在一个组循环中将细胞组合成一个答案

【问题标题】：combine cells into one in a group cycle在一个组循环中将细胞组合成一个
【发布时间】：2023-03-10 14:34:02
【问题描述】：

我的数据框架

 data <- structure(list(col1 = c(125L, 654L, 896L, 154L, 865L, 148L), 
        col2 = c(489L, 657L, 198L, 269L, 789L, 456L), col3 = c(741L, 
        852L, 963L, 987L, 951L, 632L), col4 = c(124L, 785L, 874L, 
        965L, 563L, 145L), col5 = c(963L, 146L, 259L, 367L, 365L, 
        189L), col6 = c(741L, 777L, 100L, 200L, 956L, 452L), col7 = c(456L, 
        666L, 300L, 778L, 888L, 999L), col8 = c(254L, 732L, 400L, 
        500L, 600L, 700L), col9 = c(555L, 638L, 127L, 489L, 545L, 
        54L), col10 = c(921L, 549L, 111L, 222L, 354L, 355L), GROUP = c(1L, 
        2L, 3L, 1L, 2L, 3L)), class = "data.frame", row.names = c(NA, 
    -6L))

功能：

combination <- list(c(1,2),c(3,4),c(5,6))
wilcox.fun <- function(data, id_groups, combination){
  mark.list <- list()
  result_list <- list()
  for (g in id_groups){
    df = as.matrix(data[data$GROUP %in% g,])
    df <- df[,unique(unlist(combination))]
    med <- paste(apply(df, 2, median))
    result <- data.frame(matrix(NA,ncol=length(med)+2, nrow=1))
    result[1,] <- c(g, med, NA)
    for (k in 1:(length(combination))) {
      i <- combination[[k]][1]
      j <- combination[[k]][2]
      test <- wilcox.test(df[,i], df[,j],conf.int = TRUE)
      diff.1 <- -round(test$estimate, 2)
      result[k,length(med)+2] <- paste0(i,"-", j,": ",diff.1)
    }
    result_list[[g]] <- result
  }
  result_new <- do.call(rbind, result_list)
  
  names(result_new) <- c("GROUP", as.character(unique(unlist(colnames(df)))), "dif")
  return(result_new)
}

result <- wilcox.fun(data, c("1", "2"),combination)
result

我想按组将“diff”列的值放在同一个单元格中：

我想得到什么：

Group	col1	col2	col3	col4	col5	col6	diff
1	139.5	379	864	544.5	665	470.5	1-2: 239.5, 3-4: -319.5, 5-6: -194.5
2	759.5	723	901.5	674	255.5	866.5	1-2: -36.5, 3-4: -227.5, 5-6: 611

【问题讨论】：

标签： r dataframe

【解决方案1】：

计算每个k 的dif 后，使用paste() 将所有内容组合在一起，然后删除不必要的行。

wilcox.fun <- function(data, id_groups, combination){
  mark.list <- list()
  result_list <- list()
  for (g in id_groups){
    df = as.matrix(data[data$GROUP %in% g,])
    df <- df[,unique(unlist(combination))]
    med <- paste(apply(df, 2, median))
    result <- data.frame(matrix(NA,ncol=length(med)+2, nrow=1))
    result[1,] <- c(g, med, NA)
    for (k in 1:(length(combination))) {
      i <- combination[[k]][1]
      j <- combination[[k]][2]
      test <- wilcox.test(df[,i], df[,j],conf.int = TRUE)
      diff.1 <- -round(test$estimate, 2)
      result[k,length(med)+2] <- paste0(i,"-", j,": ",diff.1)
    }
    # Merge the dif vals together
    result[1, length(med) + 2] = paste(
      result[, length(med) + 2], 
      collapse = ", "
    )
    # Only keep the first row
    result_list[[g]] <- result[1, ]
  }
  result_new <- do.call(rbind, result_list)
  
  names(result_new) <- c("GROUP", as.character(unique(unlist(colnames(df)))), "dif")
  return(result_new)
}

输出

> wilcox.fun(data, c("1", "2"),combination)
  GROUP  col1 col2  col3  col4  col5  col6                                  dif
1     1 139.5  379   864 544.5   665 470.5 1-2: 239.5, 3-4: -319.5, 5-6: -194.5
2     2 759.5  723 901.5   674 255.5 866.5    1-2: -36.5, 3-4: -227.5, 5-6: 611

【讨论】：

【解决方案2】：

diff 列似乎只是中位数的差异，所以不知道为什么要定义复杂的wilcox.fun。您可以通过以下方式更快地获得diff：

library(dplyr)
data %>% 
    group_by(GROUP) %>% 
    summarise(diff = paste0(
        "1-2: ", median(col2) - median(col1),
        ", 3-4: ", median(col4) - median(col3),
        ", 5-6: ", median(col6) - median(col5)))
## A tibble: 3 × 2
#  GROUP diff
#  <int> <chr>
#1     1 1-2: 239.5, 3-4: -319.5, 5-6: -194.5
#2     2 1-2: -36.5, 3-4: -227.5, 5-6: 611
#3     3 1-2: -195, 3-4: -288, 5-6: 52

【讨论】：