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展开两个矢量列

tidyr dplyr r

Nick Criswell · 技术社区 · 6 年前

我有一个 tibble 有两列。每列都包含有序的纬度和经度对。结构如下:

library(dplyr)
library(tidyr)

> my_df
# A tibble: 3 x 2
  V1        V2       
  <list>    <list>   
1 <dbl [2]> <dbl [2]>
2 <dbl [2]> <dbl [2]>
3 <dbl [2]> <dbl [2]>

my_df = structure(list(V1 = list(c(44.0252714, -88.1536451), c(42.9856117, 
-87.9355419), c(42.8600366, -87.9541568)), V2 = list(c(44.9535298, 
-90.9188588), c(45.4864422, -89.7339536), c(43.0743635, -87.9765372
))), row.names = c(NA, -3L), class = c("tbl_df", "tbl", "data.frame"
))

我想把它变成一个四列的数据框,看起来像:

> my_df2
        y1        x1       y2        x2
1 44.02527 -88.15365 44.95353 -90.91886
2 42.98561 -87.93554 45.48644 -89.73395
3 42.86004 -87.95416 43.07436 -87.97654

我试着用 unnest 从 tidyr 但没有成功。

> my_df %>% unnest()
# A tibble: 6 x 2
     V1    V2
  <dbl> <dbl>
1  44.0  45.0
2 -88.2 -90.9
3  43.0  45.5
4 -87.9 -89.7
5  42.9  43.1
6 -88.0 -88.0

> my_df %>% unnest(V1, V2)
# A tibble: 6 x 2
     V1    V2
  <dbl> <dbl>
1  44.0  45.0
2 -88.2 -90.9
3  43.0  45.5
4 -87.9 -89.7
5  42.9  43.1
6 -88.0 -88.0

我需要以某种方式控制不安的发生,但我不知道怎么做。

3 回复 | 直到 6 年前

s_baldur 6 年前

下面是一个技巧,首先将每个向量转换为字符串:

my_df %>%
  rowwise() %>%
  mutate_all(funs(toString(.))) %>%
  separate(V1, c("y1", "x1"), ", ") %>%
  separate(V2, c("y2", "x2"), ", ") %>%
  mutate_all(funs(as.numeric(.)))
# A tibble: 3 x 4
     y1    x1    y2    x2
  <dbl> <dbl> <dbl> <dbl>
1  44.0 -88.2  45.0 -90.9
2  43.0 -87.9  45.5 -89.7
3  42.9 -88.0  43.1 -88.0

编辑更基本的R型方法:

my_df2 <- 
  do.call(cbind, lapply(my_df, function(x) do.call(rbind, x))) %>% 
  as.tibble()
names(my_df2) <- c("y1", "x1", "y2", "x2")
my_df2
# A tibble: 3 x 4
     y1    x1    y2    x2
  <dbl> <dbl> <dbl> <dbl>
1  44.0 -88.2  45.0 -90.9
2  43.0 -87.9  45.5 -89.7
3  42.9 -88.0  43.1 -88.0

AntoniosK 6 年前

一个可能的解决方案是:

library(dplyr)
library(tidyr)

my_df %>% 
  rowwise() %>%                                  # for each row
  mutate_all(funs(list(data.frame(t(.))))) %>%   # transpose your vector and create a dataframe
  unnest()                                       # unnest

# # A tibble: 3 x 4
#      X1    X2   X11   X21
#   <dbl> <dbl> <dbl> <dbl>
# 1  44.0 -88.2  45.0 -90.9
# 2  43.0 -87.9  45.5 -89.7
# 3  42.9 -88.0  43.1 -88.0

Frank 6 年前

与 data.table::transpose ...

rn_df = function(x, suff = 0, cols = c("x","y"))
  setNames(x, paste0(cols, suff))

my_df %>% 
  lapply(data.table::transpose) %>%
  unname %>%
  Map(rn_df, ., seq_along(.)) %>%
  unlist(recursive=FALSE) %>% 
  data.frame

        x1        y1       x2        y2
1 44.02527 -88.15365 44.95353 -90.91886
2 42.98561 -87.93554 45.48644 -89.73395
3 42.86004 -87.95416 43.07436 -87.97654

这应该扩展到原来的一些列 my_df ,假设它们都变成具有相同命名模式的col。

在我看来,您最好使用长格式的数据,而不是生成以后必须解析的列名称:

library(data.table)
res = my_df %>% lapply(. %>% (data.table::transpose) %>% setDT) %>%
  rbindlist(id = "src") %>%
  setnames(-1, c("x", "y"))

   src        x         y
1:  V1 44.02527 -88.15365
2:  V1 42.98561 -87.93554
3:  V1 42.86004 -87.95416
4:  V2 44.95353 -90.91886
5:  V2 45.48644 -89.73395
6:  V2 43.07436 -87.97654