r - 计算R中每列的负值、0值和正值的数量和百分比

Question

我有一个玩具数据集，如下:

df <- structure(list(id = 1:11, price = c(40.59, 70.42, 1.8, 1.98, 
65.02, 2.23, 54.79, 54.7, 3.32, 1.77, 3.5), month_pct = structure(c(11L, 
10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 1L, 2L), .Label = c("-19.91%", 
"-8.55%", "1.22%", "1.39%", "1.41%", "1.83%", "2.02%", "2.59%", 
"2.86%", "6.58%", "8.53%"), class = "factor"), year_pct = structure(c(4L, 
9L, 5L, 3L, 10L, 1L, 11L, 8L, 6L, 7L, 2L), .Label = c("-10.44%", 
"-19.91%", "-2.46%", "-35.26%", "-4.26%", "-5.95%", "-6.35%", 
"-6.91%", "-7.95%", "1.51%", "1.54%"), class = "factor")), class = "data.frame", row.names = c(NA, 
-11L))

输出:

   id  price month_pct year_pct
0    1  40.59     8.53%  -35.26%
1    2  70.42     6.58%   -7.95%
2    3   1.80     2.86%   -4.26%
3    4   1.98     2.59%   -2.46%
4    5  65.02     2.02%    1.51%
5    6   2.23     1.83%  -10.44%
6    7  54.79     1.41%    1.54%
7    8  54.70     1.39%   -6.91%
8    9   3.32     1.22%   -5.95%
9   10   1.77   -19.91%   -6.35%
10  11   3.50    -8.55%  -19.91%

现在我想计算 month_pct 列的正数、0 和负数 的数量和百分比 year_pct，我怎样才能在 R 中做到这一点？

谢谢。

Answer 1

其中一种 dplyr 和 tidyr 可能是:

df %>%
 pivot_longer(-c(1:2)) %>%
 group_by(name, 
          value_sign = factor(sign(as.numeric(sub("%", "", value))),
                              levels = -1:1,
                              labels = c("negative", "zero", "positive")),
          .drop = FALSE) %>%
 count() %>%
 group_by(name) %>%
 mutate(prop = n/sum(n)*100) 

  name      value_sign     n  prop
  <chr>     <fct>      <int> <dbl>
1 month_pct negative       2  18.2
2 month_pct zero           0   0  
3 month_pct positive       9  81.8
4 year_pct  negative       9  81.8
5 year_pct  zero           0   0  
6 year_pct  positive       2  18.2