我们可以这样做。这相当冗长，但对于开始来说可能会有帮助：
我们在这里所做的原则上是用Q列中的每一列和其他列创建新的 Dataframe ，对于每一列，我们做同样的事情，并在最后绑定它们。
broom包中的tidy函数非常方便：

library(dplyr)
library(tidyr)
library(broom)

Q1 <- df %>% 
  select(-Q2, -Q3) %>% 
  pivot_longer(-Q1) %>% 
  group_by(name) %>% 
  nest(-name) %>% 
  mutate(stats = map(data, ~broom::tidy(chisq.test(.$Q1, .$value)))) %>% 
  select(-data) %>% 
  unnest(c(stats))

Q2 <- df %>% 
  select(-Q1, -Q3) %>% 
  pivot_longer(-Q2) %>% 
  group_by(name) %>% 
  nest(-name) %>% 
  mutate(stats = map(data, ~broom::tidy(chisq.test(.$Q2, .$value)))) %>% 
  select(-data) %>% 
  unnest(c(stats))

Q3 <- df %>% 
  select(-Q1, -Q2) %>% 
  pivot_longer(-Q3) %>% 
  group_by(name) %>% 
  nest(-name) %>% 
  mutate(stats = map(data, ~broom::tidy(chisq.test(.$Q3, .$value)))) %>% 
  select(-data) %>% 
  unnest(c(stats))

bind_rows(Q1, Q2, Q3, .id = "Q") %>% 
mutate(ID = paste0("Q",Q), .before=1, .keep="unused")

ID    name          statistic p.value parameter method                    
  <chr> <chr>             <dbl>   <dbl>     <int> <chr>                     
1 Q1    ageRange          15.6    0.209        12 Pearson's Chi-squared test
2 Q1    education         27.5    0.122        20 Pearson's Chi-squared test
3 Q1    maritalStatus      2.71   0.608         4 Pearson's Chi-squared test
4 Q2    ageRange          15.6    0.209        12 Pearson's Chi-squared test
5 Q2    education         20.8    0.407        20 Pearson's Chi-squared test
6 Q2    maritalStatus      2.71   0.608         4 Pearson's Chi-squared test
7 Q3    ageRange          14.6    0.265        12 Pearson's Chi-squared test
8 Q3    education         21.7    0.359        20 Pearson's Chi-squared test
9 Q3    maritalStatus      3.06   0.549         4 Pearson's Chi-squared test

我们也可以使用循环

library(purrr)
library(broom)
library(tidyr)
library(stringr)
library(dplyr)
str_subset(names(df), "^Q\\d+$") %>%
   map(~ df %>% 
    select(all_of(.x), ageRange:maritalStatus) %>%
    pivot_longer(cols = -1) %>% 
   group_by(ID = .x, name) %>% 
   summarise(stats = tidy(chisq.test(cur_data()[[1]], value)),
       .groups = "drop")) %>% 
   list_rbind %>%
   unnest(where(is_tibble))

• 输出

# A tibble: 9 × 6
  ID    name          statistic p.value parameter method                    
  <chr> <chr>             <dbl>   <dbl>     <int> <chr>                     
1 Q1    ageRange          15.6    0.209        12 Pearson's Chi-squared test
2 Q1    education         27.5    0.122        20 Pearson's Chi-squared test
3 Q1    maritalStatus      2.71   0.608         4 Pearson's Chi-squared test
4 Q2    ageRange          15.6    0.209        12 Pearson's Chi-squared test
5 Q2    education         20.8    0.407        20 Pearson's Chi-squared test
6 Q2    maritalStatus      2.71   0.608         4 Pearson's Chi-squared test
7 Q3    ageRange          14.6    0.265        12 Pearson's Chi-squared test
8 Q3    education         21.7    0.359        20 Pearson's Chi-squared test
9 Q3    maritalStatus      3.06   0.549         4 Pearson's Chi-squared test