фрейм данных пропорции для каждого факторного уровня на основе другого столбца

Question

Я хотел бы суммировать фрейм данных по месяцам, где каждый столбец - это доля каждого факторного уровня на основе столбца записей в фрейме данных ниже.Я пытался использовать dplyr, но не совсем понял.

library(dplyr)
set.seed(100)
df=data.frame(Month=rep(c("1/1/2017","2/1/2017","3/1/2017","4/1/2017","5/1/2017","6/1/2017","7/1/2017",

                          "8/1/2017","9/1/2017","10/1/2017","11/1/2017","12/1/2017"),10),

              Records=round(runif(120,6000,10000),0),

              V1=as.factor(sample(c("T","F"),120,replace=TRUE)),

              V2=as.factor(sample(c("A","B","C"),120,replace=TRUE)),

              V3=as.factor(sample(c("X","Y","Z","W"),120,replace=TRUE)),

              V4=as.factor(sample(c("YES","NO","Maybe"),120,replace=TRUE)))

Вот то, что я хотел бы, чтобы вывод был

> dput((resultsdf))
structure(list(Month = c("1/1/2017", "2/1/2017", "3/1/2017", 
"4/1/2017", "5/1/2017", "6/1/2017", "7/1/2017", "8/1/2017", "9/1/2017", 
"10/1/2017", "11/1/2017", "12/1/2017"), V1.F = c(0.4, 0.71, 0.63, 
0.35, 0.37, 0.41, 0.37, 0.61, 0.29, 0.5, 0.38, 0.82), V2.T = c(0.6, 
0.29, 0.37, 0.65, 0.63, 0.59, 0.63, 0.39, 0.71, 0.5, 0.62, 0.18
), V2.A = c(0.2, 0.28, 0.3, 0.31, 0.29, 0.3, 0.32, 0.45, 0.1, 
0.41, 0.3, 0.11), V2.B = c(0.59, 0.33, 0.19, 0.5, 0.51, 0.19, 
0.59, 0.22, 0.77, 0.2, 0.41, 0.16), V2.C = c(0.22, 0.38, 0.51, 
0.19, 0.21, 0.51, 0.09, 0.32, 0.12, 0.39, 0.29, 0.73), V3.W = c(0.42, 
0.11, 0, 0.21, 0.23, 0.3, 0.12, 0.45, 0.32, 0.28, 0.19, 0.19), 
    V3.X = c(0.19, 0.32, 0.18, 0.19, 0.19, 0.11, 0.19, 0, 0.27, 
    0.11, 0.23, 0.19), V3.Y = c(0.3, 0.29, 0.39, 0.4, 0.18, 0.4, 
    0.62, 0.34, 0.21, 0.33, 0.21, 0.1), V3.Z = c(0.09, 0.28, 
    0.43, 0.2, 0.4, 0.19, 0.07, 0.2, 0.2, 0.29, 0.38, 0.53), 
    V4.Maybe = c(0.4, 0.23, 0.39, 0.38, 0.62, 0.5, 0.2, 0.4, 
    0.4, 0.32, 0.3, 0.49), V4.NO = c(0.32, 0.5, 0.39, 0.31, 0.18, 
    0.29, 0.22, 0.42, 0.29, 0.3, 0.44, 0.3), V4.YES = c(0.28, 
    0.27, 0.22, 0.31, 0.2, 0.21, 0.58, 0.18, 0.3, 0.39, 0.26, 
    0.22)), row.names = c(NA, -12L), class = c("tbl_df", "tbl", 
"data.frame"), spec = structure(list(cols = list(Month = structure(list(), class = c("collector_character", 
"collector")), V1.F = structure(list(), class = c("collector_double", 
"collector")), V2.T = structure(list(), class = c("collector_double", 
"collector")), V2.A = structure(list(), class = c("collector_double", 
"collector")), V2.B = structure(list(), class = c("collector_double", 
"collector")), V2.C = structure(list(), class = c("collector_double", 
"collector")), V3.W = structure(list(), class = c("collector_double", 
"collector")), V3.X = structure(list(), class = c("collector_double", 
"collector")), V3.Y = structure(list(), class = c("collector_double", 
"collector")), V3.Z = structure(list(), class = c("collector_double", 
"collector")), V4.Maybe = structure(list(), class = c("collector_double", 
"collector")), V4.NO = structure(list(), class = c("collector_double", 
"collector")), V4.YES = structure(list(), class = c("collector_double", 
"collector"))), default = structure(list(), class = c("collector_guess", 
"collector"))), class = "col_spec"))

Answer 1

Здесь представлен альтернативный подход, который использует функции table() и prop.table() из базы R и dcast() для преобразования в широкоформатный формат.К сожалению, я не достаточно бегло говорю по dplyr, поэтому я прибегаю к data.table для группировки.

library(data.table)
library(magrittr)
setDT(df)[, lapply(.SD, function(.x) table(.x) %>% prop.table %>% as.data.table) %>% 
    rbindlist(idcol = TRUE), .SDcols = V1:V4, by = Month] %>% 
  dcast(Month ~ .id + .x)

        Month V1_F V1_T V2_A V2_B V2_C V3_W V3_X V3_Y V3_Z V4_Maybe V4_NO V4_YES
 1:  1/1/2017  0.4  0.6  0.2  0.6  0.2  0.4  0.2  0.3  0.1      0.4   0.3    0.3
 2: 10/1/2017  0.5  0.5  0.4  0.2  0.4  0.3  0.1  0.3  0.3      0.3   0.3    0.4
 3: 11/1/2017  0.4  0.6  0.3  0.4  0.3  0.2  0.2  0.2  0.4      0.3   0.4    0.3
 4: 12/1/2017  0.8  0.2  0.1  0.2  0.7  0.2  0.2  0.1  0.5      0.5   0.3    0.2
 5:  2/1/2017  0.7  0.3  0.3  0.3  0.4  0.1  0.3  0.3  0.3      0.2   0.5    0.3
 6:  3/1/2017  0.6  0.4  0.3  0.2  0.5  0.0  0.2  0.4  0.4      0.4   0.4    0.2
 7:  4/1/2017  0.4  0.6  0.3  0.5  0.2  0.2  0.2  0.4  0.2      0.4   0.3    0.3
 8:  5/1/2017  0.4  0.6  0.3  0.5  0.2  0.2  0.2  0.2  0.4      0.6   0.2    0.2
 9:  6/1/2017  0.4  0.6  0.3  0.2  0.5  0.3  0.1  0.4  0.2      0.5   0.3    0.2
10:  7/1/2017  0.4  0.6  0.3  0.6  0.1  0.1  0.2  0.6  0.1      0.2   0.2    0.6
11:  8/1/2017  0.6  0.4  0.5  0.2  0.3  0.5  0.0  0.3  0.2      0.4   0.4    0.2
12:  9/1/2017  0.3  0.7  0.1  0.8  0.1  0.3  0.3  0.2  0.2      0.4   0.3    0.3

Answer 2

Пожалуйста, проверьте ожидаемый результат.Я считаю, что есть некоторые ошибки.

Вот вариант tidyverse:

library(tidyverse)
df %>%
    gather(key, value, -Month, -Records) %>%
    group_by(Month, key, value) %>%
    summarise(freq = n()) %>%
    mutate(freq = freq / sum(freq)) %>%
    unite(col, key, value, sep = ".") %>%
    spread(col, freq)
## A tibble: 12 x 13
## Groups:   Month [12]
#   Month  V1.F  V1.T  V2.A  V2.B  V2.C  V3.W  V3.X  V3.Y  V3.Z V4.Maybe V4.NO
#   <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>    <dbl> <dbl>
# 1 1/1/…   0.4   0.6   0.2   0.6   0.2   0.4   0.2   0.3   0.1      0.4   0.3
# 2 10/1…   0.5   0.5   0.4   0.2   0.4   0.3   0.1   0.3   0.3      0.3   0.3
# 3 11/1…   0.4   0.6   0.3   0.4   0.3   0.2   0.2   0.2   0.4      0.3   0.4
# 4 12/1…   0.8   0.2   0.1   0.2   0.7   0.2   0.2   0.1   0.5      0.5   0.3
# 5 2/1/…   0.7   0.3   0.3   0.3   0.4   0.1   0.3   0.3   0.3      0.2   0.5
# 6 3/1/…   0.6   0.4   0.3   0.2   0.5  NA     0.2   0.4   0.4      0.4   0.4
# 7 4/1/…   0.4   0.6   0.3   0.5   0.2   0.2   0.2   0.4   0.2      0.4   0.3
# 8 5/1/…   0.4   0.6   0.3   0.5   0.2   0.2   0.2   0.2   0.4      0.6   0.2
# 9 6/1/…   0.4   0.6   0.3   0.2   0.5   0.3   0.1   0.4   0.2      0.5   0.3
#10 7/1/…   0.4   0.6   0.3   0.6   0.1   0.1   0.2   0.6   0.1      0.2   0.2
#11 8/1/…   0.6   0.4   0.5   0.2   0.3   0.5  NA     0.3   0.2      0.4   0.4
#12 9/1/…   0.3   0.7   0.1   0.8   0.1   0.3   0.3   0.2   0.2      0.4   0.3
## ... with 1 more variable: V4.YES <dbl>