如何产生串数在不同的样本通过的R

0

的问题

让我们说有一个数据表如下:

ID1 ID2  ID3
  a   a    b
  a   b    b
  b   b    b
  c   c    c
  c   c    d
  c   d    d
  d        e
  d        e
  e

然后我想要将其转换为如下结构:

Samples ID1 ID2 ID3
a        2  1   0
b        1  2   3
c        3  2   1
d        2  1   2
e        1  0   2

任何你,请帮助我R或庆典的代码实现这种转变?

bash r
2021-11-23 11:01:33
2

最好的答案

3

尝试R代码下面

> table(stack(df))
      ind
values ID1 ID2 ID3
     a   2   1   0
     b   1   2   3
     c   3   2   1
     d   2   1   2
     e   1   0   2

数据

> dput(df)
structure(list(ID1 = c("a", "a", "b", "c", "c", "c", "d", "d", 
"e"), ID2 = c("a", "b", "b", "c", "c", "d", NA, NA, NA), ID3 = c("b",
"b", "b", "c", "d", "d", "e", "e", NA)), class = "data.frame", row.names = c(NA,
-9L))
2021-11-23 11:04:04
1

一个选择 tidyverse -改为"长"的格式 pivot_longer,获得 count 重塑回'广泛的'格式 pivot_wider

library(dplyr)
library(tidyr)
df %>%
   pivot_longer(everything(), values_drop_na = TRUE, values_to = 'Samples') %>%
   count(name, Samples) %>% 
   pivot_wider(names_from = name, values_from = n, values_fill = 0)

-输出

# A tibble: 5 × 4
  Samples   ID1   ID2   ID3
  <chr>   <int> <int> <int>
1 a           2     1     0
2 b           1     2     3
3 c           3     2     1
4 d           2     1     2
5 e           1     0     2

数据

df <- structure(list(ID1 = c("a", "a", "b", "c", "c", "c", "d", "d", 
"e"), ID2 = c("a", "b", "b", "c", "c", "d", NA, NA, NA), ID3 = c("b",
"b", "b", "c", "d", "d", "e", "e", NA)), class = "data.frame", 
row.names = c(NA,
-9L))
2021-11-23 16:54:02

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