Question

我想从df的column1中的每个值中选择5行，这样输出对于column2中的每个唯一值至少有1个值。此外，输出中不应有任何重复

编辑：（column1，column3）对中不应该有重复项：即对于column1中的每个值，column3中的所有值都应该是唯一的

column1 = rep(c("a","b"), each = 12)
column2 = rep(c(1,2,3), each = 4)
column3 = c("x1","x2","x3","x4","x5","x3","x6","x7","x8","x1","x9","x5","x6","x2","x3","x4","x7","x5","x6","x1","x4","x1","x6","x9")

df = data.frame(column1, column2, column3)

这是一个有效的解决方案

sample_output_1 = data.frame(column1 = rep(c("a","b"), each = 5),
                         column2 = c(1,1,2,2,3,1,1,2,2,3),
                         column3 = c("x1","x2","x5","x3","x8","x6","x2","x5","x1","x9"))

Answer 1

检查

foo = function(a_df){
    inds = 1:NROW(a_df)     
    #Sample 5 indices along the rows of a_df
    my_inds = sample(inds, 5)       
    #If subset of a_df based on my_inds has duplicates
    #Or if 2nd column does not have all unique values
    while(any(duplicated(a_df[my_inds, c(1, 3)])) & 
        !identical(sort(unique(a_df[my_inds, 2])), sort(unique(a_df[[2]])))){
            #Count the number of duplicates or missing all values
            n = sum(duplicated(a_df[my_inds, c(1, 3)]))
            n = n + sum(!sort(unique(a_df[my_inds, 2])) %in% sort(unique(a_df[[2]])))
            #Remove my_inds from inds
            inds = inds[!inds %in% my_inds]
            #Remove the n indices that create duplicates from my_nds
            my_inds = my_inds[!duplicated(a_df[my_inds, c(1, 3)])]
            #Sample n more from inds and add to my_inds
            my_inds = sample(c(my_inds, sample(inds, n)))
        }
        return(a_df[my_inds,])
}

set.seed(42)
do.call(rbind, lapply(split(df, df$column1), function(a) foo(a_df = a)))
     # column1 column2 column3
# a.11       a       3      x9
# a.12       a       3      x5
# a.3        a       1      x3
# a.8        a       2      x7
# a.6        a       2      x3
# b.19       b       2      x6
# b.21       b       3      x4
# b.14       b       1      x2
# b.18       b       2      x5
# b.23       b       3      x6

固定长度的随机子集，使得每个组至少存在N次

1 个答案: