使用R计算案例控制队列中多个变体的实例数

Question

我有一张遗传变异表，每一行都代表一个病人，病人身上有这种变异，以及该变异是在一个病例中还是在一个对照中。为了执行Fisher's测试，我想输出一个包含三列的单独的矩阵--变体、事例中的数字和控件中的数字。

我使用的是R，表看起来像这样(PID -病人ID)

Variant ID      PID     Disease
2:4324:2343     FF354   Yes
2:4324:2343     FF355   Control
2:4324:2343     FF356   Control
2:4324:2343     FF357   Yes
2:4324:2343     FF358   Yes
3:346543:345    FF354   Yes
3:346543:345    FF358   Control
3:346543:345    FF390   Control
3:346543:345    FF391   Yes
6:234:34234     FF358   Yes
6:234:34234     FF390   Control
6:234:34234     FF358   Control
6:234:34234     FF213   Yes 

预期的产出将是：

Variant ID  Disease Control
2:4324:2343     3   2
3:346543:345    2   2
6:234:34234     2   2

我想我将不得不在R中使用循环，但我必须承认，这在目前我是超越我，而我可以抓住R。任何帮助将是非常感谢的！

非常感谢

Answer 1

你可以使用tapply，它给你一个很好的矩阵。

with(dat, tapply(Disease, list(Variant_ID, Disease), length))
#              Control Yes
# 2:4324:2343        2   3
# 3:346543:345       2   2
# 6:234:34234        2   2

数据：

dat <- structure(list(Variant_ID = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("2:4324:2343", "3:346543:345", 
"6:234:34234"), class = "factor"), PID = structure(c(2L, 3L, 
4L, 5L, 6L, 2L, 6L, 7L, 8L, 6L, 7L, 6L, 1L), .Label = c("FF213", 
"FF354", "FF355", "FF356", "FF357", "FF358", "FF390", "FF391"
), class = "factor"), Disease = structure(c(2L, 1L, 1L, 2L, 2L, 
2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L), .Label = c("Control", "Yes"), class = "factor")), class = "data.frame", row.names = c(NA, 
-13L))

Answer 2

我们可以得到频率count，然后把它重塑为“wide”

library(dplyr)
library(tidyr)
df1 %>% 
    count(VariantID, Disease) %>%
    pivot_wider(names_from = Disease, values_from = n)
# A tibble: 3 x 3
#  VariantID    Control   Yes
#  <chr>          <int> <int>
#1 2:4324:2343        2     3
#2 3:346543:345       2     2
#3 6:234:34234        2     2

或者是来自table的base R

table(df1[c('VariantID', 'Disease')])
#            Disease
#VariantID      Control Yes
# 2:4324:2343        2   3
# 3:346543:345       2   2
# 6:234:34234        2   2

数据

df1 <- structure(list(VariantID = c("2:4324:2343", "2:4324:2343", "2:4324:2343", 
"2:4324:2343", "2:4324:2343", "3:346543:345", "3:346543:345", 
"3:346543:345", "3:346543:345", "6:234:34234", "6:234:34234", 
"6:234:34234", "6:234:34234"), PID = c("FF354", "FF355", "FF356", 
"FF357", "FF358", "FF354", "FF358", "FF390", "FF391", "FF358", 
"FF390", "FF358", "FF213"), Disease = c("Yes", "Control", "Control", 
"Yes", "Yes", "Yes", "Control", "Control", "Yes", "Yes", "Control", 
"Control", "Yes")), class = "data.frame", row.names = c(NA, -13L
))

Answer 3

使用来自dcast的data.table

library(data.table)
setDT(df); dcast(df, VariantID ~ Disease)

#     VariantID  Control  Yes
#1  2:4324:2343       2   3
#2  3:346543:345      2   2
#3  6:234:34234       2   2

数据

df <- structure(list(VariantID = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("2:4324:2343", "3:346543:345", "6:234:34234"), class = "factor"), PID = structure(c(2L, 3L,4L, 5L, 6L, 2L, 6L, 7L, 8L, 6L, 7L, 6L, 1L), .Label = c("FF213","FF354", "FF355", "FF356", "FF357", "FF358", "FF390", "FF391"), class = "factor"), Disease = structure(c(2L, 1L, 1L, 2L, 2L,2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L), .Label = c("Control", "Yes"), class = "factor")), class = "data.frame", row.names = c(NA, -13L))