比较 R 中的两个列对

Question

我有一个像下面这样的数据框。

df <- structure(list(`chr1:110363793:G:C_A1` = c("1", "2", "2", "2"
), `chr1:110363793:G:C_A2` = c("2", "1", "2", "2"), `chr1:110363823:A:G_A1` = c("2", 
"2", "2", "2"), `chr1:110363823:A:G_A2` = c("2", "2", "2", "2"
), `chr1:110363849:A:G_A1` = c("2", "2", "2", "2"), `chr1:110363849:A:G_A2` = c("2", 
"2", "1", "2")), row.names = c("4_653_99", "4_677_99", "25_9_172", 
"27_135_62706"), class = "data.frame")

我想从前两列开始成对比较列。 对于列中的每一行，如果值相同，我想在新列Status添加 0，如果值不同，我想要 1。结果应如下所示：

             chr1:110363793:G:C_A1 chr1:110363793:G:C_A2 chr1:110363823:A:G_A1 chr1:110363823:A:G_A2 chr1:110363849:A:G_A1 chr1:110363849:A:G_A2 chr1:110363793:G:C_Status chr1:110363823:A:G_Status
4_653_99                         2                     1                     2                     2                     2                     2                         1                         0
4_677_99                         2                     2                     2                     2                     2                     2                         0                         0
25_9_172                         2                     2                     2                     2                     2                     2                         0                         0
27_135_62706                     2                     2                     2                     1                     2                     2                         0                         1
             chr1:110363849:A:G_Status
4_653_99                             0
4_677_99                             0
25_9_172                             0
27_135_62706                         0

我可以使用 for 循环来做到这一点，但是有没有更好的方法来解决这个问题？

我的 For 循环来解决这个问题：

vars <- unique(sapply(strsplit(colnames(df),"_"), `[`, 1))
    for (i in 1:length(vars)){
      df[, paste(vars[i], "Status", sep = "_")] <- as.numeric(df[,paste0(vars[i], "_A1")]!=df[,paste0(vars[i], "_A2")])
    }

Answer 1

在前面，由于重复的Status名称，这会产生一个非首选帧。

odds <- seq(1, ncol(dat), by = 2)
odds
# [1] 1 3 5
do.call(cbind, unname(lapply(split.default(dat, rep(odds, each = 2)), function(z) cbind(z, Status = +(z[[1]] != z[[2]])))))
#              chr1:110363793:G:C_A1 chr1:110363793:G:C_A2 Status chr1:110363823:A:G_A1 chr1:110363823:A:G_A2 Status chr1:110363849:A:G_A1 chr1:110363849:A:G_A2 Status
# 4_653_99                         1                     2      1                     2                     2      0                     2                     2      0
# 4_677_99                         2                     1      1                     2                     2      0                     2                     2      0
# 25_9_172                         2                     2      0                     2                     2      0                     2                     1      1
# 27_135_62706                     2                     2      0                     2                     2      0                     2                     2      0

Answer 2

我将您的数据框保存为名为xx的对象。 下面将添加一个名为status的列

xx[, "status"] <- as.numeric(xx[,1]!=xx[,2])

Answer 3

更新：改进编码！

library(dplyr)
library(tidyr)
df %>% 
  mutate(rn = row_number()) %>%
  pivot_longer(cols = -rn, names_to = c('grp', '.value'),
               names_sep = "_") %>% 
  group_by(grp) %>%
  transmute(rn, STATUS=ifelse(A1==A2, 1,0)) %>% 
  pivot_wider(names_from = grp, values_from = STATUS,        
              names_glue = "{.value}.{grp}") %>%
  select(-rn) %>%
  bind_cols(df, .) %>% 
  relocate(1,2,7,3,4,8,5,6,9)

输出：

             chr1:110363793:G:C_A1 chr1:110363793:G:C_A2 STATUS.chr1:110363793:G:C chr1:110363823:A:G_A1 chr1:110363823:A:G_A2 STATUS.chr1:110363823:A:G chr1:110363849:A:G_A1 chr1:110363849:A:G_A2 STATUS.chr1:110363849:A:G
4_653_99                         1                     2                         0                     2                     2                         1                     2                     2                         1
4_677_99                         2                     1                         0                     2                     2                         1                     2                     2                         1
25_9_172                         2                     2                         1                     2                     2                         1                     2                     1                         0
27_135_62706                     2                     2                         1                     2                     2                         1                     2                     2                         1

第一个答案：这是一个dplyr解决方案：缺点是必须手动重命名新列：

library(dplyr)
library(tidyr)
df %>% 
  mutate(rn = row_number()) %>%
  pivot_longer(cols = -rn, names_to = c('grp', '.value'),
               names_sep = "_") %>% 
  group_by(grp) %>%
  transmute(rn, new=ifelse(A1==A2, 1,0)) %>% 
  pivot_wider(names_from = grp, values_from = new) %>%
  select(-rn) %>%
  bind_cols(df, .)

             chr1:110363793:G:C_A1 chr1:110363793:G:C_A2 chr1:110363823:A:G_A1 chr1:110363823:A:G_A2 chr1:110363849:A:G_A1 chr1:110363849:A:G_A2 chr1:110363793:G:C chr1:110363823:A:G chr1:110363849:A:G
4_653_99                         1                     2                     2                     2                     2                     2                  0                  1                  1
4_677_99                         2                     1                     2                     2                     2                     2                  0                  1                  1
25_9_172                         2                     2                     2                     2                     2                     1                  1                  1                  0
27_135_62706                     2                     2                     2                     2                     2                     2                  1                  1                  1

Answer 4

带有 sapply 的简单基础 R解决方案：

new <- sapply( (1:ncol(df))[1:ncol(df)%%2==1],
                function(x) as.numeric(df[,x]!=df[,(x+1)] ) )
colnames(new) <- paste( unique(sub("_.*","",colnames(df))), "Status", sep="_" )

给出：

data.frame( df, new )
             chr1.110363793.G.C_A1 chr1.110363793.G.C_A2 chr1.110363823.A.G_A1
4_653_99                         1                     2                     2
4_677_99                         2                     1                     2
25_9_172                         2                     2                     2
27_135_62706                     2                     2                     2
             chr1.110363823.A.G_A2 chr1.110363849.A.G_A1 chr1.110363849.A.G_A2
4_653_99                         2                     2                     2
4_677_99                         2                     2                     2
25_9_172                         2                     2                     1
27_135_62706                     2                     2                     2
             chr1.110363793.G.C_Status chr1.110363823.A.G_Status
4_653_99                             1                         0
4_677_99                             1                         0
25_9_172                             0                         0
27_135_62706                         0                         0
             chr1.110363849.A.G_Status
4_653_99                             0
4_677_99                             0
25_9_172                             1
27_135_62706                         0