主页 / coding / 问题 / 77108042
Accepted
JVGen
Asked: 2023-09-15 04:03:08 +0800 CST

R：调整范围从0开始，相应调整数据帧值

  • 772

我有一个包含两列数值的数据框。我想调整值的范围，同时保持数值之间的相对距离。这个例子更好地解释了这一点。

是否有一个标准函数可以完成此任务？

df <- data.frame(
  Name = c("A", "B", "C", "D", "E"),
  Start = c(175000, 183597, 185317, 188307, 191390),
  Stop = c(182548, 184724, 187841, 188599, 192650)
)

Range_Start <- df$Start[df$Name=="C"] - 10000
Range_Stop <- df$Stop[df$Name=="C"] + 10000

Range_Start2 <- 1
Range_Stop2 <- length(Range_Start:Range_Stop)

df2 <- df %>%
  rowwise() %>%
  dplyr::mutate(Start2 = Start - Range_Start) %>%
  dplyr::mutate(Stop2 = Stop - Range_Start)

期望值：

> df2$Start2
[1]  -317  8280 10000 12990 16073

> df2$Stop2
[1]  7231  9407 12524 13282 17333

2 个回答

  1. Best Answer

    无需前往rowwise：

    df <- structure(list(Name = c("A", "B", "C", "D", "E"), Start = c(181391, 183597, 185317, 188307, 191390), Stop = c(182548, 184724, 187841, 188599, 192650)), class = "data.frame", row.names = c(NA, -5L))
df2 <- structure(list(Name = c("A", "B", "C", "D", "E"), Start = c(181391, 183597, 185317, 188307, 191390), Stop = c(182548, 184724, 187841, 188599, 192650), Start2 = c(6074, 8280, 10000, 12990, 16073), Stop2 = c(7231, 9407, 12524, 13282, 17333)), class = c("rowwise_df", "tbl_df", "tbl", "data.frame"), row.names = c(NA, -5L), groups = structure(list(.rows = structure(list(1L, 2L, 3L, 4L, 5L), ptype = integer(0), class = c("vctrs_list_of", "vctrs_vctr", "list"))), row.names = c(NA, -5L), class = c("tbl_df",  "tbl", "data.frame")))

out <- df %>%
  mutate(
    Start2 = Start - Start[Name == "C"] + 10000,
    Stop2 = Start2 + (Stop - Start)
  )
out
#   Name  Start   Stop Start2 Stop2
# 1    A 181391 182548   6074  7231
# 2    B 183597 184724   8280  9407
# 3    C 185317 187841  10000 12524
# 4    D 188307 188599  12990 13282
# 5    E 191390 192650  16073 17333

all.equal(out, df2, check.attributes = FALSE)
# [1] TRUE

    使用更新的数据重新运行：

    df <- structure(list(Name = c("A", "B", "C", "D", "E"), Start = c(175000, 183597, 185317, 188307, 191390), Stop = c(182548, 184724, 187841, 188599, 192650)), class = "data.frame", row.names = c(NA, -5L))
df2 <- structure(list(Name = c("A", "B", "C", "D", "E"), Start = c(175000, 183597, 185317, 188307, 191390), Stop = c(182548, 184724, 187841, 188599, 192650), Start2 = c(-317, 8280, 10000, 12990, 16073), Stop2 = c(7231, 9407, 12524, 13282, 17333)), class = c("rowwise_df", "tbl_df", "tbl", "data.frame"), row.names = c(NA, -5L), groups = structure(list(.rows = structure(list(1L, 2L, 3L, 4L, 5L), ptype = integer(0), class = c("vctrs_list_of", "vctrs_vctr", "list"))), row.names = c(NA, -5L), class = c("tbl_df",  "tbl", "data.frame")))

out <- df %>%
  mutate(
    Start2 = Start - Start[Name == "C"] + 10000,
    Stop2 = Start2 + (Stop - Start)
  )
out
#   Name  Start   Stop Start2 Stop2
# 1    A 175000 182548   -317  7231
# 2    B 183597 184724   8280  9407
# 3    C 185317 187841  10000 12524
# 4    D 188307 188599  12990 13282
# 5    E 191390 192650  16073 17333
all.equal(out, df2, check.attributes = FALSE)
# [1] TRUE
    • 2

  2. 要对多个列执行相同的操作（在本例中减去固定数量），我们可以使用acrossinsidemutate

    mutate(df, across(2:3, ~ .x - (Start[Name == "C"] - 1e4), .names = "{.col}2"))
#   Name  Start   Stop Start2 Stop2
# 1    A 181391 182548   6074  7231
# 2    B 183597 184724   8280  9407
# 3    C 185317 187841  10000 12524
# 4    D 188307 188599  12990 13282
# 5    E 191390 192650  16073 17333
    • 0

相关问题