r-将细胞组合成一个组周期

Question

我的数据帧

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

我想得到"；diff"；列按组在同一单元格中：

我想要什么：

col5>差异[/tr>544.5665>901.5674866.5

组	col1	col2	col3	col4	2col6
1	139.5	379	864	470.5	1-2:239.5，3-4:-31.5，5-6:-194.5
2		759.5	723	255.5	1-2:-36.5，3-4:-227.5，5-6:611

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

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

diff列似乎只是中位数的差异，所以不确定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