根據兩列獲取頻率

Question

我的大型數據框的一個片段看起來像這樣：

MARKERS.IN.HAPLOTYPES BASE           rs. alleles chrom       pos        GID marker   trial
                 1A.12    C S1A_494392059     C/G    1A 494392059 GID7173723      2 ES26-38
                 1A.13    C S1A_497201550     C/T    1A 497201550 GID7173723      0 ES26-38
                 1A.14    T S1A_499864157     C/T    1A 499864157 GID7173723      2 ES26-38
                 1B.10    A S1B_566171302     G/A    1B 566171302 GID7173723      0 ES26-38
                 1B.20    G S1B_642616640     A/G    1B 642616640 GID7173723      2 ES26-38
                 2B.10    A  S2B_24883552     A/G    2B  24883552 GID7173723      2 ES26-38

這是它的dput ：

structure(list(MARKERS.IN.HAPLOTYPES = c("1A.12", "1A.13", "1A.14", 
"1B.10", "1B.20", "2B.10"), BASE = c("C", "C", "T", "A", "G", 
"A"), rs. = c("S1A_494392059", "S1A_497201550", "S1A_499864157", 
"S1B_566171302", "S1B_642616640", "S2B_24883552"), alleles = c("C/G", 
"C/T", "C/T", "G/A", "A/G", "A/G"), chrom = c("1A", "1A", "1A", 
"1B", "1B", "2B"), pos = c(494392059L, 497201550L, 499864157L, 
566171302L, 642616640L, 24883552L), GID = c("GID7173723", "GID7173723", 
"GID7173723", "GID7173723", "GID7173723", "GID7173723"), marker = c("2", 
 "0", "2", "0", "2", "2"), trial = c("ES26-38", "ES26-38", "ES26-38", 
 "ES26-38", "ES26-38", "ES26-38")), row.names = c(NA, 6L), class = 
 "data.frame")

列rs.有22個unique值rs. 在原始數據框中，列trial有六個unique值。 我想為每個唯一的rs.計算列marker的不同值的相對頻率rs. 以及每個獨特的trial 。 例如， rs.列的第一項rs. S1A_494392059將具有試驗ES26-38的列marker頻率，依此類推，依此類推。 請注意，列marker是字符向量，而不是數字。

Answer 1

您可以嘗試以下方法：

library(dplyr)

df %>%
  add_count(rs., trial, name = "Total") %>%
  add_count(rs., trial, marker, name = "MarkerTotal") %>%
  mutate(RelativeFreq = round(MarkerTotal / Total, 2))

add_count的name列是dplyr 0.8 add_count的一項新功能，可讓您確定名稱（默認情況下為n或nn ）。 如果您沒有最新的軟件包，上面的代碼將不起作用。

您的示例中的相對頻率到處都是1，因為它並不是特別復雜。

如果您想獲取匯總的數據框（僅剩下的列將對rs. ， trial和RelativeFreq進行分組），可以采用以下方法：

df %>% 
  add_count(rs., trial, marker, name = "MarkerTotal") %>%
  group_by(rs., trial) %>%
  summarise(RelativeFreq = round(MarkerTotal / n(), 2))