
       mpg   cyl  disp    hp  drat    wt  qsec    vs    am  gear  carb
mpg     NA 0.726 0.718 0.602 0.464 0.753 0.175 0.441 0.360 0.231 0.304
cyl  0.726    NA 0.814 0.693 0.490 0.612 0.350 0.657 0.273 0.243 0.278
disp 0.718 0.814    NA 0.626 0.504 0.789 0.188 0.505 0.350 0.309 0.156
hp   0.602 0.693 0.626    NA 0.201 0.434 0.502 0.523 0.059 0.016 0.562
drat 0.464 0.490 0.504 0.201    NA 0.508 0.008 0.194 0.508 0.489 0.008
wt   0.753 0.612 0.789 0.434 0.508    NA 0.031 0.308 0.480 0.340 0.183
qsec 0.175 0.350 0.188 0.502 0.008 0.031    NA 0.554 0.053 0.045 0.431
vs   0.441 0.657 0.505 0.523 0.194 0.308 0.554    NA 0.028 0.042 0.324
am   0.360 0.273 0.350 0.059 0.508 0.480 0.053 0.028    NA 0.631 0.003
gear 0.231 0.243 0.309 0.016 0.489 0.340 0.045 0.042 0.631    NA 0.075
carb 0.304 0.278 0.156 0.562 0.008 0.183 0.431 0.324 0.003 0.075    NA

线性模型回归系数的相关矩阵

2 个答案: