Подмножество факторов

Question

Я работаю с данными "Единая преступная отчетность" по городам Небраски (щедрая классификация) и рассчитал уровень преступности по основным классификациям с 1995 по 2010 год с шагом 5 лет.

Я хотел бы построить показатели преступности по годам. Однако не все города сообщают значения за все годы из-за того, как работает отчетность в UCR.

Я довольно новичок в R, но коллега предложил мне попытаться создать цикл for, который подсчитывает уникальные значения для каждого названия города. Затем я могу использовать эти значения для отбрасывания данных или поднабора данных, чтобы у меня было как минимум три наблюдения для работы для построения графиков. Это примерно, насколько я получил и что там не работает. К сожалению, мне нужно сосредоточиться на некоторых более насущных проблемах на оставшуюся часть недели, поэтому я решил поделиться с сообществом, чтобы получить представление.

Код и названия данных приведены ниже. Благодарю.

drop = NULL
city.names <- unique(cnames)

for (i in 1:length(city.names)){
  x = sum(cnames==i)
 if (x < 3) {c(drop,i)}
} 

Есть 191 наблюдение с 64 уникальными именами. Данные CSV и импортированы как

data <- read.csv("cities.csv", header=TRUE, sep=",")

"","year","cnames" "1",1995,"Beatrice" "2",1995,"Bellevue" "3",1995,"Columbus" "4",1995,"Fremont" "5",1995,"Grand Island" "6",1995,"Hastings" "7",1995,"Kearney" "8",1995,"La Vista" "9",1995,"Lincoln" "10",1995,"Norfolk" "11",1995,"North Platte" "12",1995,"Omaha" "13",1995,"Papillion" "14",1995,"Scottsbluff" "15",1995,"South Sioux City" "16",2000,"Bellevue" "17",2000,"Columbus" "18",2000,"Fremont" "19",2000,"Grand Island" "20",2000,"Hastings" "21",2000,"Kearney" "22",2000,"La Vista" "23",2000,"Lincoln" "24",2000,"Norfolk" "25",2000,"Omaha" "26",2000,"Papillion" "27",2000,"Scottsbluff" "28",2000,"South Sioux City" "29",2005,"Alliance" "30",2005,"Ashland" "31",2005,"Auburn" "32",2005,"Bayard" "33",2005,"Beatrice" "34",2005,"Bellevue" "35",2005,"Blair" "36",2005,"Bridgeport" "37",2005,"Broken Bow" "38",2005,"Central City" "39",2005,"Chadron" "40",2005,"Columbus" "41",2005,"Cozad" "42",2005,"Crete" "43",2005,"David City" "44",2005,"Elkhorn" "45",2005,"Falls City" "46",2005,"Fremont" "47",2005,"Gering" "48",2005,"Gothenburg" "49",2005,"Grand Island" "50",2005,"Hastings" "51",2005,"Holdrege" "52",2005,"Imperial" "53",2005,"Kearney" "54",2005,"La Vista" "55",2005,"Lexington" "56",2005,"Lincoln" "57",2005,"Lyons" "58",2005,"Madison" "59",2005,"McCook" "60",2005,"Milford" "61",2005,"Minden" "62",2005,"Mitchell" "63",2005,"Nebraska City" "64",2005,"Norfolk" "65",2005,"North Platte" "66",2005,"Ogallala" "67",2005,"Omaha" "68",2005,"O'Neill" "69",2005,"Ord" "70",2005,"Papillion" "71",2005,"Plainview" "72",2005,"Plattsmouth" "73",2005,"Ralston" "74",2005,"Schuyler" "75",2005,"Scottsbluff" "76",2005,"Seward" "77",2005,"Sidney" "78",2005,"South Sioux City" "79",2005,"St. Paul" "80",2005,"Superior" "81",2005,"Valley" "82",2005,"Wahoo" "83",2005,"West Point" "84",2005,"Wymore" "85",2005,"York" "86",2010,"Alliance" "87",2010,"Ashland" "88",2010,"Auburn" "89",2010,"Aurora" "90",2010,"Bayard" "91",2010,"Beatrice" "92",2010,"Bellevue" "93",2010,"Bennington" "94",2010,"Blair" "95",2010,"Bridgeport" "96",2010,"Broken Bow" "97",2010,"Central City" "98",2010,"Chadron" "99",2010,"Columbus" "100",2010,"Cozad" "101",2010,"Crete" "102",2010,"Falls City" "103",2010,"Fremont" "104",2010,"Gering" "105",2010,"Gothenburg" "106",2010,"Grand Island" "107",2010,"Hastings" "108",2010,"Holdrege" "109",2010,"Imperial" "110",2010,"Kearney" "111",2010,"La Vista" "112",2010,"Lexington" "113",2010,"Lincoln" "114",2010,"Lyons" "115",2010,"Madison" "116",2010,"McCook" "117",2010,"Milford" "118",2010,"Minden" "119",2010,"Nebraska City" "120",2010,"Norfolk" "121",2010,"North Platte" "122",2010,"Ogallala" "123",2010,"Omaha" "124",2010,"O'Neill" "125",2010,"Papillion" "126",2010,"Plainview" "127",2010,"Plattsmouth" "128",2010,"Ralston" "129",2010,"Scottsbluff" "130",2010,"Seward" "131",2010,"Sidney" "132",2010,"South Sioux City" "133",2010,"Superior" "134",2010,"Valentine" "135",2010,"Valley" "136",2010,"Wahoo" "137",2010,"Wayne" "138",2010,"West Point" "139",2010,"Wilber" "140",2010,"York" "141",2013,"Alliance" "142",2013,"Ashland" "143",2013,"Aurora" "144",2013,"Beatrice" "145",2013,"Bellevue" "146",2013,"Bennington" "147",2013,"Blair" "148",2013,"Bridgeport" "149",2013,"Broken Bow" "150",2013,"Central City" "151",2013,"Chadron" "152",2013,"Columbus" "153",2013,"Cozad" "154",2013,"Crete" "155",2013,"Falls City" "156",2013,"Fremont" "157",2013,"Gering" "158",2013,"Gordon" "159",2013,"Gothenburg" "160",2013,"Grand Island" "161",2013,"Hastings" "162",2013,"Holdrege" "163",2013,"Imperial" "164",2013,"Kearney" "165",2013,"Kimball" "166",2013,"La Vista" "167",2013,"Lexington" "168",2013,"Lincoln" "169",2013,"Madison" "170",2013,"McCook" "171",2013,"Milford" "172",2013,"Minden" "173",2013,"Mitchell" "174",2013,"Nebraska City" "175",2013,"Norfolk" "176",2013,"Ogallala" "177",2013,"Omaha" "178",2013,"O'Neill" "179",2013,"Papillion" "180",2013,"Plattsmouth" "181",2013,"Ralston" "182",2013,"Scottsbluff" "183",2013,"Seward" "184",2013,"South Sioux City" "185",2013,"Superior" "186",2013,"Valentine" "187",2013,"Valley" "188",2013,"Wahoo" "189",2013,"West Point" "190",2013,"Wilber" "191",2013,"York"

Answer 1

Для подстановки по "частоте" столбца, есть много вариантов в base R и в других пакетах. Один вариант использует table Функция в столбце "cnames" и получить частоту. На выходе будет vector с "ключом / значениями", соответствующими names/frequency каждого уникального "cnames". Проверьте, меньше ли значения 3 (tbl <3) который дает логический индекс "ИСТИНА / ЛОЖЬ". Подмножество имен "tbl", используя этот индекс, и использовать его для индексации столбца "cnames", используя %in%, Я показываю два метода, один с отрицанием (!) и используя <, другие с >=

 tbl <- table(data$cnames)
 data[!data$cnames %in% names(tbl)[tbl <3],]

Или же

 data[data$cnames %in% names(tbl)[tbl >=3],]

Или используя ave чтобы получить length каждого уникального "cnames" и получить логический индекс по >= оператор. ave возвращает выходные данные в том же порядке, что и в исходном наборе данных. Это может быть использовано для поднабора.

 data[with(data, ave(seq_along(cnames), cnames, FUN=length)>=3),]

Если вы используете data.tableкод будет более компактным и более быстрым для больших наборов данных. Преобразуйте "data.frame" в "data.table", используя setDT, назначить счетчики (n:=.N) для каждого уникального "cnames" и, наконец, подмножество набора данных с >=

library(data.table)
setDT(data)[,n:=.N, cnames][n>=3]