Question

我对R非常陌生，来自更多的python背景。我正在一个需要使用R的项目中，结果几乎完成了，但是在最后一个问题上迷迷糊糊。

我在R中编写了一个函数，该函数以绝对糟糕的方式为我提供了所要查找的输出。

我想将其放入R数据框中，以comp_count，方差，execution_time和准确性作为列。我还需要删除空值。有什么建议吗？

编辑：要求输出dput（结果）。

    list(NULL, list(comp_count = 2L, variance = 32.6, execution_time = 2.15086028575897, 
    accuracy = 0.782779827798278), list(comp_count = 3L, variance = 43.3, 
    execution_time = 2.671033902963, accuracy = 0.832069775243207), 
    list(comp_count = 4L, variance = 52, execution_time = 3.50883383353551, 
        accuracy = 0.839405121323941), list(comp_count = 5L, 
        variance = 58.6, execution_time = 3.92320603132248, accuracy = 0.842938611204294), 
    list(comp_count = 6L, variance = 64.3, execution_time = 4.39138143459956, 
        accuracy = 0.843139885944314), list(comp_count = 7L, 
        variance = 69.2, execution_time = 5.430861667792, accuracy = 0.843609527004361), 
    list(comp_count = 8L, variance = 73.7, execution_time = 6.1574295481046, 
        accuracy = 0.844459353684446), list(comp_count = 9L, 
        variance = 77.1, execution_time = 6.21873023509979, accuracy = 0.845219724924522), 
    list(comp_count = 10L, variance = 80.5, execution_time = 7.65496598482132, 
        accuracy = 0.846069551604607), list(comp_count = 11L, 
        variance = 83.8, execution_time = 6.71301571528117, accuracy = 0.846024823884602), 
    list(comp_count = 12L, variance = 87, execution_time = 8.45073886712392, 
        accuracy = 0.846740467404674), list(comp_count = 13L, 
        variance = 89.9, execution_time = 9.64280251661936, accuracy = 0.846293190204629), 
    list(comp_count = 14L, variance = 92.5, execution_time = 9.80074710051219, 
        accuracy = 0.846382645644638), list(comp_count = 15L, 
        variance = 94.7, execution_time = 10.3888318975766, accuracy = 0.846091915464609), 
    NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 
    NULL, NULL, NULL, NULL, list(comp_count = 30L, variance = 100, 
        execution_time = 20.8339449683825, accuracy = 0.8589958626859))

我可以通过以下方式访问单独的组件：

results[[2]]

$comp_count
[1] 2

$variance
[1] 32.6

$execution_time
[1] 2.15086

$accuracy
[1] 0.7827798

解决此问题的最佳方法是什么？感谢所有帮助！

Answer 1

不幸的是，您没有以易于复制和粘贴的格式提供示例数据；对于以后的帖子，最好使用successCallback()，尤其是当数据结构是问题描述的关键组成部分时。

我将生成一些（希望）具有代表性且最少的样本数据

dput

然后您可以使用

lst <- list(
    NULL,
    list(comp_count = 2, variance = 32.6, execution_time = 2.15, accuracy = 0.78),
    list(comp_count = 3, variance = 43.3, execution_time = 2.67, accuracy = 0.83)
)

将do.call(rbind.data.frame, lst) # comp_count variance execution_time accuracy #2 2 32.6 2.15 0.78 #21 3 43.3 2.67 0.83元素行绑定到list中。

请注意，这要求所有（非{data.frame）NULL个元素都具有相同数量的元素。

另一种选择是使用list

dplyr::bind_rows

使用更新的样本数据

library(dplyr)
bind_rows(lst)
## A tibble: 2 x 4
#  comp_count variance execution_time accuracy
#       <dbl>    <dbl>          <dbl>    <dbl>
#1          2     32.6           2.15     0.78
#2          3     43.3           2.67     0.83

样本数据

do.call(rbind.data.frame, lst)
#   comp_count variance execution_time  accuracy
#2           2     32.6       2.150860 0.7827798
#21          3     43.3       2.671034 0.8320698
#3           4     52.0       3.508834 0.8394051
#4           5     58.6       3.923206 0.8429386
#5           6     64.3       4.391381 0.8431399
#6           7     69.2       5.430862 0.8436095
#7           8     73.7       6.157430 0.8444594
#8           9     77.1       6.218730 0.8452197
#9          10     80.5       7.654966 0.8460696
#10         11     83.8       6.713016 0.8460248
#11         12     87.0       8.450739 0.8467405
#12         13     89.9       9.642803 0.8462932
#13         14     92.5       9.800747 0.8463826
#14         15     94.7      10.388832 0.8460919
#15         30    100.0      20.833945 0.8589959

R怪异的输出到数据帧

1 个答案:

样本数据