Question

我正在尝试计算每个SNP名称的iets列中“Opp”的出现量（最终我想将“Opp”的出现量除以df $ MM。）

library(data.table)
df <- structure(list(SNP = structure(c(1L, 1L, 1L, 2L, 1L), .Label = c("rs80932150", "rs000001"), class = "factor"), FID = c(116601888L, 116621563L, 117253533L, 118635095L, 118943247L), IID = c(116601888L, 116621563L, 117253533L, 118635095L, 118943247L), NEW = structure(c(16L, 14L, 16L, 14L, 14L), .Label = c("A/A", "A/C", "A/G", "A/T", "C/A", "C/C", "C/G", "C/T", "G/A", "G/C", "G/G", "G/T", "T/A", "T/C", "T/G", "T/T"), class = "factor"), OLD = structure(c(6L, 6L, 6L, 6L, 6L), .Label = c("A/A", "A/C", "A/G", "A/T", "C/A", "C/C", "C/G", "C/T", "G/A", "G/C", "G/G", "G/T", "T/A", "T/C", "T/G", "T/T"), class = "factor"), count = c(1L, 1L, 1L, 1L, 1L), MM = c(4L, 4L, 4L, 1L, 4L), iets = c("Opp", "Het", "Opp", "Het", "Het")), .Names = c("SNP", "FID", "IID", "NEW", "OLD", "count", "MM", "iets"), class = "data.frame", row.names = c(NA, -5L))
setDT(df)

#         SNP       FID       IID NEW OLD count MM iets
#1 rs80932150 116601888 116601888 T/T C/C     1  4  Opp
#2 rs80932150 116621563 116621563 T/C C/C     1  4  Het
#3 rs80932150 117253533 117253533 T/T C/C     1  4  Opp
#4   rs000001 118635095 118635095 T/C C/C     1  1  Het
#5 rs80932150 118943247 118943247 T/C C/C     1  4  Het

我的预期结果如下：

df
#          SNP       FID       IID NEW OLD count MM iets oppcount percentage
#1: rs80932150 116601888 116601888 T/T C/C     1  4  Opp      2        0.5
#2: rs80932150 116621563 116621563 T/C C/C     1  4  Het      2        0.5
#3: rs80932150 117253533 117253533 T/T C/C     1  4  Opp      2        0.5
#4:   rs000001 118635095 118635095 T/C C/C     1  1  Het      0        0.0
#5: rs80932150 118943247 118943247 T/C C/C     1  4  Het      2        0.5

我一直在尝试与此类似的事情，但我似乎无法弄清楚如何将出现值分配给我的oppcount / percentage列。
首先，我必须计算每个SNP的“Opp”数量，然后将其除以MM。

as.character((sum(df$iets == "Opp")/(df[,.N, by = df$SNP][[2]])))
#[1] "0.5" "2"

如何计算每个SNP（类别）出现“Opp”的数量？

Answer 1

您可以使用data.table运算符通过引用更新:=。用：

df[, `:=` (oppcount = sum(iets=='Opp'), percentage = sum(iets=='Opp')/.N), by = SNP]

你得到：

> df
          SNP       FID       IID NEW OLD count MM iets oppcount percentage
1: rs80932150 116601888 116601888 T/T C/C     1  4  Opp        2        0.5
2: rs80932150 116621563 116621563 T/C C/C     1  4  Het        2        0.5
3: rs80932150 117253533 117253533 T/T C/C     1  4  Opp        2        0.5
4:   rs000001 118635095 118635095 T/C C/C     1  1  Het        0        0.0
5: rs80932150 118943247 118943247 T/C C/C     1  4  Het        2        0.5

或者，根据评论中@Frank的建议，您还可以使用以下两个选项之一：

# method 1
df[, c('oppcount', 'percentage') := {s = sum(iets=='Opp'); .(s, s/.N)}, by = SNP]
# method 2
df[df[, {s = sum(iets=='Opp'); .(oppcount = s, percentage = s/.N)}, by = SNP], on = 'SNP']

基础R替代方案：

transform(df,
          oppcount = ave(iets, SNP, FUN = function(x) sum(x=='Opp')),
          percentage = ave(iets, SNP, FUN = function(x) sum(x=='Opp')/length(x)))

正确的dplyr替代方案是：

library(dplyr)
df %>% 
  group_by(SNP) %>% 
  mutate(oppcount = sum(iets=='Opp'),
         percentage = oppcount/n())

Answer 2

rs8.oppcount<-length(iets[iets=='Opp' & SNP=='rs80932150'])
rs0.oppcount<-length(iets[iets=='Opp' & SNP=='rs000001'])

这可以保存snp类别的Opp出现次数！

编辑：

df1<-group_by(df, df$SNP)
df2<-summarise(df1, oppcount = length(iets[iets=='Opp']))
df1<-merge(df1, df2, by = 'SNP')

这有用吗？

Answer 3

如何使用dplyr？

library('dplyr')
df %>% group_by(iets, SNP) %>% summarize(count=sum(count)) %>% filter(iets=='Opp')

计算每个类别列的出现次数

3 个答案: