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删除未在每个时间段和数据框中出现的组

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我正在使用多个数据帧清理每个数据帧中的多个时间段的数据 . 每个数据框都有一年的数据 . 我想删除未在每个时间段（数据框内）中出现的组，并删除未出现在每个数据框中的组 . 换句话说，我希望在每个数据帧中保留每个时间段中存在的组 . 我用ID，时间变量和代表我数据的两个变量创建了数据 . 我的数据也将包含更多数据框，ID，组和变量 .

t<-c(1,1,2,2,3,3,3,4,4,4)
id<-c(200,300,200,300,100,200,300,200,300,400)
x1<-rnorm(1:10)
x2<-rnorm(1:10)
?df
df<-data.frame(id,t,x1,x2)
t<-c(1,1,1,2,2,3,3,3,4,4)
id<-c(200,300,400,200,300,200,300,400,200,300)
x1<-rnorm(1:10)
x2<-rnorm(1:10)
df2<-data.frame(id,t,x1,x2)
id<-c(200,300,200,300,600,200,300,100,200,300)
t<-c(1,1,2,2,2,3,3,4,4,4)
x1<-rnorm(1:10)
x2<-rnorm(1:10)
df3<-data.frame(id,t,x1,x2)
rb<-rbind(df,df2,df3)
rb
cb<-cbind(df,df2,df3)
cb
    id t          x1          x2  id t            x1         x2  id t          x1            x2
1  200 1  0.37223136 -0.04918183 200 1  0.6489171399 -0.1324335 200 1 -0.41387676 -0.4566678425
2  300 1 -0.22062416  0.05150952 300 1 -0.3669090613  3.0826144 300 1  0.48237987 -0.0325861333
3  200 2  0.32912208  1.03922999 400 1  0.9347859735  0.1026632 200 2 -0.31308242 -0.3021501845
4  300 2 -0.18172302 -1.41669927 200 2  0.4814364147 -0.1087465 300 2 -1.52273626  0.6357750776
5  100 3 -0.81072008  0.64522238 300 2 -0.5676866296  0.2371230 600 2 -0.09687669  2.2883585934
6  200 3  0.45175343  0.64197622 200 3  0.0006852893  0.5830704 200 3  0.01726120 -0.5905109745
7  300 3  0.40465989 -0.70796588 300 3 -0.0008717189 -1.1969493 300 3 -0.18603781  0.3722390396
8  200 4  0.09852108 -1.76958443 400 3  0.9343534507 -1.3671447 100 4 -0.57308316  0.4749221706
9  300 4 -0.53951022  0.97306346 200 4  1.9176422485  0.9879788 200 4  0.40222133  0.3278821640
10 400 4  0.24271562 -1.37269617 300 4  1.4298971045  1.6095265 300 4  0.85799186  0.0006593401

我的最终出局看起来像这样：

id  t          x1          x2
200 1  0.37223136 -0.04918183 
300 1 -0.22062416  0.05150952
200 2  0.32912208  1.03922999
300 2 -0.18172302 -1.41669927
200 3  0.45175343  0.64197622
300 3  0.40465989 -0.70796588
200 4  0.09852108 -1.76958443
300 4 -0.53951022  0.97306346
200 1  0.6489171399 -0.1324335
300 1 -0.3669090613  3.0826144
200 2  0.4814364147 -0.1087465
300 2 -0.5676866296  0.2371230
200 3  0.0006852893  0.5830704
300 3 -0.0008717189 -1.1969493
200 4  1.9176422485  0.9879788
300 4  1.4298971045  1.6095265
200 1 -0.41387676 -0.4566678425
300 1  0.48237987 -0.0325861333
200 2 -0.31308242 -0.3021501845
300 2 -1.52273626  0.6357750776
200 3  0.01726120 -0.5905109745
300 3 -0.18603781  0.3722390396
200 4  0.40222133  0.3278821640
300 4  0.85799186  0.0006593401
r dataframe time rows

2 回答

  • 0

    一种策略是计算 id 和 t 的每个组合出现的次数 . 如果这等于可能的最大值，则保留该id . （我使用 max 来获得最大可能的组合，但只有在每个 t 中出现至少一个 id 时才有效 .

    我在这里用 plyr 包中的 adply 来替换你的 rbind 步骤，因为 adply 保留了每行来自哪个数据帧的信息（在 X1 列中） .

    library(plyr)
rb <- adply(list(df, df2, df3), 1)

unique_combo <- unique(rb[, c("X1", "id", "t")])
##    X1  id t
## 1   1 200 1
## 2   1 300 1
## 3   1 200 2
## 4   1 300 2
## 5   1 100 3
## 6   1 200 3
## 7   1 300 3
## 8   1 200 4
## 9   1 300 4
## 10  1 400 4
## 11  2 200 1
## 12  2 300 1 etc.

combos_per_id <- aggregate(t ~ id, FUN = length, data = unique_combo)
##    id  t
## 1 100  2
## 2 200 12
## 3 300 12
## 4 400  3
## 5 600  1

ids_you_want <- subset(combos_per_id, t == max(t))
##    id  t
## 2 200 12
## 3 300 12

rb[rb$id %in% ids_you_want$id, ]
##    X1  id t          x1           x2
## 1   1 200 1  0.41800060 -0.729280896
## 2   1 300 1 -1.26310444  0.649438361
## 3   1 200 2  1.75130801  0.340464369
## 4   1 300 2 -0.47751518 -1.396611139
## 6   1 200 3 -0.11537438 -1.483654622
## 7   1 300 3 -1.33689508 -1.219725112 etc.

    Edit 处理另一列

    library(plyr)
t<-c(1,1,2,2,3,3,3,4,4,4)
id<-c(200,300,200,300,100,200,300,200,300,400)
x1<-rnorm(1:10)
x2<-rnorm(1:10)
r<-c("b","a","a","a","a","a","a","a","a","a")
df<-data.frame(id,t,x1,x2, r)

t<-c(1,1,1,2,2,3,3,3,4,4)
id<-c(200,300,400,200,300,200,300,400,200,300)
x1<-rnorm(1:10)
x2<-rnorm(1:10)
r<-c("a","a","a","a","a","a","a","a","a","a")
df2<-data.frame(id,t,x1,x2, r)

id<-c(200,300,200,300,600,200,300,100,200,300)
t<-c(1,1,2,2,2,3,3,4,4,4)
x1<-rnorm(1:10)
x2<-rnorm(1:10)
r<-c("a","a","a","a","a","a","a","a","a","a")
df3<-data.frame(id,t,x1,x2, r)

rb <- adply(list(df, df2, df3), 1)
unique_combo <- unique(rb[, c("X1", "id", "t", "r")])
(combos_per_id <- aggregate(t ~ id + r, FUN = length, data = unique_combo))
ids_you_want <- subset(combos_per_id, t == max(t))
rb[rb$id %in% ids_you_want$id, ]
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  • 1

    这有点暴力，但应该有效：

    frames <- list(df,df2,df3)

lu <- function(x)
    length(unique(x))

timePeriodsPerDataframe <- sapply(frames,function(x)lu(x))

for(i in seq(length(frames))){
    appearsInEveryTimePeriod <- tapply(frames[[i]]$id,
                                       frames[[i]]$t,
                                       lu) == timePeriodsPerDataframe[i]
    if(i == 1)
        IDsInEveryTimePeriod <- names(tmp[tmp])
    else
        IDsInEveryTimePeriod <- intersect(names(tmp[tmp]),IDsInEveryTimePeriod)
}
IDsInEveryTimePeriod <- as.numeric(IDsInEveryTimePeriod)
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