Question

我遇到的问题是，经过一些计算后，我的结果会出现混乱的格式。我为一年中的每个季度计算了两个不同的值，并将每个值存储在列表中，如下面的输出所示：

$`2015`
   1        2        3        4 
158.4567 165.7833 153.1233 140.8067 

$`2016`
       1        2        3        4 
140.4833 149.9200 157.9233 161.2467

我之前的解决方案最终出现了类似append(value1,value2)的内容，它为我提供了以下不需要的输出：

$`2015`
   1        2        3        4        1        2        3        4 
158.4567 165.7833 153.1233 140.8067 151.5633 155.2667 132.4667 128.4633 

$`2016`
       1        2        3        4        1        2        3        4 
140.4833 149.9200 157.9233 161.2467 127.8667 144.3600 150.8467 152.5333

我找不到将这三个列表合并为一个列表或数据框的方法，如下所示：

date   value1     value2
2015.1 158.4567   151.5633
2015.2 165.7833   155.2667
2015.3 153.1233   132.4667
2015.4 140.8067   128.4633
2016.1 ...        ...
2016.2 ...        ...
2016.3 ...        ...
2016.4 ...        ...

修改示例代码，用于创建我想在数据框中合并的两个值：

library(lubridate) # for date operations

# manual created dataframe
date <- sample(seq(as.Date('1999-01-01 00:00:00'), as.Date('2017-01-01 00:00:00'), by="day"), 500)
dataM = data.frame("first" = 1:500, "sec" = c(1:500), "date"=date)

dataM <- transform(dataM, date = ymd(dataM$date)) # date to lubridate date format

  splitByYear = split(dataM, year(dataM$date))
  splitByQuarter = sapply(splitByYear, function(y) split(y, quarter(y$date)))

  a = sapply(splitByQuarter, function(x) sapply(x, function(y) max(y$date, na.rm = TRUE)))
  b = sapply(splitByQuarter, function(x) sapply(x, function(y) min(y$sec, na.rm = TRUE)))


  res = mapply(quarterPP, a,b)
  res

  quarterPP <- function(a, b){
    value1 = a+b
    value2 = b+1900

    c(value1, value2) # this should be in a dataframe
}

Answer 1

使用matrix中的bind_rows和dplyr，

来接近它的一种方法

#create dummy data
x <- setNames(rnorm(4), 1:4)
y <- setNames(rnorm(4), 1:4)
z <- setNames(rnorm(4), 1:4)
w <- setNames(rnorm(4), 1:4)
l1 <- list(`2015` = append(x, y), `2016` = append(z, w))

l1
#$`2015`
#         1          2          3          4          1          2          3           4 
#-0.0318981 -1.1241606 -0.1040653 -0.7819973 -0.8715601 -0.2287638 -0.9092943  -0.3757804 

#$`2016`
#         1          2          3          4          1          2          3           4 
#-0.6034540 -1.1469930  0.6085236  1.2565788 -0.1020582  0.1383716  1.1358109  -0.2635427 


l2 <- lapply(l1,function(i) { 
                  ind <- max(as.numeric(names(i))); 
                  data.frame(matrix(i, nrow = ind))
               })

final_df <- dplyr::bind_rows(l2, .id = 'Year')
final_df$Year <- make.unique(final_df$Year)

final_df
#    Year         V1         V2
#1   2015 -0.0318981 -0.8715601
#2 2015.1 -1.1241606 -0.2287638
#3 2015.2 -0.1040653 -0.9092943
#4 2015.3 -0.7819973 -0.3757804
#5   2016 -0.6034540 -0.1020582
#6 2016.1 -1.1469930  0.1383716
#7 2016.2  0.6085236  1.1358109
#8 2016.3  1.2565788 -0.2635427

来自spread的{{1}}的另一种可能性，

tidyr

Answer 2

如果将所有列表合并为一个，则可以使用purrr::map_df来迭代并将结果强制转换为data.frame：

library(tidyverse)
set.seed(47)

l1 <- set_names(replicate(3, set_names(rnorm(4), 1:4), simplify = FALSE), 2015:2017)
l2 <- set_names(replicate(3, set_names(rnorm(4), 1:4), simplify = FALSE), 2015:2017)

lst(l1, l2) %>%    # make a list, pulling names from objects
    map_df(~map_df(.x,    # for each sublist, iterate over it
                   ~data_frame(val = .x,    # making a data.frame of each subelement
                               quarter = as.integer(names(.x))), 
                   .id = 'year'),    # coercing the element to a data.frame with a column of names
           .id = 'var') %>%    # and coerce both elements to a data.frame with a column of names
    spread(var, val)    # finally, reshape to wide form

#> # A tibble: 12 × 4
#>     year quarter          l1          l2
#> *  <chr>   <int>       <dbl>       <dbl>
#> 1   2015       1  1.99469634  0.49382018
#> 2   2015       2  0.71114251 -1.82822917
#> 3   2015       3  0.18540528  0.09147291
#> 4   2015       4 -0.28176501  0.67077922
#> 5   2016       1  0.10877555 -0.08107805
#> 6   2016       2 -1.08573747  1.26424109
#> 7   2016       3 -0.98548216 -0.70338819
#> 8   2016       4  0.01513086 -0.04057817
#> 9   2017       1 -0.25204590 -1.56616208
#> 10  2017       2 -1.46575030  0.24914817
#> 11  2017       3 -0.92245624 -0.34041599
#> 12  2017       4  0.03960243  0.41719084

在R中联合命名num的列表

2 个答案: