Question

假设我有这个df：

> df <- data.frame(letter = sample(letters[1:4], 15, replace=TRUE),
+                  time = c("one", "one", "one", "two", "two", "one", "two", "two", "two", "one","one","one","two","one","two"),
+                  stringsAsFactors = FALSE)
> df
   letter time
1       d  one
2       a  one
3       a  one
4       b  two
5       c  two
6       a  one
7       d  two
8       a  two
9       b  two
10      b  one
11      d  one
12      b  one
13      c  two
14      a  one
15      a  two

我要按Value对它们进行分组，并创建一个time_one列和一个名为time_two的列，其各自的计数由Value加上它们各自的计数百分比。这是我的出发点：

> x <- df %>%
+ mutate(Value = letter,
+       n = n()) %>%
+ group_by(Value) %>%
+ summarise(Quantity = length(Value),
+          Percentage = first(length(Value)/n))
> x
  Value Quantity  Percentage
1 a            6           0.4  
2 b            4           0.267
3 c            2           0.133
4 d            3           0.2

如上所述，我有每个Value的计数，但我需要每个Quantity中的每个Value用one和{{ 1}}时间列中的值。所以，我将以这样的结尾：

two

PS ：我已经检查了two by two表和creating a table for frequency的答案，它们与我正在寻找的答案很接近，但是我仍然不太了解以及Value time_one Percentage time_two Percentage 1 a 5 0.5 1 0.2 2 b 2 0.2 2 0.4 3 c 1 0.1 1 0.2 4 d 2 0.2 1 0.2，%>%，group_by，mutate的组合，使用它们来分离这些解决方案的数量和百分比是一种非常陡峭的学习曲线得到我需要的那个。

Answer 1

恐怕我不会使用现代的整洁R，但是如果您的需求可以接受的话，这是基于R的解决方案。

df <- data.frame(letter = sample(letters[1:4], 15, replace=TRUE),
                 time = c("one", "one", "one", "two", "two", "one", "two", "two", "two", "one","one","one","two","one","two"),
                 stringsAsFactors = FALSE)
# make sure your letter is a factor with all levels otherwise the subsequent cbind doesn#t work
df$letter = factor(df$letter, levels=letters[1:4])

# get the counts 
x = sapply(split(df$letter, df$time), table)

# get the percentages and cbind together 
x2 = cbind(x, apply(x, 2, function(x) x/sum(x)))

colnames(x2) = c("time_one", "time_two", "percent_one", "percent_two")


  time_one time_two percent_one percent_two
a        0        1         0.0   0.1428571
b        4        4         0.5   0.5714286
c        0        1         0.0   0.1428571
d        4        1         0.5   0.1428571

Answer 2

使用tidyverse，您可以尝试：

df %>%
 group_by(time) %>%
 mutate(n_time = n()) %>%
 group_by(time, letter) %>%
 summarise(n = n(),
           percentage = first(n()/n_time)) %>%
 ungroup() %>%
 gather(var, val, -c(time, letter)) %>%
 mutate(var = paste(var, time, sep = "_")) %>%
 select(-time) %>%
 spread(var, val) 

  letter n_one n_two percentage_one percentage_two
  <chr>  <dbl> <dbl>          <dbl>          <dbl>
1 a         3.    2.          0.375          0.286
2 b         2.   NA           0.250         NA    
3 c         2.    2.          0.250          0.286
4 d         1.    3.          0.125          0.429

在这里，首先，计算每个“时间”的计数。其次，它计算每个“时间”和“字母”的计数以及每个给定时间的总计数中给定字母的比例。第三，它将数据从宽格式转换为长格式，并创建指示时间的新变量名称。最后，它将数据恢复为所需的格式。

Answer 3

使用data.table：

library(data.table)
library(magrittr)
setDT(df)

df[, .N, by = .(letter, time)
   ][, .(N, percentage = N/sum(N), letter), by = time] %>% 
  dcast(letter ~ time, value.var = c("N", "percentage"), fill = 0)

   letter N_one N_two percentage_one percentage_two
1:      a     4     2           0.50      0.2857143
2:      b     2     2           0.25      0.2857143
3:      c     0     2           0.00      0.2857143
4:      d     2     1           0.25      0.1428571

数据

df <- structure(list(letter = c("d", "a", "a", "b", "c", "a", "d", 
"a", "b", "b", "d", "b", "c", "a", "a"), time = c("one", "one", 
"one", "two", "two", "one", "two", "two", "two", "one", "one", 
"one", "two", "one", "two")), row.names = c(NA, -15L), class = "data.frame")

带计数和百分比的多组统计方法

3 个答案: