Question

这是我的data.frame的头部，在r中的df。任何行或列都没有模式。

    {
    "keyOne": "String",
    "keyTwo": 7,
    "keyThree": {
        "subKeyOne": "String",
        "url": "String"
    },
    "arrayKey": [
        {
            "arrayKeyOne":"String",
            "url": "String"
        },
        {
            "arrayKeyOne":"String",
            "url": "String"
        }
    ],
    "url":"String"
}

对于df $ Type ==＆＃34; B＆＃34;的每一行，我想用df $ Type ==＆＃34; A＆＃34;找到上一行和下一行，然后提取他们的＆＃ 34; V1＆＃34;和＆＃34; V2＆＃34;。

期望的输出，

         Type SIZE      V1   V2    
          A    1        5    7      
          B    1        NA   NA
          B    3        NA   NA
          B    4        NA   NA     
          A    8        2    4      
          A    6        6    50      
          A    12       2    8       
          B    8        NA   NA      
          A    9        51   63       
          A    11       93   70

非常感谢有人可以提供帮助，

Answer 1

例如，首先存储type为A的索引...例如，

dat <- data.frame(type = c("A", "B", "B", "B", "A", "A", "A", "B", "A", "A"),
                  size = c(1, 1, 3, 4, 8, 6, 12, 8, 9, 11),
                  v1 = c(5, NA, NA, NA, 2, 6, 2, NA, 51, 93),
                  v2 = c(7, NA, NA, NA, 4, 50, 8, NA, 63, 70))

dat$idx <- 1:nrow(dat)
a_idx <- which(dat$type == "A")
b_idx <- which(dat$type == "B")

然后，您可以轻松找到最后一个/ {下一个B >和< ... sapply，

new <- sapply(b_idx, function(x) {
  lag_idx <- tail(a_idx[a_idx < x], 1)
  lead_idx <- head(a_idx[a_idx > x], 1)

  return (t(c(dat$v1[lag_idx], dat$v2[lag_idx], 
            dat$v1[lead_idx], dat$v2[lead_idx])))
  }
)

new <- t(new)
new <- cbind(new, b_idx)
colnames(new) <- c("V1_Lag", "V2_Lag", "V1_Lead", "V2_Lead", "idx")
merge(dat, new, all = TRUE)

   idx type size v1 v2 V1_Lag V2_Lag V1_Lead V2_Lead
1    1    A    1  5  7     NA     NA      NA      NA
2    2    B    1 NA NA      5      7       2       4
3    3    B    3 NA NA      5      7       2       4
4    4    B    4 NA NA      5      7       2       4
5    5    A    8  2  4     NA     NA      NA      NA
6    6    A    6  6 50     NA     NA      NA      NA
7    7    A   12  2  8     NA     NA      NA      NA
8    8    B    8 NA NA      2      8      51      63
9    9    A    9 51 63     NA     NA      NA      NA
10  10    A   11 93 70     NA     NA      NA      NA

Answer 2

使用此数据

dat <- data.frame(
    type = c("A", "B", "B", "B", "A", "A", "A", "B", "A", "A"),
    size = c(1, 1, 3, 4, 8, 6, 12, 8, 9, 11),
    v1 = c(5, NA, NA, NA, 2, 6, 2, NA, 51, 93),
    v2 = c(7, NA, NA, NA, 4, 50, 8, NA, 63, 70),
    stringsAsFactors = FALSE
)

计算＆＃39;游程长度编码＆＃39; type列

r <- rle(dat$type)

与

> r
Run Length Encoding
  lengths: int [1:5] 1 3 3 1 2
  values : chr [1:5] "A" "B" "A" "B" "A"

（即，1A，然后是3B，3A，1B和2A）。滞后值的指数是

lag <- setdiff(
    cumsum(r$lengths)[r$values == "A"],
    nrow(dat)   # ignore "A" value at end of column
)

需要复制每个滞后值，新值v1lag填充为

value <- rep(dat$v1[lag], r$length[r$value == "B"])

类似的故事与主导价值

一起发挥作用

lead <- pmin(
    cumsum(r$lengths)[r$values == "B"] + 1L,
    nrow(dat)   # ignore "B" value at end of column
)
value <- rep(dat$v1[lead], r$length[r$value == "B"])

具体问题的实施是

mm <- function(df) {
    r <- rle(df$type)
    lag <- setdiff(cumsum(r$lengths)[r$values == "A"], nrow(df))
    lead <- pmin(cumsum(r$lengths)[r$values == "B"] + 1L, nrow(df))
    len <- r$length[r$value == "B"]

    idx <- df$type == "B"
    df$v1_lag[idx] <- rep(df$v1[lag], len)
    df$v2_lag[idx] <- rep(df$v2[lag], len)
    df$v1_lead[idx] <- rep(df$v1[lead], len)
    df$v2_lead[idx] <- rep(df$v2[lead], len)

    df
}

这比erocoar的解决方案更快，更强大。

在r中找到上一次和下一次观察

2 个答案: