数据重排/类似于数据透视表？

Question

我很严重地处理数据重排问题。以下数据包含折叠或稳定的协议（行）（列“折叠”）以及已减少，保留，添加或不存在的特性条款（列“diff.pps_leadership”，“diff.pps_cabinet”等）

我想重新排列这些数据，以便我了解减少，保留或添加特定设置的那些协议中有多少％已折叠。这些行应该是规定（diff.pps_leadership ...），这些列应该“减少”，“保留”和“添加”，而单元格的内容应该是折叠的百分比（仅限于那些这减少，保留，或增加的规定，而不是总数）

在Excle我会在数据透视表中做到这一点，但我一直没有能够与R.到达那里我试图铸造，聚合，融化和转命令，但都没有成功。

最终，该结果应该与此类似 https://docs.google.com/spreadsheets/d/1yhIbvTQTYkkwSFVxWEnPwvSvwTc0vuTYZxa15Eh1lT8/edit?usp=sharing

希望我的问题是不是太具体。感谢有任何暗示/建议。

example <- structure(list(Agreement = structure(c(8L, 4L, 6L, 9L, 2L, 3L, 
7L, 10L, 5L, 1L), .Label = c("Abuja Agreement", "Accra Peace Agreement", 
"Arusha Agreement", "Arusha/Global Ceasefire Agreement", "Comprehensive Peace Agreement", 
"InterabsentCongolese Dialogue", "Lome Agreement", "Lusaka Protocol", 
"Ouagadougou Agreement", "Tansitional Constituion"), class = "factor"), 
    diff.pps_cabinet = structure(c(2L, 1L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L), .Label = c("kept", "reduced"), class = "factor"), 
    diff.pps_leadership = structure(c(1L, 2L, 3L, 3L, 3L, 3L, 
    3L, 3L, 2L, 3L), .Label = c("absent", "kept", "reduced"), class = "factor"), 
    diff.mps_milcmd = structure(c(3L, 2L, 3L, 3L, 3L, 3L, 1L, 
    3L, 2L, 3L), .Label = c("absent", "kept", "reduced"), class = "factor"), 
    diff.mps_armyint = structure(c(3L, 2L, 2L, 3L, 3L, 3L, 1L, 
    3L, 2L, 3L), .Label = c("absent", "kept", "reduced"), class = "factor"), 
    diff.eps_commission = structure(c(1L, 1L, 1L, 1L, 3L, 1L, 
    3L, 1L, 2L, 3L), .Label = c("absent", "kept", "reduced"), class = "factor"), 
    diff.eps_company = structure(c(1L, 2L, 1L, 1L, 3L, 1L, 1L, 
    1L, 2L, 3L), .Label = c("absent", "kept", "reduced"), class = "factor"), 
    diff.veto_leg = structure(c(1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L), .Label = c("absent", "added"), class = "factor"), 
    diff.tps_devolution = structure(c(2L, 1L, 2L, 3L, 1L, 1L, 
    1L, 2L, 2L, 1L), .Label = c("absent", "kept", "reduced"), class = "factor"), 
    diff.ca.psh = structure(c(3L, 2L, 1L, 1L, 4L, 1L, 1L, 1L, 
    4L, 1L), .Label = c("absent", "added", "kept", "reduced"), class = "factor"), 
    collapse = structure(c(1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 
    1L), .Label = c("collapse", "stable"), class = "factor")), .Names = c("Agreement", 
"diff.pps_cabinet", "diff.pps_leadership", "diff.mps_milcmd", 
"diff.mps_armyint", "diff.eps_commission", "diff.eps_company", 
"diff.veto_leg", "diff.tps_devolution", "diff.ca.psh", "collapse" 
), class = "data.frame", row.names = c(NA, -10L)) 

Answer 1

以下内容会完成工作。

library(data.table) 
setDT(example) 

mvs <- c("diff.pps_cabinet", "diff.pps_leadership", 
     "diff.mps_milcmd", "diff.mps_armyint") 

vls <- c("reduced", "kept", "added", "absent") 

melt(example, c("Agreement", "collapse"), mvs 
    )[ , setNames(vapply(
     vls, function(vv) list(paste0(
     s <- sum(collapse[idx <- value == vv] == "collapse"), 
     " out of ", sum(idx), " = ", floor(100 * s/sum(idx)), "% collapsed"), 
     paste(Agreement[idx], collapse = "\n")), 
     vector("list", 2)), 
     paste0(rep(vls, each = 2), 
       c(".percent", ".names"))), by = variable] 

当前打印NaN什么也没有;要修复此问题，请用分号(if (!any(idx)) 1 else sum(idx))替换分​​母中的sum(idx)。