【问题标题】:Select (multiple) integers with n occurrences per row选择(多个)整数,每行出现 n 次
【发布时间】:2014-04-21 17:56:35
【问题描述】:

我有一个data.frame,其中数据条目以1,2,3,10这种格式输入。也就是说,它们是逗号分隔的整数,范围为 0-20,不需要连续。目前,每一个都被认为是一个因素。我有四个包含这些值的变量,我想创建一个新变量,仅当它在四个变量中的三个中时才包含给定的整数,如果没有出现三个整数,则使用 0 .

M1    M2      M3      M4      M_NEW
1     1,2     0        1       1
3,4   3,4   1,2,3,4    4       3,4

我不确定如何处理这些逗号分隔的整数。如果它们是单个整数,我可以做类似this:

# modified from https://stackoverflow.com/a/14114085/1670053
# over each row of data.frame (or matrix)
sapply(1:nrow(df), function(idx) {
# get the number of time each entry in df occurs
    t <- table(t(g[idx, ]))
# get the maximum count (or frequency)
    if(max(t) > 2){
      t.max <- max(t)
      }else{ t.max <- 0
    }
# get all values that equate to maximum count
    t <- as.numeric(names(t[t == t.max]))

})

虽然这些多个值用逗号分隔,但我不确定从哪里开始。

# data and example output
df <- structure(list(M1 = structure(c(3L, 2L, 2L, 5L, 3L, 1L, 7L, 1L, 
8L, 1L, 3L, 4L, 3L, 6L), .Label = c("0", "1", "1,2", "1,2,3", 
"1,2,3,4", "1,2,3,4,5", "3,4,5,6,7", "6,7,8,9,10,11,12,13,14,15,16"
), class = "factor"), M2 = structure(c(5L, 2L, 2L, 4L, 4L, 1L, 
11L, 8L, 7L, 9L, 3L, 6L, 3L, 10L), .Label = c("0", "1,2", "1,2,3", 
"1,2,3,4", "1,2,3,4,5", "1,2,3,4,5,6,7", "1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16", 
"2", "2,3,4,5", "4,5,6", "4,5,6,7,8,9,10,11,12,13,14"), class = "factor"), 
    M3 = structure(c(4L, 1L, 1L, 8L, 3L, 1L, 6L, 1L, 7L, 3L, 
    2L, 5L, 9L, 3L), .Label = c("0", "1,2", "1,2,3,4", "1,2,3,4,5", 
    "1,2,3,4,5,6", "1,2,3,4,5,6,7,8", "1,2,3,4,5,6,7,8,9,10,11,12,13,14", 
    "3,4", "3,4,5"), class = "factor"), M4 = structure(c(5L, 
    1L, 2L, 8L, 2L, 1L, 6L, 3L, 4L, 1L, 3L, 3L, 7L, 9L), .Label = c("0", 
    "1", "1,2", "1,2,3,4,5,12,13,14,15,16,17", "1,2,3,4,5,6", 
    "1,2,3,4,5,6,7,8,9,10,11,12", "3,4", "4", "4,5"), class = "factor"), 
    M_NEW = structure(c(6L, 1L, 2L, 8L, 3L, 1L, 9L, 1L, 7L, 1L, 
    3L, 4L, 5L, 10L), .Label = c("0", "1", "1,2", "1,2,3", "1,2,3,", 
    "1,2,3,4,5", "1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16", "3,4", 
    "3,4,5,6,7,8", "4,5"), class = "factor")), .Names = c("M1", 
"M2", "M3", "M4", "M_NEW"), class = "data.frame", row.names = c(NA, 
-14L))

【问题讨论】:

    标签: r dataframe data.table plyr


    【解决方案1】:
    f <- function(x, n=3) {
      tab <- table(strsplit(paste(x, collapse=","), ","))
      res <- paste(names(tab[which(tab >= n)]), collapse=",")
      return(ifelse(res == "", "0", res))
    }
    (df[, 5] <- apply(df[, 1:4], 1, f))
    # [1] "1,2,3,4,5"                             
    # [2] "0"                                     
    # [3] "1"                                     
    # [4] "3,4"                                   
    # [5] "1,2"                                   
    # [6] "0"                                     
    # [7] "3,4,5,6,7,8"                           
    # [8] "0"                                     
    # [9] "1,10,11,12,13,14,15,16,2,3,4,5,6,7,8,9"
    # [10] "0"                                     
    # [11] "1,2"                                   
    # [12] "1,2,3"                                 
    # [13] "3"                                     
    # [14] "4,5"  
    

    【讨论】:

    • 代码有效,谢谢。我有一个关于resres = [1] " \"0\", \"1, \"3, 1, 2, 3, 4, 6, 7, 8,2,2\",3,4,5,6,7" 的问题,什么是非整数值?例如\"0"\,res?
    • / 转义特殊字符,例如 "。我的第一个猜测是您的真实数据中的某些列不仅包含数字,还包含引号(例如 "0",1,2,"3,4 而不是 0,1,2,3,4)。
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