如何延长轴心并将变量名与辅助 table 匹配？

Question

我正在为跨越这两个表的逻辑而苦苦挣扎：

我的数据框是：

df6 每列包含一个 id 和一个数值。 v1 v2 和 v3 将始终只有一个非零值，表示该变量的 yummyness 值。

id v1 v2 v3
1  0  0  3
2  2  0  0
3  0  1  0

df_yummy 包含 v1 v2 和 v3 的等价名称以及“美味”值。

VarId   VarName  Yummyness
v1      Apple    2
v2      Lemon    1
v3      Peach    3

我想要的结果应该是

id  Fruit  Yummyness
1   Peach  3
2   Apple  2
3   Lemon  1

编辑：错别字更正

Answer 1

这是 tidyverse 中针对您的问题的一种解决方案（确保 v1、v2 和 v3 每行仅显示一个不同于零的值非常重要，否则您将得到重复或丢失数据）：

#dummy data
df1 <- data.table::fread("id v1 v2 v3
1  0  0  3
2  2  0  0
3  0  1  0")

df2 <- data.table::fread("VarId   VarName  Yummyness
v1      Apple    2
v2      Lemon    1
v3      Peach    3")

library(tidyverse)

df1 %>%
    tidyr::pivot_longer(-id, names_to = "VarId", values_to = "vals") %>%
    dplyr::filter(vals != 0) %>%
    dplyr::right_join(df2, by = "VarId") %>%
    dplyr::select(id, Fruit = VarName,  Yummyness)

# A tibble: 3 x 3
     id Fruit Yummyness
  <int> <chr>     <int>
1     1 Peach         3
2     2 Apple         2
3     3 Lemon         1

Answer 2

不确定这是否是您要查找的内容：

library(dplyr)

df6 %>% 
  mutate(Yummyness = rowSums(across(v1:v3)), .keep = "unused") %>% 
  left_join(df_yummy, by = "Yummyness") %>% 
  select(id, Fruit = VarName, Yummyness)

returns

# A tibble: 3 x 3
     id Fruit Yummyness
  <dbl> <chr>     <dbl>
1     1 Peach         3
2     2 Apple         2
3     3 Lemon         1

数据

df6 <- structure(list(id = c(1, 2, 3), v1 = c(0, 2, 0), v2 = c(0, 0, 
1), v3 = c(3, 0, 0)), class = c("spec_tbl_df", "tbl_df", "tbl", 
"data.frame"), row.names = c(NA, -3L), spec = structure(list(
    cols = list(id = structure(list(), class = c("collector_double", 
    "collector")), v1 = structure(list(), class = c("collector_double", 
    "collector")), v2 = structure(list(), class = c("collector_double", 
    "collector")), v3 = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector")), skip = 1L), class = "col_spec"))

df_yummy <- structure(list(VarId = c("v1", "v2", "v3"), VarName = c("Apple", 
"Lemon", "Peach"), Yummyness = c(2, 1, 3)), class = c("spec_tbl_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -3L), spec = structure(list(
    cols = list(VarId = structure(list(), class = c("collector_character", 
    "collector")), VarName = structure(list(), class = c("collector_character", 
    "collector")), Yummyness = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector")), skip = 1L), class = "col_spec"))

Answer 3

这是一个使用基础 R 的简单解决方案。

# Dummy data
df6 <- data.frame(
    id = c(1, 2, 3),
    v1 = c(0, 2, 0),
    v2 = c(0, 0, 1),
    v3 = c(3, 0, 0)
)

df_yummy <- data.frame(
    VarId = c("v1", "v2", "v3"),
    VarName = c("Apple", "Lemon", "Peach"),
    Yummyness = c(2, 1, 3)
)

df6$VarId <- names(df6[-1])[apply(X = df6[-1] > 0, MARGIN = 1, FUN = which)]
df_result <- merge(df6, df_yummy)[, c("id", "VarName", "Yummyness")]
names(df_result) <- c("id", "Fruit", "Yummyness")

> df_result[order(df_result$id), ]
  id Fruit Yummyness
3  1 Peach         3
1  2 Apple         2
2  3 Lemon         1

此解决方案的关键在于 apply(X = df6[-1] > 0, MARGIN = 1, FUN = which)。这样做是获取您的 df6 并删除第一列（[-1]）部分，然后查找哪些记录大于零。然后，我们使用 apply 函数将 which 函数应用于行（MARGIN = 1）部分。这意味着对于每一行，我们都返回非零索引。我们将其传递给 names() 以获得每个名称的正确名称，然后剩下的只是一些连接。

这是一个使用 data.table 包的稍微更紧凑的版本。

library(data.table)
# Dummy data
df6 <- data.table(
    id = c(1, 2, 3),
    v1 = c(0, 2, 0),
    v2 = c(0, 0, 1),
    v3 = c(3, 0, 0)
)

df_yummy <- data.table(
    VarId = c("v1", "v2", "v3"),
    VarName = c("Apple", "Lemon", "Peach"),
    Yummyness = c(2, 1, 3)
)

df6[, VarId := names(df6)[-1][which(.SD > 0)], by = id]
df_result <- merge(df6, df_yummy)[order(id), list(id, Fruit = VarName, Yummyness)]

这在很大程度上是相同的原理；我们通过 id 在 data.table 特定的 .SD 上使用 which 函数，然后使用它来选择正确的名称。由于 DT 语法，这最终会稍微紧凑一些，但本质上并没有什么不同。

Answer 4

一个data.table选项

setorder(
  melt(setDT(df6), id.var = "id")[
    value != 0
  ][setDT(df_yummy),
    on = ("variable == VarId")
  ],
  by = "id"
)[, .(id, VarName, Yummyness)]

给予

   id VarName Yummyness
1:  1   Peach         3
2:  2   Apple         2
3:  3   Lemon         1

如何延长轴心并将变量名与辅助 table 匹配？

How pivot longer and match the variable name with a secondary table?

r

dataframe

match

recode

数据