Процент 2 ответов в переменной с использованием Dplyr - PullRequest
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Процент 2 ответов в переменной с использованием Dplyr

0 голосов
/ 23 ноября 2018

Для задания я хотел бы видеть количество субъектов, у которых 0 для переменной CIU против 1 для CIU. 

structure(list(Last_Name = c("Banks", "Beamon", "Dandridge", 
"Deakle, Jr.", "Doyle", "Drinkard", "Ellis", "Embry", "Gaines", 
"Gurley", "Hinton", "Holemon", "Holsomback", "Hunt", "Jones", 
"Mahan", "Mahan", "McMillian", "Moore", "Padgett"), First_Name = c("Medell", 
"Melvin Todd", "Beniah Alton", "Evan Lee", "Robert E.", "Gary", 
"Andre", "Anthony", "Freddie Lee", "Timothy", "Anthony", "Jeffrey", 
"John", "H. Guy", "Lydia Diane", "Dale", "Ronnie", "Walter", 
"Daniel Wade", "Larry Randal"), Age = c("27", "24", "29", "59", 
"44", "37", "35", "23", "22", "22", "29", "23", "33", "54", "40", 
"22", "26", "45", "24", "40"), Race = c("Black", "Black", "Caucasian", 
"Caucasian", "Caucasian", "Caucasian", "Black", "Black", "Black", 
"Caucasian", "Black", "Caucasian", "Caucasian", "Caucasian", 
"Black", "Caucasian", "Caucasian", "Black", "Caucasian", "Caucasian"
), Sex = c("Male", "Male", "Male", "Male", "Male", "Male", "Male", 
"Male", "Male", "Male", "Male", "Male", "Male", "Male", "Female", 
"Male", "Male", "Male", "Male", "Male"), State = c("Alabama", 
"Alabama", "Alabama", "Alabama", "Alabama", "Alabama", "Alabama", 
"Alabama", "Alabama", "Alabama", "Alabama", "Alabama", "Alabama", 
"Alabama", "Alabama", "Alabama", "Alabama", "Alabama", "Alabama", 
"Alabama"), CIU = c(0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 
0, 0, 0, 0, 1, 0), Guilty_Plea = c(1, 0, 0, 0, 0, 0, 0, 1, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), IO = c(0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Worst_Crime = c(6, 1, 
1, 4, 4, 1, 2, 1, 1, 6, 1, 2, 4, 6, 3, 2, 2, 1, 1, 1), Occurred = c(1999, 
1988, 1994, 2014, 1991, 1993, 2012, 1992, 1972, 1999, 1985, 1987, 
1987, 1987, 1997, 1983, 1983, 1986, 1999, 1990), Convicted = c(2001, 
1989, 1996, 2015, 1992, 1995, 2013, 1993, 1974, 2000, 1986, 1988, 
1988, 1993, 2000, 1986, 1986, 1988, 2002, 1992), Exonerated = c(2003, 
1990, 2015, 2015, 2001, 2001, 2014, 1997, 1991, 2002, 2015, 1999, 
2000, 1998, 2006, 1998, 1998, 1993, 2009, 1997), Sentence = c("15", 
"25", "Life", "Not sentenced", "20", "Death", "85", "20", "30", 
"35", "Death", "Life", "25", "Probation", "Life without parole", 
"35", "Life without parole", "Death", "Death", "Death"), Death_Penalty = c(0, 
0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1), DNA_Only = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0), FC = c(1, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), MWID = c(0, 
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0), F_MFE = c(0, 
0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1), P_FA = c(1, 
1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0), OM = c(1, 
1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1), ILD = c(0, 
0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0), State_Statute = c("Y", 
"Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", 
"Y", "Y", "Y", "Y", "Y", "Y"), State_Claim_Made = c(0, 0, 1, 
0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0), Zero_time = c(0, 
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), Prem = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Pending = c(0, 
0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0), Denied = c(0, 
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), State_Award = c("0", 
"0", "2", "0", "1", "0", "0", "0", "1", "0", "2", "0", "0", "0", 
"0", "0", "0", "0", "0", "0"), Amount = c("0", "0", NA, "0", 
"129041.88", "0", "0", "0", "1000000", "0", NA, "0", "0", "0", 
"0", "0", "0", "0", "0", "0"), `Non-Statutory_Case_Filed` = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0), No_Time = c(0, 
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), Unfiled = c(1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1), Dismissed = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0), Pending__1 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Award = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0), Premature = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Amount__1 = c("0", 
"0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", 
"0", "0", "0", "$ undisclosed", "0", "0"), Years_Lost = c(1.7, 
0.1, 19.5, 0, 2.6, 5.7, 1.8, 4, 10.7, 1.5, 28.5, 10.6, 10.1, 
0, 5.8, 11.4, 11.4, 4.5, 5.4, 5.5), State_Award2 = c("0", "0", 
"0", "0", "1", "0", "0", "0", "1", "0", "0", "0", "0", "0", "0", 
"0", "0", "0", "0", "0")), row.names = c(NA, -20L), class = c("tbl_df", 
"tbl", "data.frame"))

Используя пакет dplyr, я достиг этого многого: 

CUI <- jail %>%
  group_by(CIU)  %>% 
  summarize(count = n())

Теперь я хотел бы создать таблицу, показывающую процент каждой группы в категории «State_Claim_Made», но я не уверен, что делать дальше.В конце я хотел бы видеть процент CUI = 0, который имеет State_Claim_Made = 0 против State_Claim_Made = 1 и такой же для CUI = 1;таблица 2-2 сортов.Я также предпочитаю продолжать использовать пакет dplyr, но не обязательно.

Ответы [ 2 ]

0 голосов
/ 23 ноября 2018

Ваш пример на самом деле не позволяет увидеть полную картину, поэтому давайте 

df <- data.frame(CIU = rep(0:1, times = c(20, 30)),
                 State_Claim_Made = rep(1:0, times = c(15, 35)))

Тогда 

table(CIU = df$CIU, State_Claim_Made = df$State_Claim_Made)
#    State_Claim_Made
# CIU  0  1
#   0  5 15
#   1 30  0
table(CIU = df$CIU, State_Claim_Made = df$State_Claim_Made) / c(table(df$CIU))
#    State_Claim_Made
# CIU    0    1
#   0 0.25 0.75
#   1 1.00 0.00
0 голосов
/ 23 ноября 2018

Используя базу R, вы можете просто использовать команду таблицы: 

 table(data$CIU, data$State_Claim_Made)

Вывод: 

     0  1
  0 15  5

Если у вас есть данные, включая CUI = 1, то результатом будет таблица 2x2как тебе нужно

...