Question

我正在使用合并的df，并使用summarise(n = count(var))了解重复项。然后，我获得了该对象调用x：

      n.x         n.freq
1 001121011522      1
2 001121711014      1
3 001121711015      1
4 001121711722      1
5 001121711723      1
6 001131811722      1

我知道n.freq从1变到6，所以我只想抓取那些等于或大于2的那些。

y <- x %>% filter(n$freq >= 2)

但是我明白了：

Error: Column `n` must be a 1d atomic vector or a list
Call `rlang::last_error()` to see a backtrace

所以我检查了它的结构并得到了它：

    > str(x)
'data.frame':   485843 obs. of  1 variable:
 $ n:'data.frame':  485843 obs. of  2 variables:
  ..$ x   : Factor w/ 485843 levels "001121011522",..: 1 2 3 4 5 6 7 8 9 10 ...
  ..$ freq: int  1 1 1 1 1 1 1 1 1 1 ...

据我所知，它看起来在另一个内部有一个df。我想知道如何提取第二个（内部一个）df来使用它。

[UPDATE]：使用dput(head(x, 20))我得到了：

"ABC253478", "ABC983659", "ABC654911", "ABC882310", 
..... # there were at least 100 rows of results
"ABC665892", "ABC441276", "ABC906138", "ABC679967"
), class = "factor"), freq = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L)), row.names = c(NA, 
20L), class = "data.frame")), row.names = c(NA, 20L), class = "data.frame")

Answer 1

有 got 可以使这些数据变得有些混乱...但是让我们看看我是否可以尝试再现足够。（前期：并不完美，但是我们没有足够的信息来改进这一点。）

一些示例数据，手动编写以模拟我认为您缺少的行：

x <- structure(
  list(n = structure(
    list(x = structure(c(3L, 7L, 6L, 2L, 10L, 9L, 1L, 8L, 4L, 5L),
                       .Label = c("112916", "116806", "118489", "146802", "154999",
                                  "157333", "170238", "183345", "194348", "194384"),
                       class = "factor"),
         freq = c(2L, 1L, 3L, 1L, 2L, 3L, 3L, 1L, 2L, 1L)),
    class = "data.frame", row.names = c(NA, -10L) ) ),
  class = "data.frame", row.names = c(NA, -10L)
)
str(x)
# 'data.frame': 10 obs. of  1 variable:
#  $ n:'data.frame':    10 obs. of  2 variables:
#   ..$ x   : Factor w/ 10 levels "112916","116806",..: 3 7 6 2 10 9 1 8 4 5
#   ..$ freq: int  2 1 3 1 2 3 3 1 2 1

这不会复制您的错误：

str( x %>% filter(n$freq >= 2) )
# 'data.frame': 6 obs. of  1 variable:
#  $ n:'data.frame':    6 obs. of  2 variables:
#   ..$ x   : Factor w/ 10 levels "112916","116806",..: 3 6 10 9 1 4
#   ..$ freq: int  2 3 2 3 3 2

也许这对您有用吗？

str( subset(x, n$freq >= 2) )
# 'data.frame': 6 obs. of  1 variable:
#  $ n:'data.frame':    6 obs. of  2 variables:
#   ..$ x   : Factor w/ 10 levels "112916","116806",..: 3 6 10 9 1 4
#   ..$ freq: int  2 3 2 3 3 2

不幸的是，这可能对您有用，但具有平整数据库（无意）的作用：

str( x[ x$n$freq >= 2, ] )
# 'data.frame': 6 obs. of  2 variables:
#  $ x   : Factor w/ 10 levels "112916","116806",..: 3 6 10 9 1 4
#  $ freq: int  2 3 2 3 3 2

看来，此嵌套框架很容易解散：

str( x[[1]] )
# 'data.frame': 10 obs. of  2 variables:
#  $ x   : Factor w/ 10 levels "112916","116806",..: 3 7 6 2 10 9 1 8 4 5
#  $ freq: int  2 1 3 1 2 3 3 1 2 1

因此，我们尝试将嵌套框架替换为“正常”框架：

xflat <- x[[1]]
str( filter(xflat, freq >= 2) )
# 'data.frame': 6 obs. of  2 variables:
#  $ x   : Factor w/ 10 levels "112916","116806",..: 3 6 10 9 1 4
#  $ freq: int  2 3 2 3 3 2
str( subset(xflat, freq >= 2) )
# 'data.frame': 6 obs. of  2 variables:
#  $ x   : Factor w/ 10 levels "112916","116806",..: 3 6 10 9 1 4
#  $ freq: int  2 3 2 3 3 2
str( xflat[ xflat$freq >= 2, ] )
# 'data.frame': 6 obs. of  2 variables:
#  $ x   : Factor w/ 10 levels "112916","116806",..: 3 6 10 9 1 4
#  $ freq: int  2 3 2 3 3 2

如何从R中的另一个数据帧中提取一个数据帧

1 个答案: