我可以比较一列中值的频率吗？

该函数为您提供计数。如果需要，可以将表转换为 ，并通过将每个计数除以总数来获得比例。以下是一些虚拟数据，其中组被过度代表：tabledata.framee

df <- data.frame(var = ordered(c(rep('a', 2), rep('b', 4),
                                 rep('c', 4), rep('d', 3), rep('e', 10)))) %>% print()
table(df$var)

 a  b  c  d  e 
 2  4  4  3 10

然后我们可以计算出每组的频率：

df_counts <- as.data.frame(table(df$var))
df_counts$prop <- df_counts$Freq/sum(df_counts$Freq)
print(df_counts)

  Var1 Freq       prop
1    a    2 0.08695652
2    b    4 0.17391304
3    c    4 0.17391304
4    d    3 0.13043478
5    e   10 0.43478261

对于统计分析，我们可以使用卡方来确定分布是否可能与随机/零分布相同：

chisq.test(df_counts$Freq)

Chi-squared test for given probabilities

data:  df_counts$Freq
X-squared = 8.5217, df = 4, p-value = 0.0742

差一点！此外，这并不能告诉我们哪个群体的代表性过高。为此，我们可以做一个非常愚蠢的暴力排列测试：在与原始数据一样多的试验中随机抽样组变量，1000 次，并计算出每个组的模拟计数大于观察到的计数的频率。如果随机化给出的给定组的计数大于实际数据中显示的计数，则该组的代表性可能未过高。

# initialize permutation count columns
df_counts$n_greater <- rep(0, nrow(df_counts))
df_counts$n_lesser <- rep(0, nrow(df_counts))
set.seed(123)  # for reproducible "randomness"
# simulate 1000 random apportionments of group memberships to the observed number of trials
n_permut <- 1000
for(i in 1:n_permut) {
  # random "draw" of group variables
  sim <- sample(df_counts$Var1, nrow(df), replace=T)
  sim_df <- as.data.frame(table(sim))
  # for each group, was the number of randomized calls greater or lesser than observed?
  # increment counters accordingly
  df_counts$n_greater <- df_counts$n_greater + as.numeric(sim_df$Freq > df_counts$Freq)
  df_counts$n_lesser <- df_counts$n_lesser + as.numeric(sim_df$Freq < df_counts$Freq)
}
# the permutation test p-values are simply the proportion of simulations with greater or lesser counts
df_counts$p_greater <- df_counts$n_greater/n_permut
df_counts$p_lesser <- df_counts$n_lesser/n_permut
# we will use Bonferroni correction on the p-values, because of the multiple comparisons that we've performed
df_counts$p_greater <- p.adjust(df_counts$p_greater, method='bonferroni', n=nrow(df_counts) * 2)
df_counts$p_lesser <- p.adjust(df_counts$p_lesser, method='bonferroni', n=nrow(df_counts) * 2)
print(df_counts)

  Var1 Freq       prop n_greater n_lesser p_greater p_lesser
1    a    2 0.08695652       867       49      1.00     0.49
2    b    4 0.17391304       521      287      1.00     1.00
3    c    4 0.17391304       514      292      1.00     1.00
4    d    3 0.13043478       672      157      1.00     1.00
5    e   10 0.43478261         1      990      0.01     1.00

因此，通过这种相当基本的方法，组具有高度显着的过度表示 p 值，而其他组无论哪种方式都不显著。e