R根据添加条件的特定列合并两个数据集

时间:2020-05-28 14:16:02

标签: r dataframe join merge

Uwe和GKi的答案都是正确的。 Gki收到赏金是因为Uwe迟到了,但是Uwe的解决方案运行速度快了15倍

我有两个数据集,其中包含针对多个测量时刻的不同患者的评分,如下所示:

df1 <- data.frame("ID" = c("patient1","patient1","patient1","patient1","patient2","patient3"),
                  "Days" = c(0,25,235,353,100,538),
                  "Score" = c(NA,2,3,4,5,6), 
                  stringsAsFactors = FALSE)
df2 <- data.frame("ID" = c("patient1","patient1","patient1","patient1","patient2","patient2","patient3"),
                  "Days" = c(0,25,248,353,100,150,503),
                  "Score" = c(1,10,3,4,5,7,6), 
                  stringsAsFactors = FALSE)
> df1
        ID Days Score
1 patient1    0    NA
2 patient1   25     2
3 patient1  235     3
4 patient1  353     4
5 patient2  100     5
6 patient3  538     6

> df2
        ID Days Score
1 patient1    0     1
2 patient1   25    10
3 patient1  248     3
4 patient1  353     4
5 patient2  100     5
6 patient2  150     7
7 patient3  503     6

ID列显示患者ID,第Days列显示测量时刻(自患者入院以来的天数),第Score列显示测量的分数。这两个数据集都显示相同的数据,但是显示的时间不同(df1是2年前,df2具有相同的数据,但今年以来的更新)。

我必须比较每个患者以及两个数据集之间每个时刻的得分。但是,在某些情况下,Days变量会随时间变化很小,因此无法通过简单的联接比较数据集。示例:

library(dplyr)

> full_join(df1, df2, by=c("ID","Days")) %>% 
+   arrange(.[[1]], as.numeric(.[[2]]))

        ID Days Score.x Score.y
1 patient1    0      NA       1
2 patient1   25       2      10
3 patient1  235       3      NA
4 patient1  248      NA       3
5 patient1  353       4       4
6 patient2  100       5       5
7 patient2  150      NA       7
8 patient3  503      NA       6
9 patient3  538       6      NA

此处,第3行和第4行包含用于同一度量(得分为3)的数据,但由于Days列的值不同(235对248)而没有合并。

问题:我正在寻找一种方法来设置第二列(例如30天)的阈值,该阈值将导致以下输出:

> threshold <- 30
> *** insert join code ***

        ID Days Score.x Score.y
1 patient1    0      NA       1
2 patient1   25       2      10
3 patient1  248       3       3
4 patient1  353       4       4
5 patient2  100       5       5
6 patient2  150      NA       7
7 patient3  503      NA       6
8 patient3  538       6      NA

此输出显示先前输出的第3行和第4行已合并(因为248-235 <30),并且已采用第二个df的Days的值(248)。

要记住的三个主要条件是:

  • 在同一df(第1行和第2行)之内的阈值之内的连续日期不会合并
  • 在某些情况下,Days变量中最多有四个值存在于同一数据帧中,因此不应合并。这些值之一可能确实存在于另一个数据帧的阈值中,并且这些值必须合并。请参见下面示例中的第3行。
  • 每个分数/天数/患者组合只能使用一次。如果合并满足所有条件,但仍然有可能进行两次合并,则应使用第一个合并。
> df1
        ID Days Score
1 patient1    0     1
2 patient1    5     2
3 patient1   10     3
4 patient1   15     4
5 patient1   50     5

> df2
        ID Days Score
1 patient1    0     1
2 patient1    5     2
3 patient1   12     3
4 patient1   15     4
5 patient1   50     5

> df_combined
        ID Days Score.x Score.y
1 patient1    0       1       1
2 patient1    5       2       2
3 patient1   12       3       3
4 patient1   15       4       4
5 patient1   50       5       5

编辑CHINSOON12

> df1
          ID Days Score
 1: patient1    0     1
 2: patient1  116     2
 3: patient1  225     3
 4: patient1  309     4
 5: patient1  351     5
 6: patient2    0     6
 7: patient2   49     7
> df2
          ID Days Score
 1: patient1    0    11
 2: patient1   86    12
 3: patient1  195    13
 4: patient1  279    14
 5: patient1  315    15
 6: patient2    0    16
 7: patient2   91    17
 8: patient2  117    18

我将您的解决方案包装成如下功能:

testSO2 <- function(DT1,DT2) {
    setDT(DT1);setDT(DT2)
    names(DT1) <- c("ID","Days","X")
    names(DT2) <- c("ID","Days","Y")
    DT1$Days <- as.numeric(DT1$Days)
    DT2$Days <- as.numeric(DT2$Days)
    DT1[, c("s1", "e1", "s2", "e2") := .(Days - 30L, Days + 30L, Days, Days)]
    DT2[, c("s1", "e1", "s2", "e2") := .(Days, Days, Days - 30L, Days + 30L)]
    byk <- c("ID", "s1", "e1")
    setkeyv(DT1, byk)
    setkeyv(DT2, byk)
    o1 <- foverlaps(DT1, DT2)

    byk <- c("ID", "s2", "e2")
    setkeyv(DT1, byk)
    setkeyv(DT2, byk)
    o2 <- foverlaps(DT2, DT1)

    olaps <- funion(o1, setcolorder(o2, names(o1)))[
        is.na(Days), Days := i.Days]

    outcome <- olaps[, {
        if (all(!is.na(Days)) && any(Days == i.Days)) {
            s <- .SD[Days == i.Days, .(Days = Days[1L],
                                       X = X[1L],
                                       Y = Y[1L])]
        } else {
            s <- .SD[, .(Days = max(Days, i.Days), X, Y)]
        }
        unique(s)
    },
    keyby = .(ID, md = pmax(Days, i.Days))][, md := NULL][]
    return(outcome)
}

这将导致:

> testSO2(df1,df2)
          ID Days  X  Y
 1: patient1    0  1 11
 2: patient1  116  2 12
 3: patient1  225  3 13
 4: patient1  309  4 14
 5: patient1  315  4 15
 6: patient1  351  5 NA
 7: patient2    0  6 16
 8: patient2   49  7 NA
 9: patient2   91 NA 17
10: patient2  117 NA 18

如您所见,第4行和第5行是错误的。 df1中Score的值使用了两次(4)。这些行周围的正确输出应如下所示,因为每个分数(在这种情况下为X或Y)只能使用一次:

          ID Days  X  Y
 4: patient1  309  4 14
 5: patient1  315 NA 15
 6: patient1  351  5 NA

下面的数据框代码。

> dput(df1)
structure(list(ID = c("patient1", "patient1", "patient1", "patient1", 
"patient1", "patient2", "patient2"), Days = c("0", "116", "225", 
"309", "351", "0", "49"), Score = 1:7), row.names = c(NA, 7L), class = "data.frame")
> dput(df2)
structure(list(ID = c("patient1", "patient1", "patient1", "patient1", 
"patient1", "patient2", "patient2", "patient2"), Days = c("0", 
"86", "195", "279", "315", "0", "91", "117"), Score = 11:18), row.names = c(NA, 
8L), class = "data.frame")

6 个答案:

答案 0 :(得分:5)

遗憾的是,听起来像是对真实但杂乱的数据集进行数据清理的练习,不幸的是,我们大多数人以前都曾有过经验。这是另一个data.table选项:

DT1[, c("Xrn", "s1", "e1", "s2", "e2") := .(.I, Days - 30L, Days + 30L, Days, Days)]
DT2[, c("Yrn", "s1", "e1", "s2", "e2") := .(.I, Days, Days, Days - 30L, Days + 30L)]
byk <- c("ID", "s1", "e1")
setkeyv(DT1, byk)
setkeyv(DT2, byk)
o1 <- foverlaps(DT1, DT2)

byk <- c("ID", "s2", "e2")
setkeyv(DT1, byk)
setkeyv(DT2, byk)
o2 <- foverlaps(DT2, DT1)

olaps <- funion(o1, setcolorder(o2, names(o1)))[
    is.na(Days), Days := i.Days]

ans <- olaps[, {
        if (any(Days == i.Days)) {
            .SD[Days == i.Days, 
                .(Days=Days[1L], Xrn=Xrn[1L], Yrn=Yrn[1L], X=X[1L], Y=Y[1L])]
        } else {
            .SD[, .(Days=md, Xrn=Xrn[1L], Yrn=Yrn[1L], X=X[1L], Y=Y[1L])]
        }
    },
    keyby = .(ID, md = pmax(Days, i.Days))]

#or also ans[duplicated(Xrn), X := NA_integer_][duplicated(Yrn), Y := NA_integer_]
ans[rowid(Xrn) > 1L, X := NA_integer_]
ans[rowid(Yrn) > 1L, Y := NA_integer_]
ans[, c("md", "Xrn", "Yrn") := NULL][]

以下数据集的输出:

   ID Days  X  Y
1:  1    0  1 11
2:  1   10  2 12
3:  1   25  3 13
4:  1  248  4 14
5:  1  353  5 15
6:  2  100  6 16
7:  2  150 NA 17
8:  3  503 NA 18
9:  3  538  7 NA

OP编辑中第二个数据集的输出:

          ID Days  X  Y
 1: patient1    0  1 11
 2: patient1  116  2 12
 3: patient1  225  3 13
 4: patient1  309  4 14
 5: patient1  315 NA 15
 6: patient1  351  5 NA
 7: patient2    0  6 16
 8: patient2   49  7 NA
 9: patient2   91 NA 17
10: patient2  117 NA 18

数据(我从其他链接的帖子中添加了更多数据,并且还简化了数据以便于查看):

library(data.table)
DT1 <- data.table(ID = c(1,1,1,1,1,2,3),
    Days = c(0,10,25,235,353,100,538))[, X := .I]
DT2 <- data.table(ID = c(1,1,1,1,1,2,2,3),
    Days = c(0,10,25,248,353,100,150,503))[, Y := .I + 10L]

说明:

  1. 依次使用每个表作为左表执行2个重叠联接。

  2. 将2个结果从右表中的NA天设置为左表中的天。

  3. 按患者和重叠日期分组。如果存在相同的日期,则保留记录。否则,使用最大日期。

  4. 每个分数只能使用一次,因此请删除重复项。

如果您发现这种方法无法提供正确结果的情况,请告诉我。

答案 1 :(得分:5)

使用lapply base 解决方案,找出中的差异在阈值以下的位置,并得出expand.grid获得所有可能的组合。然后删除那些会两次采摘或正在另一次采摘的那些。从这些中计算出日差并选择具有连续最低差的行。然后rbind与df2不匹配。

threshold <- 30
nmScore <- threshold
x <- do.call(rbind, lapply(unique(c(df1$ID, df2$ID)), function(ID) {
  x <- df1[df1$ID == ID,]
  y <- df2[df2$ID == ID,]
  if(nrow(x) == 0) {return(data.frame(ID=ID, y[1,-1][NA,], y[,-1]))}
  if(nrow(y) == 0) {return(data.frame(ID=ID, x[,-1], x[1,-1][NA,]))}
  x <- x[order(x$Days),]
  y <- y[order(y$Days),]
  z <- do.call(expand.grid, lapply(x$Days, function(z) c(NA,
         which(abs(z - y$Days) < threshold))))
  z <- z[!apply(z, 1, function(z) {anyDuplicated(z[!is.na(z)]) > 0 ||
         any(diff(z[!is.na(z)]) < 1)}), , drop = FALSE]
  s <- as.data.frame(sapply(seq_len(ncol(z)), function(j) {
         abs(x$Days[j] - y$Days[z[,j]])}))
  s[is.na(s)] <- nmScore
  s <- matrix(apply(s, 1, sort), nrow(s), byrow = TRUE)
  i <- rep(TRUE, nrow(s))
  for(j in seq_len(ncol(s))) {i[i]  <- s[i,j] == min(s[i,j])}
  i <- unlist(z[which.max(i),])
  j <- setdiff(seq_len(nrow(y)), i)
  rbind(data.frame(ID=ID, x[,-1], y[i, -1]),
  if(length(j) > 0) data.frame(ID=ID, x[1,-1][NA,], y[j, -1], row.names=NULL))
}))
x <- x[order(x[,1], ifelse(is.na(x[,2]), x[,4], x[,2])),]

数据:

0..Boris Ruwe的第一个测试用例, 1..2来自Boris Ruwe的测试用例, 2..3 Boris Ruwe的测试用例, 3 ..来自Uwe的测试用例, 4 ..来自R rolling join two data.tables with error margin on join的Boris Ruwe的测试用例, 5..GKi的测试用例。

df1 <- structure(list(ID = c("0patient1", "0patient1", "0patient1", 
"0patient1", "0patient2", "0patient3", "1patient1", "1patient1", 
"1patient1", "1patient1", "1patient1", "2patient1", "2patient1", 
"2patient1", "2patient1", "2patient1", "2patient2", "2patient2", 
"3patient1", "3patient1", "3patient1", "3patient1", "3patient1", 
"3patient1", "3patient2", "3patient3", "4patient1", "4patient1", 
"4patient1", "4patient1", "4patient2", "4patient3", "5patient1", 
"5patient1", "5patient1", "5patient2"), Days = c(0, 25, 235, 
353, 100, 538, 0, 5, 10, 15, 50, 0, 116, 225, 309, 351, 0, 49, 
0, 1, 25, 235, 237, 353, 100, 538, 0, 10, 25, 340, 100, 538, 
3, 6, 10, 1), Score = c(NA, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 1, 
2, 3, 4, 5, 6, 7, NA, 2, 3, 4, 5, 6, 7, 8, NA, 2, 3, 99, 5, 6, 
1, 2, 3, 1)), row.names = c(NA, -36L), class = "data.frame")
df2 <- structure(list(ID = c("0patient1", "0patient1", "0patient1", 
"0patient1", "0patient2", "0patient2", "0patient3", "1patient1", 
"1patient1", "1patient1", "1patient1", "1patient1", "2patient1", 
"2patient1", "2patient1", "2patient1", "2patient1", "2patient2", 
"2patient2", "2patient2", "3patient1", "3patient1", "3patient1", 
"3patient1", "3patient1", "3patient1", "3patient2", "3patient2", 
"3patient3", "4patient1", "4patient1", "4patient1", "4patient1", 
"4patient2", "4patient2", "4patient3", "5patient1", "5patient1", 
"5patient1", "5patient3"), Days = c(0, 25, 248, 353, 100, 150, 
503, 0, 5, 12, 15, 50, 0, 86, 195, 279, 315, 0, 91, 117, 0, 25, 
233, 234, 248, 353, 100, 150, 503, 0, 10, 25, 353, 100, 150, 
503, 1, 4, 8, 1), Score = c(1, 10, 3, 4, 5, 7, 6, 1, 2, 3, 4, 
5, 11, 12, 13, 14, 15, 16, 17, 18, 11, 12, 13, 14, 15, 16, 17, 
18, 19, 1, 10, 3, 4, 5, 7, 6, 11, 12, 13, 1)), row.names = c(NA, 
-40L), class = "data.frame")
df1
#          ID Days Score
#1  0patient1    0    NA
#2  0patient1   25     2
#3  0patient1  235     3
#4  0patient1  353     4
#5  0patient2  100     5
#6  0patient3  538     6
#7  1patient1    0     1
#8  1patient1    5     2
#9  1patient1   10     3
#10 1patient1   15     4
#11 1patient1   50     5
#12 2patient1    0     1
#13 2patient1  116     2
#14 2patient1  225     3
#15 2patient1  309     4
#16 2patient1  351     5
#17 2patient2    0     6
#18 2patient2   49     7
#19 3patient1    0    NA
#20 3patient1    1     2
#21 3patient1   25     3
#22 3patient1  235     4
#23 3patient1  237     5
#24 3patient1  353     6
#25 3patient2  100     7
#26 3patient3  538     8
#27 4patient1    0    NA
#28 4patient1   10     2
#29 4patient1   25     3
#30 4patient1  340    99
#31 4patient2  100     5
#32 4patient3  538     6
#33 5patient1    3     1
#34 5patient1    6     2
#35 5patient1   10     3
#36 5patient2    1     1

df2
#          ID Days Score
#1  0patient1    0     1
#2  0patient1   25    10
#3  0patient1  248     3
#4  0patient1  353     4
#5  0patient2  100     5
#6  0patient2  150     7
#7  0patient3  503     6
#8  1patient1    0     1
#9  1patient1    5     2
#10 1patient1   12     3
#11 1patient1   15     4
#12 1patient1   50     5
#13 2patient1    0    11
#14 2patient1   86    12
#15 2patient1  195    13
#16 2patient1  279    14
#17 2patient1  315    15
#18 2patient2    0    16
#19 2patient2   91    17
#20 2patient2  117    18
#21 3patient1    0    11
#22 3patient1   25    12
#23 3patient1  233    13
#24 3patient1  234    14
#25 3patient1  248    15
#26 3patient1  353    16
#27 3patient2  100    17
#28 3patient2  150    18
#29 3patient3  503    19
#30 4patient1    0     1
#31 4patient1   10    10
#32 4patient1   25     3
#33 4patient1  353     4
#34 4patient2  100     5
#35 4patient2  150     7
#36 4patient3  503     6
#37 5patient1    1    11
#38 5patient1    4    12
#39 5patient1    8    13
#40 5patient3    1     1

结果:

#           ID Days Score Days.1 Score.1
#1   0patient1    0    NA      0       1
#2   0patient1   25     2     25      10
#3   0patient1  235     3    248       3
#4   0patient1  353     4    353       4
#5   0patient2  100     5    100       5
#110 0patient2   NA    NA    150       7
#111 0patient3   NA    NA    503       6
#6   0patient3  538     6     NA      NA
#7   1patient1    0     1      0       1
#8   1patient1    5     2      5       2
#9   1patient1   10     3     12       3
#10  1patient1   15     4     15       4
#11  1patient1   50     5     50       5
#12  2patient1    0     1      0      11
#112 2patient1   NA    NA     86      12
#13  2patient1  116     2     NA      NA
#210 2patient1   NA    NA    195      13
#14  2patient1  225     3     NA      NA
#37  2patient1   NA    NA    279      14
#15  2patient1  309     4    315      15
#16  2patient1  351     5     NA      NA
#17  2patient2    0     6      0      16
#18  2patient2   49     7     NA      NA
#113 2patient2   NA    NA     91      17
#211 2patient2   NA    NA    117      18
#19  3patient1    0    NA      0      11
#20  3patient1    1     2     NA      NA
#21  3patient1   25     3     25      12
#114 3patient1   NA    NA    233      13
#22  3patient1  235     4    234      14
#23  3patient1  237     5    248      15
#24  3patient1  353     6    353      16
#25  3patient2  100     7    100      17
#115 3patient2   NA    NA    150      18
#116 3patient3   NA    NA    503      19
#26  3patient3  538     8     NA      NA
#27  4patient1    0    NA      0       1
#28  4patient1   10     2     10      10
#29  4patient1   25     3     25       3
#30  4patient1  340    99    353       4
#31  4patient2  100     5    100       5
#117 4patient2   NA    NA    150       7
#118 4patient3   NA    NA    503       6
#32  4patient3  538     6     NA      NA
#119 5patient1   NA    NA      1      11
#33  5patient1    3     1      4      12
#34  5patient1    6     2      8      13
#35  5patient1   10     3     NA      NA
#36  5patient2    1     1     NA      NA
#NA  5patient3   NA    NA      1       1

格式化结果:

data.frame(ID=x[,1], Days=ifelse(is.na(x[,2]), x[,4], x[,2]),
 Score.x=x[,3], Score.y=x[,5])
#          ID Days Score.x Score.y
#1  0patient1    0      NA       1
#2  0patient1   25       2      10
#3  0patient1  235       3       3
#4  0patient1  353       4       4
#5  0patient2  100       5       5
#6  0patient2  150      NA       7
#7  0patient3  503      NA       6
#8  0patient3  538       6      NA
#9  1patient1    0       1       1
#10 1patient1    5       2       2
#11 1patient1   10       3       3
#12 1patient1   15       4       4
#13 1patient1   50       5       5
#14 2patient1    0       1      11
#15 2patient1   86      NA      12
#16 2patient1  116       2      NA
#17 2patient1  195      NA      13
#18 2patient1  225       3      NA
#19 2patient1  279      NA      14
#20 2patient1  309       4      15
#21 2patient1  351       5      NA
#22 2patient2    0       6      16
#23 2patient2   49       7      NA
#24 2patient2   91      NA      17
#25 2patient2  117      NA      18
#26 3patient1    0      NA      11
#27 3patient1    1       2      NA
#28 3patient1   25       3      12
#29 3patient1  233      NA      13
#30 3patient1  235       4      14
#31 3patient1  237       5      15
#32 3patient1  353       6      16
#33 3patient2  100       7      17
#34 3patient2  150      NA      18
#35 3patient3  503      NA      19
#36 3patient3  538       8      NA
#37 4patient1    0      NA       1
#38 4patient1   10       2      10
#39 4patient1   25       3       3
#40 4patient1  340      99       4
#41 4patient2  100       5       5
#42 4patient2  150      NA       7
#43 4patient3  503      NA       6
#44 4patient3  538       6      NA
#45 5patient1    1      NA      11
#46 5patient1    3       1      12
#47 5patient1    6       2      13
#48 5patient1   10       3      NA
#49 5patient2    1       1      NA
#50 5patient3    1      NA       1

获得Days的替代方法:

#From df1 and in case it is NA I took it from df2
data.frame(ID=x[,1], Days=ifelse(is.na(x[,2]), x[,4], x[,2]),
 Score.x=x[,3], Score.y=x[,5])

#From df2 and in case it is NA I took it from df1
data.frame(ID=x[,1], Days=ifelse(is.na(x[,4]), x[,2], x[,4]),
 Score.x=x[,3], Score.y=x[,5])

#Mean
data.frame(ID=x[,1], Days=rowMeans(x[,c(2,4)], na.rm=TRUE),
 Score.x=x[,3], Score.y=x[,5])

如果应将天的差异减至最小,以免取最近的数值,则可能的方法是:

threshold <- 30
nmScore <- threshold
x <- do.call(rbind, lapply(unique(c(df1$ID, df2$ID)), function(ID) {
  x <- df1[df1$ID == ID,]
  y <- df2[df2$ID == ID,]
  x <- x[order(x$Days),]
  y <- y[order(y$Days),]
  if(nrow(x) == 0) {return(data.frame(ID=ID, y[1,-1][NA,], y[,-1]))}
  if(nrow(y) == 0) {return(data.frame(ID=ID, x[,-1], x[1,-1][NA,]))}
  z <- do.call(expand.grid, lapply(x$Days, function(z) c(NA,
         which(abs(z - y$Days) < threshold))))
  z <- z[!apply(z, 1, function(z) {anyDuplicated(z[!is.na(z)]) > 0 ||
         any(diff(z[!is.na(z)]) < 1)}), , drop = FALSE]
  s <- as.data.frame(sapply(seq_len(ncol(z)), function(j) {
         abs(x$Days[j] - y$Days[z[,j]])}))
  s[is.na(s)] <- nmScore
  i <- unlist(z[which.min(rowSums(s)),])
  j <- setdiff(seq_len(nrow(y)), i)
  rbind(data.frame(ID=ID, x[,-1], y[i, -1]),
  if(length(j) > 0) data.frame(ID=ID, x[1,-1][NA,], y[j, -1], row.names=NULL))
}))
x <- x[order(x[,1], ifelse(is.na(x[,2]), x[,4], x[,2])),]

答案 2 :(得分:4)

迟到聚会了,这是一个解决方案,它使用完全外部联接,并根据OP的规则对行进行随后的分组和聚合

library(data.table)
threshold <- 30
# full outer join
m <- merge(setDT(df1)[, o := 1L], setDT(df2)[, o := 2L], 
           by = c("ID", "Days"), all = TRUE)
# reorder rows
setorder(m, ID, Days)
# create grouping variable
m[, g := rleid(ID,
               cumsum(c(TRUE, diff(Days) > threshold)),
               !is.na(o.x) & !is.na(o.y),
               cumsum(c(TRUE, diff(fcoalesce(o.x, o.y)) == 0L))
)][, g := rleid(g, (rowid(g) - 1L) %/% 2)][]
# collapse rows where required
m[, .(ID = last(ID), Days = last(Days), 
      Score.x = last(na.omit(Score.x)), 
      Score.y = last(na.omit(Score.y)))
  , by = g][, g := NULL][]

对于OP的第一个测试用例,我们得到

         ID Days Score.x Score.y
1: patient1    0      NA       1
2: patient1   25       2      10
3: patient1  248       3       3
4: patient1  353       4       4
5: patient2  100       5       5
6: patient2  150      NA       7
7: patient3  503      NA       6
8: patient3  538       6      NA

符合预期。

与其他用例的验证

使用OP的第二个测试用例

df1 <- data.table(ID = rep("patient1", 5L), Days = c(0, 5, 10, 15, 50), Score = 1:5)
df2 <- data.table(ID = rep("patient1", 5L), Days = c(0, 5, 12, 15, 50), Score = 1:5)

我们得到

         ID Days Score.x Score.y
1: patient1    0       1       1
2: patient1    5       2       2
3: patient1   12       3       3
4: patient1   15       4       4
5: patient1   50       5       5

使用OP的第3个测试用例(用于讨论chinsoon12's answer

df1 <- data.table(ID = paste0("patient", c(rep(1, 5L), 2, 2)), 
                  Days = c(0, 116, 225, 309, 351, 0, 49), Score = 1:7)
df2 <- data.table(ID = paste0("patient", c(rep(1, 5L), 2, 2, 2)), 
                  Days = c(0, 86, 195, 279, 315, 0, 91, 117), Score = 11:18)

我们得到

          ID Days Score.x Score.y
 1: patient1    0       1      11
 2: patient1  116       2      12
 3: patient1  225       3      13
 4: patient1  309       4      14
 5: patient1  315      NA      15
 6: patient1  351       5      NA
 7: patient2    0       6      16
 8: patient2   49       7      NA
 9: patient2   91      NA      17
10: patient2  117      NA      18

OP所期望的(特别是参见第5行)

最后,我自己的测试用例在233到248之间有5个“重叠天”,以验证是否可以处理此案

df1 <- data.table(ID = paste0("patient", c(rep(1, 6L), 2, 3)),
                  Days = c(0,1,25,235,237,353,100,538),
                  Score = c(NA, 2:8))
df2 <- data.table(ID = paste0("patient", c(rep(1, 6L), 2, 2, 3)),
                  Days = c(0, 25, 233, 234, 248, 353, 100, 150, 503),
                  Score = 11:19)

我们得到

          ID Days Score.x Score.y
 1: patient1    0      NA      11    # exact match
 2: patient1    1       2      NA    # overlapping, not collapsed
 3: patient1   25       3      12    # exact match
 4: patient1  233      NA      13    # overlapping, not collapsed
 5: patient1  235       4      14    # overlapping, collapsed
 6: patient1  248       5      15    # overlapping, collapsed
 7: patient1  353       6      16    # exact match
 8: patient2  100       7      17    # exact match
 9: patient2  150      NA      18    # not overlapping
10: patient3  503      NA      19    # not overlapping
11: patient3  538       8      NA    # not overlapping

说明

完整的外部联接merge(..., all = TRUE)可以找到相同ID和日期的完全匹配项,但包括两个数据集中没有匹配项的所有其他行。

在加入之前,每个数据集都会获得一个附加列o,以指示每个Score起源

对结果进行排序是因为后续操作取决于正确的行顺序。

所以,有了我自己的测试用例,我们得到了

m <- merge(setDT(df1)[, o := 1L], setDT(df2)[, o := 2L], 
           by = c("ID", "Days"), all = TRUE)
setorder(m, ID, Days)[]
          ID Days Score.x o.x Score.y o.y
 1: patient1    0      NA   1      11   2
 2: patient1    1       2   1      NA  NA
 3: patient1   25       3   1      12   2
 4: patient1  233      NA  NA      13   2
 5: patient1  234      NA  NA      14   2
 6: patient1  235       4   1      NA  NA
 7: patient1  237       5   1      NA  NA
 8: patient1  248      NA  NA      15   2
 9: patient1  353       6   1      16   2
10: patient2  100       7   1      17   2
11: patient2  150      NA  NA      18   2
12: patient3  503      NA  NA      19   2
13: patient3  538       8   1      NA  NA

现在,使用rleid()创建分组变量:

m[, g := rleid(ID,
               cumsum(c(TRUE, diff(Days) > threshold)),
               !is.na(o.x) & !is.na(o.y),
               cumsum(c(TRUE, diff(fcoalesce(o.x, o.y)) == 0L))
)][, g := rleid(g, (rowid(g) - 1L) %/% 2)][]

满足以下条件之一时,组计数器将前进:

  • ID发生变化
  • ID内,如果连续Days之间的间隔超过30天(因此ID中间隔30天或更短的行属于一个组或“重叠” )
  • 当一行直接匹配时,
  • 当连续的行具有相同的原点时,从而标识出不同原点的行的条纹,例如1, 2, 1, 2, ...2, 1, 2, 1, ...
  • ,最后,在上述条纹内,计数成对交替来源的行,例如,df1的一行,然后是df2的一行,或者df2的一行,然后是df1的一行。

OP尚未明确说明最后一个条件,但这是我对

的解释

每个分数/天数/患者组合只能使用一次。如果合并满足所有条件,但仍有可能进行双重合并, 应该使用第一个。

它确保最多折叠两行,每行分别来自不同的数据集

分组后,我们得到

          ID Days Score.x o.x Score.y o.y  g
 1: patient1    0      NA   1      11   2  1
 2: patient1    1       2   1      NA  NA  2
 3: patient1   25       3   1      12   2  3
 4: patient1  233      NA  NA      13   2  4
 5: patient1  234      NA  NA      14   2  5
 6: patient1  235       4   1      NA  NA  5
 7: patient1  237       5   1      NA  NA  6
 8: patient1  248      NA  NA      15   2  6
 9: patient1  353       6   1      16   2  7
10: patient2  100       7   1      17   2  8
11: patient2  150      NA  NA      18   2  9
12: patient3  503      NA  NA      19   2 10
13: patient3  538       8   1      NA  NA 11

大多数组仅包含一行,少数包含2行,这些行在最后一步中被折叠(按组聚合,返回所需的列并删除分组变量g)。

改进的代码

按组聚合要求每个组仅返回一个值(长度为1的向量)。 (否则,分组结果将由多行组成。)为了简化起见,上述实现在所有4列上均使用last()

last(Days)等效于max(Days),因为数据集是有序的。

但是,如果我理解正确,OP倾向于从Days返回df2值(尽管OP提到max(Days)也可以接受)。

为了从Days返回df2值,需要修改聚合步骤:如果组大小.N大于1,我们选择Days值来自df2的行,即o.y == 2所在的行。

# collapse rows where required
m[, .(ID = last(ID), 
      Days = last(if (.N > 1) Days[which(o.y == 2)] else Days), 
      Score.x = last(na.omit(Score.x)), 
      Score.y = last(na.omit(Score.y)))
  , by = g][, g := NULL][]

这将返回

          ID Days Score.x Score.y
 1: patient1    0      NA      11
 2: patient1    1       2      NA
 3: patient1   25       3      12
 4: patient1  233      NA      13
 5: patient1  234       4      14
 6: patient1  248       5      15
 7: patient1  353       6      16
 8: patient2  100       7      17
 9: patient2  150      NA      18
10: patient3  503      NA      19
11: patient3  538       8      NA

现在已从Days中选取了折叠的第5行中的df2值234。

对于Score列,last()的使用根本不重要,因为在2行的组中应该只有一个非NA值。因此,na.omit()应该只返回一个值,而last()可能只是为了保持一致性。

答案 3 :(得分:3)

此代码可让您提供阈值,然后将df1中的分数合并为df1作为新列。它只会添加落入df2 +/-阈值的单个分数范围内的分数。请注意,由于没有分数可以唯一匹配的阈值,因此不可能将所有分数连接在一起。

threshold <- 40
WhereDF1inDF2 <- apply(sapply(lapply(df2$Days, function(x) (x+threshold):(x-threshold)), function(y) df1$Days %in% y),1,which)
useable <- sapply(WhereDF1inDF2, function(x) length(x) ==1 )
df2$Score1 <- NA
df2$Score1[unlist(WhereDF1inDF2[useable])] <- df1$Score[useable]

> df2
        ID Days Score Score1
1 patient1    0     1     NA
2 patient1   25    10     NA
3 patient1  248     3      3
4 patient1  353     4      4
5 patient2  100     5      5
6 patient2  150     7     NA
7 patient3  503     6      6

答案 4 :(得分:3)

这是一个可能的data.table解决方案

library(data.table)
#convert df1 and df2 to data.table format
setDT(df1);setDT(df2)
#set colnames for later on 
#  (add .df1/.df2 suffix after Days and Score-colnamaes)
cols <- c("Days", "Score")
setnames(df1, cols, paste0( cols, ".df1" ) )
setnames(df2, cols, paste0( cols, ".df2" ) )
#update df1 with new measures from df2 (and df2 with df1)
# copies are made, to prevent changes in df1 and df2
dt1 <- copy(df1)[ df2, `:=`(Days.df2 = i.Days.df2, Score.df2 = i.Score.df2), on = .(ID, Days.df1 = Days.df2), roll = 30]
dt2 <- copy(df2)[ df1, `:=`(Days.df1 = i.Days.df1, Score.df1 = i.Score.df1), on = .(ID, Days.df2 = Days.df1), roll = -30]
#rowbind by columnnames (here the .df1/.df2 suffix is needed!), only keep unique rows
ans <- unique( rbindlist( list( dt1, dt2), use.names = TRUE ) )
#wrangle data to get to desired output
ans[, Days := ifelse( is.na(Days.df2), Days.df1, Days.df2 ) ]
ans <- ans[, .(Days, Score.x = Score.df1, Score.y = Score.df2 ), by = .(ID) ]
setkey( ans, ID, Days )  #for sorting; setorder() can also be used.
#          ID Days Score.x Score.y
# 1: patient1    0      NA       1
# 2: patient1   25       2      10
# 3: patient1  248       3       3
# 4: patient1  353       4       4
# 5: patient2  100       5       5
# 6: patient2  150      NA       7
# 7: patient3  503      NA       6
# 8: patient3  538       6      NA

答案 5 :(得分:2)

以下代码适用于您的示例数据。根据您的条件,它应该可以处理您的全部数据。对于其他例外情况,您可以调整df31df32

df1 <- data.frame("ID" = c("patient1","patient1","patient1","patient1","patient2","patient3"),
                  "Days1" = c(0,25,235,353,100,538),
                  "Score1" = c(NA,2,3,4,5,6), 
                  stringsAsFactors = FALSE)
df2 <- data.frame("ID" = c("patient1","patient1","patient1","patient1","patient2","patient2","patient3"),
                  "Days2" = c(0,25,248,353,100,150,503),
                  "Score2" = c(1,10,3,4,5,7,6), 
                  stringsAsFactors = FALSE)

##  define a dummy sequence for each patient
df11 <- df1 %>% group_by(ID) %>% mutate(ptseq = row_number())
df21 <- df2 %>% group_by(ID) %>% mutate(ptseq = row_number())

df3 <- dplyr::full_join(df11, df21, by=c("ID","ptseq")) %>% 
         arrange(.[[1]], as.numeric(.[[2]]))

df31 <- df3 %>% mutate(Days=Days2, diff=Days1-Days2) %>% 
    mutate(Score1=ifelse(abs(diff)>30, NA, Score1))
df32 <- df3 %>% mutate(diff=Days1-Days2) %>%
     mutate(Days = case_when(abs(diff)>30 ~ Days1), Score2=c(NA), Days2=c(NA)) %>% 
     subset(!is.na(Days))

df <- rbind(df31,df32) %>%  select(ID, ptseq, Days, Score1, Score2) %>% 
         arrange(.[[1]], as.numeric(.[[2]])) %>% select(-2)

>df

ID        Days Score1 Score2
  <chr>    <dbl>  <dbl>  <dbl>
1 patient1     0     NA      1
2 patient1    25      2     10
3 patient1   248      3      3
4 patient1   353      4      4
5 patient2   100      5      5
6 patient2   150     NA      7
7 patient3   503     NA      6
8 patient3   538      6     NA