Question

有没有办法比较一行的“任何值”是否与上面一行的“任何值”相同 - 无论订单是？下面是一个非常随机的输入数据表。

DT <- data.table(A=c("a","a","b","d","e","f","h","i","j"),
                 B=c("a","b","c","c","f","g",NA,"j",NA),
                 C=c("a","b","c","b","g","h",NA,NA,NA))

> DT
   A  B  C
1: a  a  a
2: a  b  b
3: b  c  c
4: d  c  b
5: e  f  g
6: f  g  h
7: h NA NA
8: i  j NA
9: j NA NA

我想添加一个D行，用于将行与上面的行进行比较，并比较两行的任何值是否相同（无论顺序如何）。所以期望的输出是：

 > DT
   A  B  C  D
1: a  a  a  0 #No row above to compare; could be either NA or 0
2: a  b  b  1 #row 2 has "a", which is in row 1; returns 1
3: b  c  c  1 #row 3 has "b", which is in row 2; returns 1
4: d  c  b  1 #row 4 has "b" and "c", which are in row 3; returns 1
5: e  f  g  0 #row 5 has nothing that is in row 4; returns 0
6: f  g  h  1 #row 6 has "f" and "g", which are in row 5; returns 1
7: h NA NA  1 #row 7 has "h", which is in row 6; returns 1
8: i  j NA  0 #row 8 has nothing that is in row 7 (NA doesn't count)
9: j NA NA  1 #row 9 has "j", which is in row 8; returns 1 (NA doesn't count)

主要思想是我想将行（或向量）与另一行（向量）进行比较，并且如果每行（向量）中的任何元素都是相同的，则将两行定义为相同。（不重复比较每个元素）

Answer 1

我们可以通过获取数据集的lead行，每行paste，使用{{1}检查原始数据集的paste行的任何模式来执行此操作}和grepl，然后Map并转换为unlist

integer

或者我们可以使用DT[, D := { v1 <- do.call(paste, .SD) v2 <- do.call(paste, c(shift(.SD, type = "lead"), sep="|")) v2N <- gsub("NA\\|*|\\|*NA", "", v2) v3 <- unlist(Map(grepl, v2N, v1), use.names = FALSE) as.integer(head(c(FALSE, v3), -1)) }] DT # A B C D #1: a a a 0 #2: a b b 1 #3: b c c 1 #4: d c b 1 #5: e f g 0 #6: f g h 1 #7: h NA NA 1 #8: i j NA 0 #9: j NA NA 1

进行split并进行比较

Map

Answer 2

这是另一种方法。它可能不适用于大型data.tables，因为它使用的by=1:nrow(DT)往往很慢。

DT[, D:= sign(DT[, c(.SD, shift(.SD))][,
   sum(!is.na(intersect(unlist(.SD[, .(A, B, C)]), unlist(.SD[, .(V4, V5, V6)])))),
   by=1:nrow(DT)]$V1)]

此处，[, c(.SD, shift(.SD))]创建data.frame的副本，包含滞后变量（cbinded）。然后第二个链与原始data.table中的未列出变量和移位的data.table相交。 NA被分配0并且非NA被分配1并且这些结果被求和。对复制的data.table的每一行都会执行此操作。总和使用$v1提取，并使用sign转换为二进制（0和1）。

返回

DT
   A  B  C D
1: a  a  a 0
2: a  b  b 1
3: b  c  c 1
4: d  c  b 1
5: e  f  g 0
6: f  g  h 1
7: h NA NA 1
8: i  j NA 0
9: j NA NA 1

Answer 3

我会按照表格的索引（减去最后一个）来做一个顺利的事情：

compare <- function(i) {
    row1 <- as.character(DT[i,])
    row2 <- as.character(DT[i+1,])
    return(length(intersect(row1[!is.na(row1)], row2[!is.na(row2)])) > 0)
}

result <- sapply(1:(nrow(DT) - 1), compare)

这将返回逻辑向量，因此如果您希望获得整数，请将compare的输出包装在as.numeric()

中

Answer 4

以下是使用base的{{1}} R解决方案：

intersect

Answer 5

这是一个使用data.table连接的无循环方法：

DT[, id := 1:.N]
dt <- melt(DT, id.vars = "id")
dt[, id2 := id-1]
dt <- dt[!is.na(value)]
idx <- dt[dt, on = .(id2 = id, value), nomatch=0][, unique(id)]
DT[, `:=`(D = as.integer(id %in% idx), id = NULL)]

它看起来有些复杂，但对于包含三列的100万行数据集，id确实表现得相当不错。

Answer 6

此解决方案将两行与%in%（unlist()之后）进行比较：

DT[, result:=as.integer(c(NA, sapply(2:DT[,.N], function(i) any(na.omit(unlist(DT[i])) %in% unlist(DT[i-1])))))]
#> DT
#   A  B  C result
#1: a  a  a     NA
#2: a  b  b      1
#3: b  c  c      1
#4: d  c  b      1
#5: e  f  g      0
#6: f  g  h      1
#7: h NA NA      1
#8: i  j NA      0
#9: j NA NA      1

Answer 7

使用intersect和mapply的组合，您可以：

#list of unique elements in each row
tableList = apply(DT,1,function(x) unique(na.omit(x)))

#a lagged list to be compared with above list
tableListLag = c(NA,tableList[2:length(tableList)-1])

#find common elements using intersect function
#if length > 0 implies common elements hence set value as 1 else 0
DT$D = mapply(function(x,y) ifelse(length(intersect(x,y))>0,1,0) ,tableList,tableListLag,
             SIMPLIFY = TRUE)


DT
#   A  B  C D
#1: a  a  a 0
#2: a  b  b 1
#3: b  c  c 1
#4: d  c  b 1
#5: e  f  g 0
#6: f  g  h 1
#7: h NA NA 1
#8: i  j NA 0
#9: j NA NA 1

如何比较一行中的任何元素是否相同

7 个答案: