匹配来自不同来源的数据框中的多个列答案

【问题标题】：Match multiple columns in dataframe from different sources匹配来自不同来源的数据框中的多个列
【发布时间】：2013-07-06 11:18:24
【问题描述】：

我有多个玩家在不同时间点进行游戏的结果。我从两个不同的来源获得了这些信息，它们为每个玩家分配了不同的唯一 ID。我想找到一种雄辩的方式来通过玩家 id 匹配两个数据源。两个数据源

sourcex <- structure(list(outcomedate = structure(c(12637, 12637, 12637, 
12637, 12637, 12637, 12637, 12637, 12637, 12637, 12637, 12638, 
12639, 12640, 12640, 12640, 12640, 12640, 12640, 12640, 12640, 
12641, 12641, 12641, 12643, 12644, 12644, 12644, 12644, 12644, 
12644, 12644, 12644, 12644, 12644, 12645), class = "Date"), xid1 = c(206L, 
208L, 209L, 216L, 233L, 235L, 239L, 241L, 250L, 253L, 259L, 238L, 
236L, 211L, 221L, 234L, 249L, 254L, 255L, 257L, 258L, 207L, 230L, 
248L, 258L, 207L, 211L, 230L, 234L, 236L, 248L, 249L, 254L, 255L, 
257L, 221L), xid2 = c(211L, 207L, 221L, 249L, 248L, 257L, 234L, 
255L, 236L, 258L, 254L, 230L, 241L, 253L, 235L, 238L, 208L, 233L, 
239L, 259L, 206L, 209L, 250L, 216L, 259L, 216L, 241L, 208L, 235L, 
239L, 253L, 250L, 209L, 238L, 206L, 233L), outcome1 = c(2L, 1L, 
0L, 2L, 1L, 3L, 1L, 1L, 2L, 2L, 0L, 2L, 3L, 3L, 1L, 0L, 2L, 0L, 
0L, 0L, 2L, 1L, 2L, 1L, 0L, 3L, 2L, 0L, 0L, 0L, 2L, 2L, 2L, 1L, 
1L, 1L), outcome2 = c(0L, 0L, 0L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 
0L, 1L, 0L, 1L, 0L, 0L, 1L, 2L, 0L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 
2L, 0L, 1L, 1L, 2L, 1L, 0L, 1L, 1L, 3L)), .Names = c("outcomedate", 
"xid1", "xid2", "outcome1", "outcome2"), row.names = c(NA, 36L
), class = "data.frame")

sourcey <- structure(list(outcomedate = structure(c(12637, 12637, 12637, 
12637, 12637, 12637, 12637, 12637, 12637, 12637, 12637, 12638, 
12639, 12640, 12640, 12640, 12640, 12640, 12640, 12640, 12640, 
12641, 12641, 12641, 12643, 12644, 12644, 12644, 12644, 12644, 
12644, 12644, 12644, 12644, 12644, 12645), class = "Date"), yid1 = c(56, 
46, 67, 68, 59, 63, 55, 50, 66, 61, 57, 58, 53, 60, 64, 48, 69, 
54, 51, 65, 62, 47, 49, 52, 64, 60, 47, 48, 69, 49, 54, 51, 65, 
53, 52, 62), yid2 = c(47, 51, 64, 48, 62, 69, 53, 54, 60, 49, 
65, 52, 50, 63, 57, 56, 61, 46, 58, 67, 66, 59, 68, 55, 63, 57, 
68, 55, 59, 67, 58, 66, 50, 46, 56, 61), outcome1 = structure(c(1L, 
1L, 2L, 2L, 3L, 3L, 2L, 1L, 4L, 1L, 2L, 2L, 4L, 3L, 2L, 2L, 3L, 
3L, 3L, 4L, 1L, 1L, 1L, 2L, 3L, 1L, 4L, 2L, 2L, 2L, 1L, 3L, 2L, 
3L, 3L, 1L), .Label = c("1", "2", "0", "3", "4", "5", "6"), class = "factor"), 
    outcome2 = structure(c(1L, 2L, 3L, 2L, 1L, 1L, 2L, 2L, 3L, 
    2L, 1L, 2L, 1L, 3L, 2L, 2L, 3L, 1L, 1L, 2L, 1L, 3L, 2L, 3L, 
    2L, 2L, 3L, 2L, 1L, 3L, 2L, 2L, 3L, 2L, 1L, 4L), .Label = c("0", 
    "1", "2", "3", "4"), class = "factor")), .Names = c("outcomedate", 
"yid1", "yid2", "outcome1", "outcome2"), row.names = c(NA, 36L
), class = "data.frame")

两个来源有一个共同的outcomedate、outcome1、outcome2。他们为游戏中的各个玩家分配不同的 ID。我已完成以下操作来查找 id 之间的匹配项。

sourcex$ID <- with(sourcex, paste0(outcomedate, outcome1, outcome2))
sourcey$ID <- with(sourcey, paste0(outcomedate, outcome1, outcome2))

uPlayersx <- with(sourcex, unique(c(xid1, xid2)))
uPlayersy <- with(sourcey, unique(c(yid1, yid2)))

comparex <- sapply(uPlayersx, function(x){
  paste0(with(sourcex, ID[xid1 == x| xid2 == x]), collapse = '~')
})

comparey <- sapply(uPlayersy, function(x){
  paste0(with(sourcey, ID[yid1 == x| yid2 == x]), collapse = '~')
})

dumMatch <- data.frame(xid = uPlayersx, yid = uPlayersy[match(comparex, comparey)])

在这个测试数据集上它工作正常，但实际应用程序更大，这感觉就像一个 hack。此外，真实数据集在报告等方面可能存在错误，因此可能需要部分匹配。任何帮助将不胜感激。

【问题讨论】：

有趣的问题。你的真实数据有多大？我在考虑比赛的数量、日期的数量、团队的数量。
@flodel 我可能有 15 年的数据和数百名玩家。还有多达 10 个数据源，可能缺少数据和错误等。我一直在寻找一个 R api 来谷歌优化（打开优化）。如果没有类似的东西，我可能会考虑写一个。

标签： r dataframe

【解决方案1】：

这将（至少）帮助过滤掉完美匹配的日期：

match.day <- function(d)
{
    tempx <- sourcex[sourcex$outcomedate==d,]
    tempy <- sourcey[sourcey$outcomedate==d,]

    if(nrow(tempx)!=nrow(tempy)) stop("matching failed: number of rows differ.")

    P <- outer(tempx$outcome1, tempy$outcome1, `==`) &
         outer(tempx$outcome2, tempy$outcome2, `==`)

    if(any(rowSums(P)!=1)) stop("maching failed: ambiguous or impossible assignment.")

    map <- P %*% seq_len(nrow(tempy))

    cbind(tempx[,c("xid1","xid2")], tempy[map,c("yid1","yid2")])
}

days <- unique(c(sourcex$outcomedate, sourcey$outcomedate))

do.call(rbind, lapply(days[-c(1,4,7)], match.day))

请注意，它在第 1、4 和 7 天都失败了（请参阅 days[c(1,4,7)]）。

其他日子的结果：

   xid1 xid2 yid1 yid2
12  238  230   58   52
13  236  241   53   50
22  207  209   47   59
23  230  250   52   55
24  248  216   49   68
25  258  259   64   63
36  221  233   62   61

【讨论】：