如何从数据创建包含零计数的类列表?

Chr*_*oph 4 r dplyr

给定一些随机数据,例如x = 10.08506, 10.32809, ...,如何以有效的方式创建分类列表?结果(请参阅下面的可重现示例)应如下所示

classes         n
(10,10.1]       3
(10.1,10.2]     1
(10.2,10.3]     0
(10.3,10.4]     2
(10.4,10.5]     3
(10.5,10.6]     0
(10.6,10.7]     0
(10.7,10.8]     1
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这是一个可重现的示例,它显示了迄今为止最简单的方法:我可以摆脱 data.framedffull_join? 也许,我也可以摆脱br, h

library(dplyr)
set.seed(1)
number_of_observations <- 10
nbr <- 10

x <- rnorm(n = number_of_observations, mean = 10.273, sd = 0.3)
br <- seq(from = ceiling(min(nbr*x)-1)/nbr, 
          to = floor(max(nbr*x)+1)/nbr, by = 1/nbr)
h <- hist(x, breaks = br)

df <- tibble(
  classes = h$mids)
df <- df %>% 
  mutate(classes = cut(classes, breaks = br)) %>%
  group_by(classes) %>%
  mutate(n = n()) %>%
  ungroup() %>%
  mutate(freq = n / sum(n)) %>%
  arrange(classes)

df2 <- tibble(
  classes = x)
df2 <- df2 %>% 
  mutate(classes = cut(classes, breaks = br)) %>%
  group_by(classes) %>%
  mutate(n = n()) %>%
  ungroup() %>%
  mutate(freq = n / sum(n)) %>%
  arrange(classes) %>%
  distinct()

df <- df %>% full_join(df2, by = "classes")
df$n.y[is.na(df$n.y)] <- 0

result <- df[, c("classes", "n.y")]
colnames(result) <- c("classes", "n")
result
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All*_*ron 7

您可以使用一个班轮做到这一点seqcuttable,和as.data.frame

setNames(as.data.frame(table(cut(x, seq(10, 10.8, 0.1)))), c("classes", "n"))
#>       classes n
#> 1   (10,10.1] 3
#> 2 (10.1,10.2] 1
#> 3 (10.2,10.3] 0
#> 4 (10.3,10.4] 2
#> 5 (10.4,10.5] 3
#> 6 (10.5,10.6] 0
#> 7 (10.6,10.7] 0
#> 8 (10.7,10.8] 1

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  • @Christoph,你几乎总是可以只使用基本 R,但 tidyverse 使许多常见的数据操作任务变得更容易。我认为这对于枢轴和连接以及交互式使用等功能非常有用。它还通常使代码更具可读性。然而,我经常看到人们陷入困境,试图让 tidyverse 做一些在 R 基础上微不足道的事情。我想了解两者并且不害怕在两者之间切换是关键。 (5认同)

Tho*_*ing 5

的方法cut+table通过@Allan卡梅伦是有效的。这是通过的另一个选项hist

> list2DF(hist(x,breaks = seq(10, 10.8, 0.1), plot = FALSE))
  breaks counts density  mids xname equidist
1   10.0      3       3 10.05     x     TRUE
2   10.1      1       1 10.15     x     TRUE
3   10.2      0       0 10.25     x     TRUE
4   10.3      2       2 10.35     x     TRUE
5   10.4      3       3 10.45     x     TRUE
6   10.5      0       0 10.55     x     TRUE
7   10.6      0       0 10.65     x     TRUE
8   10.7      1       1 10.75     x     TRUE
9   10.8      3       3 10.05     x     TRUE
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