Как извлечь указанный c dataframe из вывода в r?

Question

У меня есть вывод, как показано ниже;

В моем выводе есть несколько таких кластеров. Как видите, есть 2 столбца (день 1 и день 2). Я хочу извлечь (или экспортировать) кадры данных, которые начинаются с Hour_1 для каждого дня.

Итак, для Cluster 1 я хочу иметь 2 кадра данных, один начинается с 0.00, 0, 0... и продолжается до Hour_24, а другой начинается с 1.07, 0, 0... и продолжается до Hour_24, тоже.

Вот структура моих данных;

 v1<-structure(list(`Cluster 1` = structure(list(Day = c(1, 2), Season = c(1, 
    1), Year = c(2011, 2011), Daily = c(1.48, 2.53), Hour_1 = c(0, 
    1.07), Hour_2 = c(0, 0), Hour_3 = c(0, 0), Hour_4 = c(0, 0), 
        Hour_5 = c(0, 0), Hour_6 = c(0, 1.27), Hour_7 = c(0, 0.19
        ), Hour_8 = c(0, 0), Hour_9 = c(0, 0), Hour_10 = c(0, 0), 
        Hour_11 = c(0, 0), Hour_12 = c(0, 0), Hour_13 = c(0, 0), 
        Hour_14 = c(0, 0), Hour_15 = c(0, 0), Hour_16 = c(0, 0), 
        Hour_17 = c(0, 0), Hour_18 = c(0.041, 0), Hour_19 = c(0.673, 
        0), Hour_20 = c(0, 0), Hour_21 = c(0, 0), Hour_22 = c(0, 
        0), Hour_23 = c(0.319, 0), Hour_24 = c(0.447, 0)), .Names = c("Day", 
    "Season", "Year", "Daily", "Hour_1", "Hour_2", "Hour_3", "Hour_4", 
    "Hour_5", "Hour_6", "Hour_7", "Hour_8", "Hour_9", "Hour_10", 
    "Hour_11", "Hour_12", "Hour_13", "Hour_14", "Hour_15", "Hour_16", 
    "Hour_17", "Hour_18", "Hour_19", "Hour_20", "Hour_21", "Hour_22", 
    "Hour_23", "Hour_24"), row.names = c("Day_1", "Day_2"), class = "data.frame"), 
        `Cluster 2` = structure(list(Day = c(7, 8), Season = c(1, 
        1), Year = c(2011, 2011), Daily = c(0.19, 0.75), Hour_1 = c(0, 
        0), Hour_2 = c(0, 0.521), Hour_3 = c(0, 0.229), Hour_4 = c(0, 
        0), Hour_5 = c(0, 0), Hour_6 = c(0, 0), Hour_7 = c(0, 0), 
            Hour_8 = c(0, 0), Hour_9 = c(0, 0), Hour_10 = c(0, 0), 
            Hour_11 = c(0, 0), Hour_12 = c(0, 0), Hour_13 = c(0, 
            0), Hour_14 = c(0, 0), Hour_15 = c(0, 0), Hour_16 = c(0, 
            0), Hour_17 = c(0.189, 0), Hour_18 = c(0.001, 0), Hour_19 = c(0, 
            0), Hour_20 = c(0, 0), Hour_21 = c(0, 0), Hour_22 = c(0, 
            0), Hour_23 = c(0, 0), Hour_24 = c(0, 0)), .Names = c("Day", 
        "Season", "Year", "Daily", "Hour_1", "Hour_2", "Hour_3", 
        "Hour_4", "Hour_5", "Hour_6", "Hour_7", "Hour_8", "Hour_9", 
        "Hour_10", "Hour_11", "Hour_12", "Hour_13", "Hour_14", "Hour_15", 
        "Hour_16", "Hour_17", "Hour_18", "Hour_19", "Hour_20", "Hour_21", 
        "Hour_22", "Hour_23", "Hour_24"), row.names = c("Day_3", 
        "Day_4"), class = "data.frame")), .Names = c("Cluster 1", 
    "Cluster 2"))

Answer 1

Опция с tidyverse

library(purrr)
library(dplyr)
map(v1, ~ .x %>%
             select(contains('Hour')))
#$`Cluster 1`
#      Hour_1 Hour_2 Hour_3 Hour_4 Hour_5 Hour_6 Hour_7 Hour_8 Hour_9 Hour_10 #Hour_11 Hour_12 Hour_13
#Day_1   0.00      0      0      0      0   0.00   0.00      0      0       0       0       0       0
#Day_2   1.07      0      0      0      0   1.27   0.19      0      0       0       0       0       0
#      Hour_14 Hour_15 Hour_16 Hour_17 Hour_18 Hour_19 Hour_20 Hour_21 Hour_22 #Hour_23 Hour_24
#Day_1       0       0       0       0   0.041   0.673       0       0       0   0.319   0.447
#Day_2       0       0       0       0   0.000   0.000       0       0       0   0.000   0.000

#$`Cluster 2`
#      Hour_1 Hour_2 Hour_3 Hour_4 Hour_5 Hour_6 Hour_7 Hour_8 Hour_9 Hour_10 #Hour_11 Hour_12 Hour_13
#Day_3      0  0.000  0.000      0      0      0      0      0      0       0       0       0       0
#Day_4      0  0.521  0.229      0      0      0      0      0      0       0       0       0       0
#      Hour_14 Hour_15 Hour_16 Hour_17 Hour_18 Hour_19 Hour_20 Hour_21 Hour_22 #Hour_23 Hour_24
#Day_3       0       0       0   0.189   0.001       0       0       0       0       0       0
#Day_4       0       0       0   0.000   0.000       0       0       0       0       0       0

Answer 2

Вы можете собрать все столбцы, которые начинаются с «Часа», вместе.

lapply(v1, function(x) x[grep('Hour', names(x))])

#$`Cluster 1`
#      Hour_1 Hour_2 Hour_3 Hour_4 Hour_5 Hour_6 Hour_7 Hour_8...
#Day_1   0.00      0      0      0      0   0.00   0.00      0...
#Day_2   1.07      0      0      0      0   1.27   0.19      0...

#$`Cluster 2`
#      Hour_1 Hour_2 Hour_3 Hour_4 Hour_5 Hour_6 Hour_7 Hour_8...
#Day_3      0  0.000  0.000      0      0      0      0      0...
#Day_4      0  0.521  0.229      0      0      0      0      0...

Если вы хотите каждый день отдельно, вы можете использовать asplit

lapply(v1, function(x) asplit(x[grep('Hour', names(x))], 1))