复杂的长到宽整形算法

Complex long to wide reshape algorithm

我遇到一个问题,我需要根据 ID1 和 ID2 将长格式数据 table 重新整形为具有非重叠条目的宽格式。逻辑相当复杂,取决于 3 列(“Seq”、“ID1”和“ID2”)。

属于 ID1 的

Value_1 如果 'overlaps' 与 ID2 相加,反之亦然,但仅适用于不同的 ID。

请参阅下面的输入示例和输出,希望能说明问题。

输入:

df <- structure(list(Seq = c(9143L, 916L, 9293L, 9301L, 9302L, 9304L, 
9305L, 9306L, 9307L, 931L, 9311L), ID1 = c("ID1_1", "ID1_1", 
NA, "ID1_2", "ID1_2", NA, "ID1_3", "ID1_3", "ID1_3", "ID1_4", 
"ID1_4"), value_1 = c(30L, 30L, NA, 30L, 30L, NA, 30L, 30L, 30L, 
50L, 50L), ID2 = c(NA, NA, "ID2_1", "ID2_2", "ID2_3", "ID2_4", 
"ID2_4", "ID2_4", "ID2_4", "ID2_4", "ID2_5"), value_2 = c(NA, 
NA, 33L, 200L, 46L, 58L, 58L, 58L, 58L, 58L, 46L)), class = "data.frame", row.names = c(NA, 
-11L))

输出:

(例如,注意最后一行,value_1 = 80,因为 30+50 属于 ID1_3 和 ID1_4 的值之和)

我使用了 data.table 包中的 rleid() 函数,这是一个计算 运行 长度编码的有趣函数。这样做

library(data.table)
library(dplyr)
df %>% 
  mutate(d = cumsum( c(0, diff(rleid(ID1))) != 0 & c(0, diff(rleid(ID2))) != 0),
         value_1 = value_1 * c(1, diff(rleid(ID1))),
         value_2 = value_2 * c(1, diff(rleid(ID2)))) %>% group_by(d) %>%
  summarise(Seq = toString(Seq),
            value_1 = sum(value_1, na.rm = T),
            value_2 = sum(value_2, na.rm = T)) %>%
  ungroup() %>% select(-d)

# A tibble: 4 x 3
  Seq                               value_1 value_2
  <chr>                               <int>   <int>
1 9143, 916                              30       0
2 9293                                    0      33
3 9301, 9302                             30     246
4 9304, 9305, 9306, 9307, 931, 9311      80     104

旧答案

df %>% group_by(d = cumsum( c(0, diff(rleid(ID1))) != 0 & c(0, diff(rleid(ID2))) != 0)) %>%
  summarise(Seq = toString(Seq),
            value_1 = sum(unique(value_1), na.rm = T),
            value_2 = sum(unique(value_2), na.rm = T)) %>%
  ungroup() %>% select(-d)

首先,我非常喜欢 AnilGoyal 的解决方案。我可以看到我需要开始使用 data.table 包。

话虽如此,我正在研究一种没有 data.table 的 dplyr 方法,这显然更加冗长。此外,我花了一段时间才弄清楚如何处理重复值。乘以 changei 列(0 或 1)删除了重复项。以下是我的方法:

df %>% 
  mutate_if(is.numeric, replace_na, 0) %>% 
  mutate_if(is.character, replace_na, "NA") %>% 
  mutate(
    change1 = ID1 != lag(ID1, default = "Start"),
    value_1 = value_1 * change1,

    change2 = ID2 != lag(ID2, default = "Start"),
    value_2 = value_2 * change2,

    change = cumsum(change1 & change2)
  ) %>% 
  group_by(change) %>% 
  summarise(
    Seq = toString(Seq),
    value_1 = sum(value_1, na.rm = T),
    value_2 = sum(value_2, na.rm = T)
  ) %>% 
  ungroup()

结果是:

df
#   change Seq                               value_1 value_2
#    <int> <chr>                               <dbl>   <dbl>
# 1      1 9143, 916                              30       0
# 2      2 9293                                    0      33
# 3      3 9301, 9302                             30     246
# 4      4 9304, 9305, 9306, 9307, 931, 9311      80     104

没有上面那么简洁,而是一个Base R的解决方案none the less:

# Function to calculate the aggregate value: .agg_func => function() 
.agg_func <- function(df, id_col, value_col){
  sbst <- subset(
    df, 
    !(is.na(df[,id_col])) & !(duplicated(df[,id_col])),
    select = c(id_col, value_col)
  )
  return(sum(sbst[,value_col], na.rm = TRUE))
}

# Function to group data by ids: .grouping_func => function() 
.grouping_func <- function(df, id_col){
  r_l_e <- rle(df[,id_col])
  rle_id <- rep(seq_along(r_l_e$values), times = r_l_e$lengths)
  return(c(0, diff(rle_id)) != 0)
}

# Group the data: grpd_df => data.frame 
grpd_df <- transform(
  df, 
  grp = cumsum(
    apply(
      vapply(
        names(df)[startsWith(names(df), "ID")],
        function(x).grouping_func(df, x),
        logical(nrow(df))
        ), 
      1,
      all
    )
  )
)  

# Split-apply-combine the aggregate function to the grouped data: 
data.frame(do.call(rbind, lapply(with(grpd_df, split(grpd_df, grp)), function(s){
        data.frame(
          Seq = toString(s$Seq), 
          value_1 = .agg_func(s, "ID1", "value_1"), 
          value_2 = .agg_func(s, "ID2", "value_2")
        )
      }
    )
  ), row.names = NULL, stringsAsFactors = FALSE
)