R：从列表和解包列表创建 data.frame 或 data.table答案

【问题标题】：R: create a data.frame or data.table from a list and unpacked list of listsR：从列表和解包列表创建 data.frame 或 data.table
【发布时间】：2021-05-02 09:23:25
【问题描述】：

我有一个 list 和一个 list of lists，并且想创建一个 data.frame 或 data.table。

这是列表：

 head(stadte_namen)
[1] "Berlin"            "Hamburg"           "München"

和一个列表列表

> head(result)
[[1]]
       min      max
x 13.22886 13.54886
y 52.35704 52.67704

[[2]]
        min      max
x  9.840654 10.16065
y 53.390341 53.71034

[[3]]
       min      max
x 11.36078 11.72291
y 48.06162 48.24812

如何创建具有以下结构的 data.frame 或 data.table？

name                xmin        ymin        xmax       ymax

Berlin              13.22886    52.35704    13.54886   52.67704
Hamburg             9.840654    53.390341   10.16065   53.71034
München             11.36078    48.06162    11.72291   48.24812
...

这是数据：

stadte_namen<-c("Berlin", "Hamburg", "München", "Köln", "Frankfurt am Main", 
"Stuttgart")

result<-list(structure(c(13.2288599, 52.3570365, 13.5488599, 52.6770365
), .Dim = c(2L, 2L), .Dimnames = list(c("x", "y"), c("min", "max"
))), structure(c(9.840654, 53.390341, 10.160654, 53.710341), .Dim = c(2L, 
2L), .Dimnames = list(c("x", "y"), c("min", "max"))), structure(c(11.360777, 
48.0616244, 11.7229083, 48.2481162), .Dim = c(2L, 2L), .Dimnames = list(
    c("x", "y"), c("min", "max"))), structure(c(6.7725303, 50.8304399, 
7.162028, 51.0849743), .Dim = c(2L, 2L), .Dimnames = list(c("x", 
"y"), c("min", "max"))), structure(c(8.4727933, 50.0155435, 8.8004716, 
50.2271408), .Dim = c(2L, 2L), .Dimnames = list(c("x", "y"), 
    c("min", "max"))), structure(c(9.0386007, 48.6920188, 9.3160228, 
48.8663994), .Dim = c(2L, 2L), .Dimnames = list(c("x", "y"), 
    c("min", "max"))))

【问题讨论】：

标签： r dataframe data.table purrr

【解决方案1】：

你也可以试试

l <- result
df <- data.frame(t(sapply(l,c)))
colnames(df) <- c("minX", "minY", "maxX", "maxY"); df
df$stadte_namen <- c("Berlin", "Hamburg", "München", "Köln", "Frankfurt am Main", 
                     "Stuttgart");df

回答

       minX     minY      maxX     maxY      stadte_namen
1 13.228860 52.35704 13.548860 52.67704            Berlin
2  9.840654 53.39034 10.160654 53.71034           Hamburg
3 11.360777 48.06162 11.722908 48.24812           München
4  6.772530 50.83044  7.162028 51.08497              Köln
5  8.472793 50.01554  8.800472 50.22714 Frankfurt am Main
6  9.038601 48.69202  9.316023 48.86640         Stuttgart

【讨论】：

谢谢！一个很好的解决方案，因为它不使用依赖项。

【解决方案2】：

使用lapply 和purrr：

library(dplyr)
library(purrr)

data <- lapply(result, function(x) c(xmin = x[1,1],
                                     xmax = x[1,2],
                                     ymin = x[2,1],
                                     ymax = x[2,2])) %>%
        purrr::map_dfr(~.x)

data$stadte_namen <- stadte_namen

# A tibble: 6 x 5
   xmin  xmax  ymin  ymax stadte_namen     
  <dbl> <dbl> <dbl> <dbl> <chr>            
1 13.2  13.5   52.4  52.7 Berlin           
2  9.84 10.2   53.4  53.7 Hamburg          
3 11.4  11.7   48.1  48.2 München          
4  6.77  7.16  50.8  51.1 Köln             
5  8.47  8.80  50.0  50.2 Frankfurt am Main
6  9.04  9.32  48.7  48.9 Stuttgart

【讨论】：

【解决方案3】：

将stadte_namen 指定为result 的名称，并将数据帧绑定到一个数据帧中。您可以使用pivot_wider 获取宽格式数据。

library(tidyverse)

map_df(setNames(result, stadte_namen), ~.x %>% 
         as.data.frame %>%
         rownames_to_column('row'), .id = 'name') %>%
  pivot_wider(names_from = row, values_from = c(min, max))

#  name              min_x min_y max_x max_y
#  <chr>             <dbl> <dbl> <dbl> <dbl>
#1 Berlin            13.2   52.4 13.5   52.7
#2 Hamburg            9.84  53.4 10.2   53.7
#3 München           11.4   48.1 11.7   48.2
#4 Köln               6.77  50.8  7.16  51.1
#5 Frankfurt am Main  8.47  50.0  8.80  50.2
#6 Stuttgart          9.04  48.7  9.32  48.9

【讨论】：