Question

使用dplyr获取多列独立频率计数的好方法是什么？我想从价值表中找到：

# A tibble: 7 x 4
      a     b     c     d
  <int> <int> <int> <int>
1     1     2     1     3
2     1     2     1     3
3     2     2     5     3
4     3     2     4     3
5     3     3     2     3
6     5     3     4     3
7     5     4     2     1

到这样的频率表：

# A tibble: 5 x 5
      x   a_n   b_n   c_n   d_n
  <int> <int> <int> <int> <int>
1     1     2     0     2     1
2     2     1     4     2     0
3     3     2     2     0     6
4     4     0     1     2     0
5     5     2     0     1     0

我仍然试图绕过dplyr，但似乎这是可以做到的。如果使用附加库更容易，那也没关系。

Answer 1

对于您在问题中提供的相同数据集，这将是另一种解决方案（base-R）：

myfreq <- sapply(df, function(x) table(factor(x, levels=unique(unlist(df)), ordered=TRUE)))

输出将是：

> myfreq

#   a b c d 
# 1 2 0 2 1 
# 2 1 4 2 0 
# 3 2 2 0 6 
# 5 2 0 1 0 
# 4 0 1 2 0

Answer 2

library(tidyverse)

dt <- data.frame(a = c(1L, 1L, 2L, 3L, 3L, 5L, 5L), b = c(2L, 2L, 2L, 2L, 3L, 3L, 4L),
                 c = c(1L, 1L, 5L, 4L, 2L, 4L, 2L), d = c(3L, 3L, 3L, 3L, 3L, 3L, 1L))

dt2 <- dt %>%
  mutate(ID = 1:n()) %>%
  gather(Group, x, -ID) %>%
  select(-ID) %>%
  mutate(Group = paste(Group, "n", sep = "_")) %>%
  count(Group, x) %>%
  spread(Group, n, fill = 0L)

Answer 3

library(dplyr)
library(reshape2)
df %>%
  melt() %>%
  dcast(value ~ variable, fun.aggregate=length)

#   value a b c d
# 1     1 2 0 2 1
# 2     2 1 4 2 0
# 3     3 2 2 0 6
# 4     4 0 1 2 0
# 5     5 2 0 1 0

数据

df <- structure(list(a = c(1L, 1L, 2L, 3L, 3L, 5L, 5L), b = c(2L, 2L, 
2L, 2L, 3L, 3L, 4L), c = c(1L, 1L, 5L, 4L, 2L, 4L, 2L), d = c(3L, 
3L, 3L, 3L, 3L, 3L, 1L)), .Names = c("a", "b", "c", "d"), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7"))

Answer 4

在基础R中使用tabulate：

apply(df,2,function(x) tabulate(x)[min(df):max(df)])

#     a  b c  d
#[1,] 2  0 2  1
#[2,] 1  4 2  0
#[3,] 2  2 0  6
#[4,] 0  1 2 NA
#[5,] 2 NA 1 NA

表

4 个答案:

数据