如何使用 R 将排行榜重新塑造成不同的格式？

Question

我有一个看起来像这样的排名表（由于 netmeta 包中的 netleague 函数）（本示例中的值是为了简化它的外观）：

A	0.3 (-0.4 to 0.6)	.
0.1 (-0.9 to 0.3)	B	-0.6 (-0.9 to 0.0
0.2 (-0.8 to 0.4)	0.3 (-0.6 to 0.1)	C

下面的三角形代表间接比较（我对此示例感兴趣的那些）并读取列与行。例如，0.1 (-0.9 to 0.3) 是 A vs B。

这些值作为data.frame 存储在R 的环境中list 的元素中。

结果我需要的是：

char1	char2	value1	value2	value3
A	B	0.1	0.9	0.3
A	C	0.2	-0.8	0.4
B	A	-0.1	-0.3	0.9
B	C	0.3	-0.6	0.1
C	A	-0.2	-0.4	0.8
C	B	-0.3	-0.1	0.6

请注意，为了获得 B 与 A，我颠倒了这些值。根据结果​​的性质，我可能会改为使用 1/value。

我试过看at the code behind the netleague function，但揭穿它对我来说是相当先进的。

有人知道如何使用 R 自动执行此任务吗？

到目前为止，我最好的选择是手动执行此操作（耗时超过 8000 个值并且可能引入人为拼写错误）或至少使用 Excel 中的一些公式（仍然很耗时，我必须逐行调整公式）。

可重现格式的数据：（来源：dput(league[["fixed"]][1:4, ])）

structure(list(V1 = c("dalia", "0.08 (-0.9 to  0.26)", 
"-0.15 (-0.40 to  0.06)", "0.37 ( 0.00 to  0.78)"), V2 = c("-0.05 (-0.33 to  0.22)", 
"camelia", "-0.24 (-0.49 to -0.01)", "0.31 (-0.09 to  0.75)"
), V3 = c("-0.14 (-0.64 to  0.32)", "-0.37 (-0.66 to -0.05)", 
"margher", "0.54 ( 0.12 to  0.95)"), V4 = c(".", ".", ".", 
"rosa_can"), V5 = c(".", ".", ".", "."), V6 = c(".", ".", ".", 
"."), V7 = c(".", ".", ".", "."), V8 = c("0.65 ( 0.54 to  0.87)", 
"0.54 ( 0.38 to  0.78)", "0.77 ( 0.2 to  1.28)", "0.29 (-0.08 to  0.67)"
), V9 = c(".", ".", ".", "."), V10 = c(".", ".", ".", "."), V11 = c(".", 
".", ".", "."), V12 = c("0.23 (-0.52 to  0.99)", ".", "-0.05 (-0.56 to  0.47)", 
"."), V13 = c(".", "0.07 (-0.25 to  0.33)", ".", ".")), row.names = c(NA, 
4L), class = "data.frame")

Answer 1

你可以试试：

ff <- function(x, y, z){
  sprintf("%.1f (%.1f to %.1f)",  -as.numeric(x),
          -as.numeric(z), -as.numeric(y))
}
DT <- as.matrix(DT)
rownames(DT) <- colnames(DT) <- diag(DT)

pattern <- r"((\S+)\s+\(\s*(\S+)\s*to\s*(\S+)\))"
DT[upper.tri(DT)] <- gsubfn::gsubfn(pattern,ff, t(DT)[upper.tri(DT)])

with(subset(as.data.frame.table(DT), Var1!=Var2),
cbind(Var2, Var1, read.table(text=gsub("[()to]", " ", Freq))))                          
      

 Var2 Var1   V1   V2  V3
1    A    B  0.1 -0.9 0.3
2    A    C  0.2 -0.8 0.4
3    B    A -0.1 -0.3 0.9
4    B    C  0.3 -0.6 0.1
5    C    A -0.2 -0.4 0.8
6    C    B -0.3 -0.1 0.6

Answer 2

这是您的问题的第一次尝试，但我无法完全匹配您想要的输出。 所以请指出错误，这样我就可以再打一轮了..

library(data.table)
# Sample data
DT <- fread('"A"    "0.3 (-0.4 to 0.6)"     "."
"0.1 (-0.9 to 0.3)"     "B"     "-0.6 (-0.9 to 0.0)"
"0.2 (-0.8 to 0.4)"     "0.3 (-0.6 to 0.1)"     "C"', header = FALSE)

# Code
# Get names, set as row and colnames
names.v <- diag(as.matrix(DT))
setnames(DT, new = names.v)
DT[, char2 := names.v]
#                    A                 B                  C rowname
# 1:                 A 0.3 (-0.4 to 0.6)                  .       A
# 2: 0.1 (-0.9 to 0.3)                 B -0.6 (-0.9 to 0.0)       B
# 3: 0.2 (-0.8 to 0.4) 0.3 (-0.6 to 0.1)                  C       C

# Melt to long
ans <- setcolorder(melt(DT, id.vars = "char2", variable.name = "char1"), c(2,1,3))
#    char1 char2              value
# 1:     A     A                  A
# 2:     A     B  0.1 (-0.9 to 0.3)
# 3:     A     C  0.2 (-0.8 to 0.4)
# 4:     B     A  0.3 (-0.4 to 0.6)
# 5:     B     B                  B
# 6:     B     C  0.3 (-0.6 to 0.1)
# 7:     C     A                  .
# 8:     C     B -0.6 (-0.9 to 0.0)
# 9:     C     C                  C
                
# keep relevant rows
ans <- ans[!char1 == char2, ]
# extract numeric values
ans[, paste0("val", 1:length(tstrsplit(ans$value, "[^0-9-\\.]+", perl = TRUE))) := 
      tstrsplit(ans$value, "[^0-9-\\.]+", perl = TRUE)][]
#    char1 char2              value val1 val2 val3
# 1:     A     B  0.1 (-0.9 to 0.3)  0.1 -0.9  0.3
# 2:     A     C  0.2 (-0.8 to 0.4)  0.2 -0.8  0.4
# 3:     B     A  0.3 (-0.4 to 0.6)  0.3 -0.4  0.6
# 4:     B     C  0.3 (-0.6 to 0.1)  0.3 -0.6  0.1
# 5:     C     A                  .    . <NA> <NA>
# 6:     C     B -0.6 (-0.9 to 0.0) -0.6 -0.9  0.0