的R - 由
问题描述：
组
通过休息和出现计数砍我有一个数据帧，看起来像这样：该城市属于的R - 由
dat <- structure(list(Geocode = c("1100015", "1100023", "1100031", "1100049",
"1100056", "1100064", "1100072", "1100080", "1100098", "1100106",
"1100114", "1100122", "1100130", "1100148", "1100155", "1100189",
"1100205", "1100254", "1100262", "1100288", "1100296", "1100304",
"1100320", "1100338", "1100346", "1100379", "1100403", "1100452",
"1100502", "1100601"), Location = c("Alta Floresta D'oeste, RO",
"Ariquemes, RO", "Cabixi, RO", "Cacoal, RO", "Cerejeiras, RO",
"Colorado Do Oeste, RO", "Corumbiara, RO", "Costa Marques, RO",
"Espigo D'oeste, RO", "Guajar-Mirim, RO", "Jaru, RO", "Ji-Paran, RO",
"Machadinho D'oeste, RO", "Nova Brasilndia D'oeste, RO", "Ouro Preto Do Oeste, RO",
"Pimenta Bueno, RO", "Porto Velho, RO", "Presidente Mdici, RO",
"Rio Crespo, RO", "Rolim De Moura, RO", "Santa Luzia D'oeste, RO",
"Vilhena, RO", "So Miguel Do Guapor, RO", "Nova Mamor, RO", "Alvorada D'oeste, RO",
"Alto Alegre Dos Parecis, RO", "Alto Paraso, RO", "Buritis, RO",
"Novo Horizonte Do Oeste, RO", "Cacaulandia, RO"), Region = c("Norte",
"Norte", "Norte", "Norte", "Norte", "Norte", "Norte", "Norte",
"Norte", "Norte", "Sul", "Sul", "Sul", "Sul", "Sul",
"Sul", "Sul", "Sul", "Sul", "Sul", "Nordeste", "Nordeste",
"Nordeste", "Nordeste", "Nordeste", "Nordeste", "Nordeste", "Nordeste", "Nordeste",
"Nordeste"), Population = c(25578L, 104401L, 6355L, 87226L, 17986L,
18817L, 8842L, 16651L, 32385L, 46632L, 55738L, 130419L, 37167L,
21592L, 39924L, 37512L, 502748L, 22557L, 3750L, 56242L, 8532L,
91801L, 23933L, 27600L, 17063L, 13940L, 20210L, 37838L, 10276L,
6367L)), .Names = c("Geocode", "Location", "Region", "Population"
), row.names = c(NA, 30L), class = "data.frame")
这表明一些城市的人口，以及该地区。
我需要将人口分为两部分（breaks=c(0,50000,100000)），然后根据整体（所有地区）的休息时间以及按地区分开来查找城市的数量。
所得数据帧应该是这样的（无规，假设值）：
Class Region Count
[0-50000] Norte 7
[50000-100000] Norte 3
[>100000] Norte 0
[0-50000] Sul 5
[50000-100000] Sul 4
[>100000] Sul 1
[0-50000] Nordeste 4
[50000-100000] Nordeste 5
[>100000] Nordeste 1
[0-50000] All 16
[50000-100000] All 12
[>100000] All 2
任何帮助理解。
答
通过使用cut和dplyr
dat$Class=cut(dat$Population,c(0,50000,100000,Inf),labels=c('0-50000','50000-100000','>100000'))
library(dplyr)
d1=dat%>%group_by(Class,Region)%>%summarise(count=n())
d2=dat%>%group_by(Class)%>%summarise(count=n(),Region='All')
bind_rows(d1,d2)
Class Region count
<fctr> <chr> <int>
1 0-50000 Nordeste 9
2 0-50000 Norte 8
3 0-50000 Sul 6
4 50000-100000 Nordeste 1
5 50000-100000 Norte 1
6 50000-100000 Sul 2
7 >100000 Norte 1
8 >100000 Sul 2
9 0-50000 All 23
10 50000-100000 All 4
11 >100000 All 3
答
这里是一个快速和肮脏的方法，可能会更新此之后，使其更加清洁，避免不得不bind_rows()
尝试以下操作：
library(tidyverse)
dat_1 <- dat %>%
mutate(population_breaks = case_when(Population <= 50000 ~ "0-50000",
Population >= 50000 & Population <= 100000 ~ "50000-100000",
Population >= 100000 ~ ">100000")) %>%
group_by(population_breaks) %>%
count(Region)
dat_2 <- dat %>%
mutate(population_breaks = case_when(Population <= 50000 ~ "0-50000",
Population >= 50000 & Population <= 100000 ~ "50000-100000",
Population >= 100000 ~ ">100000")) %>%
group_by(population_breaks) %>%
count(population_breaks) %>%
mutate(Region = "All")
bind_rows(dat_1, dat_2)
其中返回：
# A tibble: 11 x 3
# Groups: population_breaks [3]
population_breaks Region n
<chr> <chr> <int>
1 0-50000 Nordeste 9
2 50000-100000 Nordeste 1
3 >100000 Norte 1
4 0-50000 Norte 8
5 50000-100000 Norte 1
6 >100000 Sul 2
7 0-50000 Sul 6
8 50000-100000 Sul 2
9 >100000 All 3
10 0-50000 All 23
11 50000-100000 All 4
或者，在创建'Class'后，bind_rows（count（dat，Region，Class），count（dat，Class，Region =“all”））'会做。 – jazzurro
@jazzurro很棒:-)谢谢 – Wen
不客气。 :) – jazzurro