根据来自不同数据帧的多个百分比从 df 中选择值

Question

我正在尝试根据另一个 df 的百分比将一个 df 中的值存储到一个新的数据框中。

df1

   "seq"  "loc"  "ball" "square" "triangle"
1    abcd   b      65      12        9
2    abcd   a      0       75        0
3    ght5   a      5       10        25
4    dfsf   b      12      9         75

df2

    "Type"    "n"    "n/2"    "1/n" 
1    ball      10     50       10
2    square    9      55.5     11.1
3    triangle  6      50       16.6`

我想按“Type”存储“seq”和“loc”，让我知道它是高于（>=）“n/2”值还是低于（

output df3

     "seq" "loc" "ball" "square" "triangle"
1    abcd   b     True    NA       False
2    abcd   a     False   True     False
3    ght5   a     False   False    NA
4    dfsf   b     NA      False    True`

我使用 True/NA/False 来表示 >=/between/

当我输入此内容时，我想我可能会先复制 df1 以用作 df3，然后只存储新值来代替百分比。这似乎比选择所有内容并制作新表更容易？在那之后，我不知道该去哪里。

编辑： 目前正在玩这个，但没有运气在数据框中存储一个值。

df3 <- df1
for(i in seq_len(nrow(df2)))
{
df3[, df2$Type[i]] <- 
sapply(1:length(df2$Type), 
   function(y) sapply(df3[, df2$Type][,y], 
                 function(x) ifelse(x < df2[y, c("1/n")], "1", 
                             ifelse(x > df2[y, c("n/2")], "3", "2")) ))
}

编辑：（添加 df1 和 df2 的 dput(head)）

df1
structure(list(Seq = structure(c(1L, 2L, 2L, 3L, 3L), .Label = 
c("AAAAAACCAGTCCCAGTTCGGATTG", 
"AAAAAACCAGTCTCAGTTCGGATTG", "AAAAAACCGGTCACAGTTCAGATTG"), class = 
"factor"), 
loc = structure(c(2L, 1L, 2L, 1L, 2L), .Label = c("b", 
"t"), class = "factor"), Ball = c(0, 0, 0, 0, 0), Cat = c(0, 
0, 0, 16.6666666666667, 16.6666666666667), Square = c(0, 
0, 0, 0, 0), Water = c(0, 0, 0, 33.3333333333333, 33.3333333333333
)), row.names = c(NA, -5L), class = c("grouped_df", "tbl_df", 
"tbl", "data.frame"), vars = c("Seq", "loc"), drop = TRUE, indices = 
list(
0L, 1L, 2L, 3L, 4L), group_sizes = c(1L, 1L, 1L, 1L, 1L), 
biggest_group_size = 1L, labels = structure(list(
Seq = structure(c(1L, 2L, 2L, 3L, 3L), .Label = 
c("AAAAAACCAGTCCCAGTTCGGATTG", 
"AAAAAACCAGTCTCAGTTCGGATTG", "AAAAAACCGGTCACAGTTCAGATTG"), class = 
"factor"), 
loc = structure(c(2L, 1L, 2L, 1L, 2L), .Label = c("b", 
"t"), class = "factor")), row.names = c(NA, -5L), class = "data.frame", 
vars = c("Seq", 
"loc"), drop = TRUE))

df2
dput(head(df2))
structure(list(Type = c("Ball", "Cat", "Square", "Water"), n = c(4L, 
6L, 3L, 6L), `n/2` = c(50, 50, 66.6666666666667, 50), `1/n` = c(25, 
16.6666666666667, 33.3333333333333, 16.6666666666667)), row.names = c(NA, 
-4L), class = c("tbl_df", "tbl", "data.frame"))
> 

Answer 1

您提供的dputs：

df1 <- structure(list(Seq = structure(c(1L, 2L, 2L, 3L, 3L), .Label = 
                                 c("AAAAAACCAGTCCCAGTTCGGATTG", 
                                   "AAAAAACCAGTCTCAGTTCGGATTG", "AAAAAACCGGTCACAGTTCAGATTG"), class = 
                                 "factor"), 
               loc = structure(c(2L, 1L, 2L, 1L, 2L), .Label = c("b", 
                                                                 "t"), class = "factor"), Ball = c(0, 0, 0, 0, 0), Cat = c(0, 
                                                                                                                           0, 0, 16.6666666666667, 16.6666666666667), Square = c(0, 
                                                                                                                                                                                 0, 0, 0, 0), Water = c(0, 0, 0, 33.3333333333333, 33.3333333333333
                                                                                                                                                                                 )), row.names = c(NA, -5L), class = c("grouped_df", "tbl_df", 
                                                                                                                                                                                                                       "tbl", "data.frame"), vars = c("Seq", "loc"), drop = TRUE, indices = 
            list(
              0L, 1L, 2L, 3L, 4L), group_sizes = c(1L, 1L, 1L, 1L, 1L), 
          biggest_group_size = 1L, labels = structure(list(
            Seq = structure(c(1L, 2L, 2L, 3L, 3L), .Label = 
                              c("AAAAAACCAGTCCCAGTTCGGATTG", 
                                "AAAAAACCAGTCTCAGTTCGGATTG", "AAAAAACCGGTCACAGTTCAGATTG"), class = 
                              "factor"), 
            loc = structure(c(2L, 1L, 2L, 1L, 2L), .Label = c("b", 
                                                              "t"), class = "factor")), row.names = c(NA, -5L), class = "data.frame", 
            vars = c("Seq", 
                     "loc"), drop = TRUE))


df2 <- structure(list(Type = c("Ball", "Cat", "Square", "Water"), n = c(4L, 
                                                                 6L, 3L, 6L), `n/2` = c(50, 50, 66.6666666666667, 50), `1/n` = c(25, 
                                                                                                                                 16.6666666666667, 33.3333333333333, 16.6666666666667)), row.names = c(NA, 
                                                                                                                                                                                                       -4L), class = c("tbl_df", "tbl", "data.frame"))

代码之前无法正常工作，原因我不知道，但与数据框对象的class 有关。在这里，我转换为data.frame，然后sapply 代码工作正常。答案：

library(tidyverse)
df1 <- as.data.frame(df1)
df2 <- as.data.frame(df2)

df3 <- sapply(1:length(df2$Type), 
       function(y) sapply(df1[, df2$Type][,y], 
                          function(x) ifelse(x < df2[y, c("1/n")], "1", 
                                             ifelse(x > df2[y, c("n/2")], "3", 
                                                    "2")) ))

df3 <- df3 %>% as_data_frame %>% 
  setNames(df2$Type) %>% 
  add_column(Seq = df1[,c("Seq")], loc = df1[,c("loc")], .before = 1)

df3
# A tibble: 5 x 6
  Seq                       loc   Ball  Cat   Square Water
  <fct>                     <fct> <chr> <chr> <chr>  <chr>
1 AAAAAACCAGTCCCAGTTCGGATTG t     1     1     1      1    
2 AAAAAACCAGTCTCAGTTCGGATTG b     1     1     1      1    
3 AAAAAACCAGTCTCAGTTCGGATTG t     1     1     1      1    
4 AAAAAACCGGTCACAGTTCAGATTG b     1     2     1      2    
5 AAAAAACCGGTCACAGTTCAGATTG t     1     2     1      2