我想识别并标记基于2列的重复行。我想为每个重复项创建唯一的标识符，因此我不仅知道该行是重复项，而且还知道它与哪一行是重复项。我有一个如下所示的数据框，其中包含一些重复的项对（适合和坐着）和其他不重复的对。当项目对重复时，它们包含的信息是唯一的（例如，一行将在Value1中保留1行的值，但不包含Value2和Value 3，第二行或“重复”行仅具有Value2和Value3的数字）不是Value1）

当前数据框

     value1 value2 value3 fit   sit  
[1,] "1"    NA     NA     "it1" "it2"
[2,] NA     "3"    "2"    "it2" "it1"
[3,] "2"    "3"    "4"    "it3" "it4"
[4,] NA     NA     NA     "it4" "it3"
[5,] "5"    NA     NA     "it5" "it6"
[6,] NA     NA     "2"    "it6" "it5"
[7,] NA     "4"    NA     "it7" "it9"

代码生成示例数据框

value1<-c(1,NA,2,NA,5,NA,NA)
value2<-c(NA,3,3,NA,NA,NA, 4)
value3<-c(NA,2,4,NA,NA,2, NA)
fit<-c("it1","it2","it3","it4", "it5", "it6","it7")
sit<-c("it2","it1","it4","it3", "it6", "it5", "it9")
df.now<-cbind(value1,value2,value3, fit, sit)

我想要的是将其转换为如下所示的数据框：

所需的数据框

     val1 val2 val3 it1   it2  
[1,] "1"  "3"  "2"  "it1" "it2"
[2,] "2"  "3"  "4"  "it3" "it4"
[3,] "5"  NA   "2"  "it5" "it6"
[4,] NA   "4"  NA   "it7" "it9"

我正在考虑执行以下步骤：1.使用fit创建新变量，并与最低项目和最高项目坐在一起以识别重复的对2.识别重复的项目对3.使用ifelse选择并填写唯一信息。

我知道如何执行第1步和第3步，但仍停留在第2步。我想我需要做的不仅是识别TRUE / FALSE重复项，还需要为每个项目对创建一个具有唯一标识符的列，如下所示：由于我的步骤1）多了2行：

     value1 value2 value3 fit   sit   lit   hit    dup
[1,] "1"    NA     NA     "it1" "it2" "it1" "it2"   1
[2,] NA     "3"    "2"    "it2" "it1" "it1" "it2"   1
[3,] "2"    "3"    "4"    "it3" "it4" "it3" "it4"   2
[4,] NA     NA     NA     "it4" "it3" "it3" "it4"   2
[5,] "5"    NA     NA     "it5" "it6" "it5" "it6"   3
[6,] NA     NA     "2"    "it6" "it5" "it5" "it6"   3
[7,] NA     "4"    NA     "it7" "it9" "it7" "it9"   NA

我不确定该怎么做。

我要的是对步骤2有所帮助，或者有比我概述的步骤更好的解决方法。

r dataframe

— 希瑟·克拉克
source

6

一种dplyr选择是：

df.now %>%
 group_by(pair = paste(pmax(fit, sit), pmin(fit, sit), sep = "_")) %>%
 summarise_at(vars(starts_with("value")), ~ ifelse(all(is.na(.)), 
                                                   NA,
                                                   first(na.omit(.))))

  pair    value1 value2 value3
  <chr>    <dbl>  <dbl>  <dbl>
1 it2_it1      1      3      2
2 it4_it3      2      3      4
3 it6_it5      5     NA      2
4 it9_it7     NA      4     NA

而且，如果您还需要单独的列中的对，则tidyr可以添加：

df.now %>%
 group_by(pair = paste(pmax(fit, sit), pmin(fit, sit), sep = "_")) %>%
 summarise_at(vars(starts_with("value")), ~ ifelse(all(is.na(.)), 
                                                   NA,
                                                   first(na.omit(.)))) %>%
 separate(pair, into = c("fit", "hit"), sep = "_", remove = FALSE)

  pair    fit   hit   value1 value2 value3
  <chr>   <chr> <chr>  <dbl>  <dbl>  <dbl>
1 it2_it1 it2   it1        1      3      2
2 it4_it3 it4   it3        2      3      4
3 it6_it5 it6   it5        5     NA      2
4 it9_it7 it9   it7       NA      4     NA

— tmfmnk
source

谢谢！这很好。我很高兴添加选项来分隔项目。

— 希瑟·克拉克

3

ing !duplicated()后使用sort。

df.now[!duplicated(t(apply(df.now[, c("fit", "sit")], 1, sort))), ]
#       value1 value2 value3 fit   sit  
# [1,] "1"    NA     NA     "it1" "it2"
# [2,] "2"    "3"    "4"    "it3" "it4"
# [3,] "5"    NA     NA     "it5" "it6"
# [4,] NA     "4"    NA     "it7" "it9"

— 杰伊
source

谢谢你的快速反应。但是，此解决方案删除了我需要保留的信息。我想合并来自在同一项目对的2行中找到的3个值列中的信息。让我知道是否不清楚

— Heather Clark

2

melt/dcast从使用data.table

library(data.table)
dcast(melt(setDT(df.now)[, c('fit1', 'sit1') := .(pmin(fit, sit), 
    pmax(fit, sit))], measure = patterns("^value"), na.rm = TRUE),
     fit1 + sit1 ~ variable, value.var = 'value')
#   fit1 sit1 value1 value2 value3
#1:  it1  it2      1      3      2
#2:  it3  it4      2      3      4
#3:  it5  it6      5     NA      2
#4:  it7  it9     NA      4     NA

数据

df.now <- data.frame(value1,value2,value3, fit, sit, stringsAsFactors = FALSE)

— 阿克伦
source

2

另一种data.table选择：

library(data.table)
as.data.table(df.now)[, lapply(.SD, function(x) first(x[!is.na(x)])), 
    .(it1=pmin(fit, sit), it2=pmax(fit, sit)), 
    .SDcols=value1:value3]

输出：

   it1 it2 value1 value2 value3
1: it1 it2      1      3      2
2: it3 it4      2      3      4
3: it5 it6      5   <NA>      2
4: it7 it9   <NA>      4   <NA>

— chinsoon12
source

1

这是我使用data.table的尝试。您的数据称为mydf。首先，我对每一行进行排序fit并sit创建了一个新变量group。然后，对于每个组，我在三个值列（即value1，value2和value3）中对值进行了排序。最后，我为每个组提取了第一行。

library(data.table)

mydt <- setDT(mydf)[, group := paste(sort(.SD), collapse = "_"),
                    .SD = c("fit", "sit"), by = 1:nrow(mydf)][,
                        c("value1", "value2", "value3") := lapply(.SD, sort),
                        .SDcols = value1:value3, by = group][, .SD[1], by = group]

mydt[]

#     group value1 value2 value3 fit sit
#1: it1_it2      1      3      2 it1 it2
#2: it3_it4      2      3      4 it3 it4
#3: it5_it6      5     NA      2 it5 it6
#4: it7_it9     NA      4     NA it7 it9

数据

mydf <- structure(list(value1 = c(1L, NA, 2L, NA, 5L, NA, NA), value2 = c(NA, 
3L, 3L, NA, NA, NA, 4L), value3 = c(NA, 2L, 4L, NA, NA, 2L, NA
), fit = c("it1", "it2", "it3", "it4", "it5", "it6", "it7"), 
sit = c("it2", "it1", "it4", "it3", "it6", "it5", "it9")), class = "data.frame", row.names = c(NA, 
-7L))

— 爵士乐
source

1

也可以使用与结合使用tidyr的来完成此操作：pivot_longervalues_drop_na = TRUEpivot_wider

library(tidyverse)

mydf %>%
   mutate(it1 = pmin(fit, sit), it2 = pmax(fit, sit)) %>%
   pivot_longer(cols = starts_with("value"), values_drop_na = TRUE) %>%
   pivot_wider(id_cols = c("it1", "it2"))

#> # A tibble: 4 x 5
#>   it1   it2   value1 value2 value3
#>   <chr> <chr>  <int>  <int>  <int>
#> 1 it1   it2        1      3      2
#> 2 it3   it4        2      3      4
#> 3 it5   it6        5     NA      2
#> 4 it7   it9       NA      4     NA

数据

mydf <- structure(list(value1 = c(1L, NA, 2L, NA, 5L, NA, NA), value2 = c(NA, 
3L, 3L, NA, NA, NA, 4L), value3 = c(NA, 2L, 4L, NA, NA, 2L, NA
), fit = c("it1", "it2", "it3", "it4", "it5", "it6", "it7"), 
sit = c("it2", "it1", "it4", "it3", "it6", "it5", "it9")), class = "data.frame", row.names = c(NA, 
-7L))

— 周
source

识别并标记r中的重复行

当前数据框

所需的数据框

数据