在包含许多ID的数据集中，我只是试图操纵id为7或9的行，并保持其他所有内容不受影响。
我试图在没有与之对应的变量的所有实例中有条件地从7或9中删除一行。所以，如果在下面的dput示例的情况下，我想从id = 9中删除第九行，因为id = 7没有itemcode = 2。反之亦然id = 7，我试图删除它的itemcode = 9，因为id = 9没有它。
id client item itemcode unit X2001 X2002 X2003 X2004 X2005 X2006 X2007
...
7 7 Bob eighth 8 100 13 18 15 NA NA NA NA
8 7 Bob ninth 9 100 11 21 10 NA NA NA NA
9 9 Bob_new first 1 100 NA NA NA 23 18 25 18
代码：
structure(list(id = c(7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 9L, 9L,
9L, 9L, 9L, 9L, 9L, 9L, 10L), client = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L), .Label = c("Bob",
"Bob_new", "Mark"), class = "factor"), item = structure(c(3L,
9L, 4L, 2L, 8L, 7L, 1L, 5L, 3L, 6L, 9L, 4L, 2L, 8L, 7L, 1L, 3L
), .Label = c("eighth", "fifth", "first", "fourth", "ninth",
"second", "seventh", "sixth", "third"), class = "factor"), itemcode = c(1L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L
), unit = c(100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L,
100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L), X2001 = structure(c(5L,
6L, 1L, 4L, 2L, 5L, 3L, 1L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L
), .Label = c("11", "12", "13", "22", "24", "25", "NA"), class = "factor"),
X2002 = structure(c(4L, 8L, 1L, 3L, 7L, 2L, 5L, 6L, 9L, 9L,
9L, 9L, 9L, 9L, 9L, 9L, 9L), .Label = c("13", "14", "15",
"17", "18", "21", "22", "24", "NA"), class = "factor"), X2003 = structure(c(5L,
1L, 4L, 2L, 6L, 1L, 3L, 1L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L), .Label = c("10", "11", "15", "19", "23", "24", "NA"), class = "factor"),
X2004 = structure(c(7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 5L, 4L,
2L, 6L, 1L, 3L, 4L, 3L, 4L), .Label = c("11", "14", "15",
"20", "23", "25", "NA"), class = "factor"), X2005 = structure(c(6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 3L, 2L, 4L, 3L, 5L, 3L, 1L, 4L,
3L), .Label = c("11", "13", "18", "19", "25", "NA"), class = "factor"),
X2006 = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 8L, 6L,
1L, 2L, 5L, 3L, 7L, 8L, 4L), .Label = c("10", "15", "18",
"19", "20", "22", "23", "25", "NA"), class = "factor"), X2007 = structure(c(8L,
8L, 8L, 8L, 8L, 8L, 8L, 8L, 4L, 7L, 6L, 2L, 4L, 1L, 5L, 5L,
3L), .Label = c("12", "13", "16", "18", "19", "21", "24",
"NA"), class = "factor")), .Names = c("id", "client", "item",
"itemcode", "unit", "X2001", "X2002", "X2003", "X2004", "X2005",
"X2006", "X2007"), class = "data.frame", row.names = c(NA, -17L
))
---------------------------------------- 另一个情景：
之前：
structure(list(id = c(7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L), client = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 3L), .Label = c("Bob", "Bob_new", "Mark"), class = "factor"),
item = structure(c(3L, 9L, 10L, 9L, 4L, 2L, 8L, 7L, 7L, 1L,
5L, 3L, 6L, 9L, 4L, 2L, 8L, 7L, 1L, 3L), .Label = c("eighth",
"fifth", "first", "fourth", "ninth", "second", "seventh",
"sixth", "third", "third "), class = "factor"), itemcode = c(1L,
3L, 3L, 3L, 4L, 5L, 6L, 7L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L,
6L, 7L, 8L, 1L), type = structure(c(1L, 1L, 2L, 3L, 1L, 1L,
1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("A",
"B", "C"), class = "factor"), unit = c(100L, 100L, 100L,
100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L,
100L, 100L, 100L, 100L, 100L, 100L, 100L), X2001 = c(24L,
25L, 30L, 26L, 11L, 22L, 12L, 25L, 24L, 13L, 11L, NA, NA,
NA, NA, NA, NA, NA, NA, NA), X2002 = c(17L, 24L, 12L, 96L,
13L, 15L, 22L, 21L, 14L, 18L, 21L, NA, NA, NA, NA, NA, NA,
NA, NA, NA), X2003 = c(23L, 10L, 46L, 94L, 19L, 11L, 24L,
19L, 10L, 15L, 10L, NA, NA, NA, NA, NA, NA, NA, NA, NA),
X2004 = c(NA, NA, 43L, 83L, NA, NA, NA, 6L, NA, NA, NA, 23L,
20L, 14L, 25L, 11L, 15L, 20L, 15L, 20L), X2005 = c(NA, NA,
97L, 86L, NA, NA, NA, 17L, NA, NA, NA, 18L, 13L, 19L, 18L,
25L, 18L, 11L, 19L, 18L), X2006 = c(NA, NA, 11L, 91L, NA,
NA, NA, 11L, NA, NA, NA, 25L, 22L, 10L, 15L, 20L, 18L, 23L,
25L, 19L), X2007 = c(NA, NA, 19L, 27L, NA, NA, NA, 15L, NA,
NA, NA, 18L, 24L, 21L, 13L, 18L, 12L, 19L, 19L, 16L)), .Names = c("id",
"client", "item", "itemcode", "type", "unit", "X2001", "X2002",
"X2003", "X2004", "X2005", "X2006", "X2007"), class = "data.frame", row.names = c(NA,
-20L))
后：
structure(list(id = c(7L, 7L, 7L, 7L, 7L, 7L, 9L, 9L, 9L, 9L,
9L, 9L, 10L), client = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 3L), .Label = c("Bob", "Bob_new", "Mark"), class = "factor"),
item = structure(c(2L, 7L, 3L, 1L, 5L, 4L, 2L, 6L, 3L, 1L,
5L, 4L, 2L), .Label = c("fifth", "first", "fourth", "seventh",
"sixth", "third", "third "), class = "factor"), itemcode = c(1L,
3L, 4L, 5L, 6L, 7L, 1L, 3L, 4L, 5L, 6L, 7L, 1L), type = structure(c(1L,
2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L), .Label = c("A",
"B"), class = "factor"), unit = c(100L, 100L, 100L, 100L,
100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L), X2001 = c(24L,
10L, 11L, 22L, 12L, 17L, NA, NA, NA, NA, NA, NA, NA), X2002 = c(17L,
87L, 13L, 15L, 22L, 19L, NA, NA, NA, NA, NA, NA, NA), X2003 = c(23L,
47L, 19L, 11L, 24L, 17L, NA, NA, NA, NA, NA, NA, NA), X2004 = c(NA,
28L, NA, NA, NA, 28L, 23L, 14L, 25L, 11L, 15L, 20L, 20L),
X2005 = c(NA, 43L, NA, NA, NA, 16L, 18L, 19L, 18L, 25L, 18L,
11L, 18L), X2006 = c(NA, 69L, NA, NA, NA, 5L, 25L, 10L, 15L,
20L, 18L, 23L, 19L), X2007 = c(NA, 72L, NA, NA, NA, 20L,
18L, 21L, 13L, 18L, 12L, 19L, 16L)), .Names = c("id", "client",
"item", "itemcode", "type", "unit", "X2001", "X2002", "X2003",
"X2004", "X2005", "X2006", "X2007"), class = "data.frame", row.names = c(NA,
-13L))
我可以实现上述过滤器代码来删除相应位置（id 7和9）中不存在的项目。
但是如果项目有子级别，例如项目类型。我也试图删除项目，如果他们在相应的字段中没有类似的类型。
答案 0 :(得分：1)
如果我理解了问题： id = 9子集中的每个itemcode必须在 id = 7子集中具有相同的itemcode（并且反向）。如果不是这种情况，那么我们使用非配对的商品代码过滤掉行，但保留id不在7或9中的所有内容。这是一种方法：
首先获取常用商品代码：
items_9 <- df_all$itemcode[ df_all$id==9 ]
items_7 <- df_all$itemcode[ df_all$id==7 ]
items_common <- items_9[ items_9 %in% items_7 ]
使用7和9的常用项目代码选择所有内容，其余部分：
df_new <- df_all[
which(
( df_all$id %in% c(7, 9) &
df_all$itemcode %in% items_common
) |
!df_all$id %in% c(7,9)
)
,]
答案 1 :(得分：1)
您可以使用filter
dplyr
library(dplyr)
filter(df_all, itemcode %in% intersect(itemcode[id==7],
itemcode[id==9])|!id %in% c(7,9) )
# id client item itemcode unit X2001 X2002 X2003 X2004 X2005 X2006 X2007
#1 7 Bob first 1 100 24 17 23 NA NA NA NA
#2 7 Bob third 3 100 25 24 10 NA NA NA NA
#3 7 Bob fourth 4 100 11 13 19 NA NA NA NA
#4 7 Bob fifth 5 100 22 15 11 NA NA NA NA
#5 7 Bob sixth 6 100 12 22 24 NA NA NA NA
#6 7 Bob seventh 7 100 24 14 10 NA NA NA NA
#7 7 Bob eighth 8 100 13 18 15 NA NA NA NA
#8 9 Bob_new first 1 100 NA NA NA 23 18 25 18
#9 9 Bob_new third 3 100 NA NA NA 14 19 10 21
#10 9 Bob_new fourth 4 100 NA NA NA 25 18 15 13
#11 9 Bob_new fifth 5 100 NA NA NA 11 25 20 18
#12 9 Bob_new sixth 6 100 NA NA NA 15 18 18 12
#13 9 Bob_new seventh 7 100 NA NA NA 20 11 23 19
#14 9 Bob_new eighth 8 100 NA NA NA 15 19 25 19
#15 10 Mark first 1 100 NA NA NA 20 18 19 16
基于新数据集，或许这有助于
library(dplyr)
library(tidyr)
dfnew %>%
unite(itemtype, itemcode,type) %>%
filter(itemtype %in% intersect(itemtype[id==7],
itemtype[id==9])|!id %in% c(7,9)) %>%
separate(itemtype, c('itemcode', 'type'))
# id client item itemcode type unit X2001 X2002 X2003 X2004 X2005 X2006
# 1 7 Bob first 1 A 100 24 17 23 NA NA NA
# 2 7 Bob third 3 B 100 30 12 46 43 97 11
# 3 7 Bob fourth 4 A 100 11 13 19 NA NA NA
# 4 7 Bob fifth 5 A 100 22 15 11 NA NA NA
# 5 7 Bob sixth 6 A 100 12 22 24 NA NA NA
# 6 7 Bob seventh 7 A 100 25 21 19 6 17 11
# 7 9 Bob_new first 1 A 100 NA NA NA 23 18 25
# 8 9 Bob_new third 3 B 100 NA NA NA 14 19 10
# 9 9 Bob_new fourth 4 A 100 NA NA NA 25 18 15
# 10 9 Bob_new fifth 5 A 100 NA NA NA 11 25 20
# 11 9 Bob_new sixth 6 A 100 NA NA NA 15 18 18
# 12 9 Bob_new seventh 7 A 100 NA NA NA 20 11 23
# 13 10 Mark first 1 A 100 NA NA NA 20 18 19
# X2007
#1 NA
#2 19
#3 NA
#4 NA
#5 NA
#6 15
#7 18
#8 21
#9 13
#10 18
#11 12
#12 19
#13 16
答案 2 :(得分：1)
library(dplyr)
df$remove <- paste(df$itemcode, df$type)
df<-invisible(filter(df,
remove %in% intersect(remove[type==7],
remove[type==9])|!type %in% c(7,9) ))
#Remove the additional column after filter
df$remove <- NULL
答案 3 :(得分：0)
您可以执行以下操作：在两个方向上运行setdiff。 cl()函数并不是必需的，但我真的不喜欢一遍又一遍地写同一个表达式。
f <- function(x, y) setdiff(union(x, y), x)
cl <- function(var) substitute(df$itemcode[df$id == x], list(x = var))
现在，您可以在f()上致电c(id7, id9)，然后将其反转并获得c(id9, id7)结果。
do.call(f, x <- list(cl(7), cl(9)))
# [1] 2
do.call(f, rev(x))
# [1] 9