如何删除唯一条目并在R中保留重复项

ID     Cat1  Cat2    Cat3   Cat4
A0001   358 11.25   37428   0
A0001   279 14.6875 38605   0
A0013   367 5.125   40152   1
A0014   337 16.3125 38624   0
A0020   367 8.875   37797   0
A0020   339 9.625   39324   0

I need help learning to how remove the unique rows in my file while keeping the duplicates or triplicates. For example, output should look like below:

我需要帮助学习如何删除文件中的唯一行,同时保持重复或重复。例如,输出应如下所示:

ID     Cat1  Cat2    Cat3   Cat4
A0001   358 11.25   37428   0
A0001   279 14.6875 38605   0
A0020   367 8.875   37797   0
A0020   339 9.625   39324   0

If you can give me advice how to approach this problem, much appreciated.

如果你能给我建议如何解决这个问题,非常感谢。

Thanks for everyone's suggestions. I wanted to calculate the difference in value in the different Categories (i.e. Cat2, Cat 3) between the repeated measures (by unique ID). Would appreciate any suggestions.

谢谢大家的建议。我想计算重复测量之间的不同类别(即Cat2,Cat 3)的值差异(通过唯一ID)。将不胜感激任何建议。

2 个解决方案

#1

Another option in base R Using duplicated

基础R中的另一个选项使用重复

dx[dx$ID %in% dx$ID[duplicated(dx$ID)],]

#      ID Cat1    Cat2  Cat3 Cat4
# 1 A0001  358 11.2500 37428    0
# 2 A0001  279 14.6875 38605    0
# 5 A0020  367  8.8750 37797    0
# 6 A0020  339  9.6250 39324    0

data.table using duplicated

using duplicated and fromLast version you get :

使用duplicated和fromLast版本,您将获得:

library(data.table)
setkey(setDT(dx),ID) # or with data.table 1.9.5+: setDT(dx,key="ID")
dx[duplicated(dx) |duplicated(dx,fromLast=T)]

#       ID Cat1    Cat2  Cat3 Cat4
# 1: A0001  358 11.2500 37428    0
# 2: A0001  279 14.6875 38605    0
# 3: A0020  367  8.8750 37797    0
# 4: A0020  339  9.6250 39324    0

This can be applied to base R also but I prefer data.table here for syntax sugar.

这也可以应用于基数R但我更喜欢data.table这里的语法糖。

#2

General comments.

• The ave approach is the only one here that preserves the data's initial row ordering.

这里唯一的方法是保留数据的初始行排序。

• The by approach should be very slow. I suspect that data.table and dplyr are not much faster than ave and tapply (yet) at selecting groups. Benchmarks to prove me wrong welcome!

by方法应该非常慢。我怀疑data.table和dplyr在选择组时并不比ave和tapply(还)快。基准来证明我的错误欢迎!

---

base R (Thanks to @thelatemail for both of the first two approaches.)

base R(感谢前两种方法的@thelatemail。)

1) Each row is assigned the length of its df$ID group, and we filter based on the vector of lengths.

1)为每行分配其df $ ID组的长度,并根据长度向量进行过滤。

df[ ave(1:nrow(df), df$ID, FUN=length) > 1 , ]

2) Alternately, we split row names or numbers by df$ID, selecting which groups' rows to keep. tapply returns a list of groups of rows, so we must unlist them into a single vector of rows.

2)或者,我们用df $ ID分割行名或数字,选择要保留的组的行。 tapply返回一组行列表,因此我们必须将它们取消列为单个行向量。

df[ unlist(tapply(1:nrow(df), df$ID, function(x) if (length(x) > 1) x)) , ]

What follows is a worse approach, but better parallels what you see with data.table and dplyr:

接下来是一种更糟糕的方法,但与data.table和dplyr所看到的更好的相似之处:

3) The data is split by df$ID, keeping each subset of data, SD if if has more than one row. by returns a list, so we must rbind them back together.

3)数据按df $ ID分割,保留每个数据子集,如果有多行,则保留SD。通过返回一个列表,所以我们必须将它们重新组合在一起。

do.call( rbind, c(list(make.row.names = FALSE),
    by(df, df$ID, FUN=function(SD) if (nrow(SD) > 1) SD )))

---

data.table .N corresponds to nrow within a by=ID group; and .SD is the subset of data.

data.table .N对应于by = ID组中的nrow;和.SD是数据的子集。

library(data.table)
setDT(df)[, if (.N>1) .SD, by=ID]

#       ID Cat1    Cat2  Cat3 Cat4
# 1: A0001  358 11.2500 37428    0
# 2: A0001  279 14.6875 38605    0
# 3: A0020  367  8.8750 37797    0
# 4: A0020  339  9.6250 39324    0

---

dplyr n() corresponds to nrow within a group_by(ID) group.

dplyr n()对应于group_by(ID)组中的nrow。

library(dplyr)
df %>% group_by(ID) %>% filter( n() > 1 )

# Source: local data frame [4 x 5]
# Groups: ID
# 
#      ID Cat1    Cat2  Cat3 Cat4
# 1 A0001  358 11.2500 37428    0
# 2 A0001  279 14.6875 38605    0
# 3 A0020  367  8.8750 37797    0
# 4 A0020  339  9.6250 39324    0