Question

我想在 R 中创建人口金字塔。我知道StackOverflow上有很多例子，但我想创建一个也包括人口预测的例子，即按性别划分每个年龄组的栏，按性别和年龄组划分预测。

您可以在此处查看示例：http://geographyblog.eu/wp/the-worlds-population-pyramid-is-changing-shape/

enter image description here

如果有一些关于如何更好地说明这一点的建议（例如使用平滑的线条），也欢迎它们，但我想指出当前的情况和预测。示例数据可在联合国网站上找到：http://esa.un.org/wpp/population-pyramids/population-pyramids_absolute.htm

非常感谢任何帮助。

Answer 1

使用ggplot2和geom_bar以及geom_step可能会少一点临时方法。

如果您更喜欢较旧的版本，可以从wpp2015包（或wpp2012，wpp2010或wpp2008中提取数据。

library("dplyr")
library("tidyr")
library("wpp2015")

#load data in wpp2015
data(popF)
data(popM)
data(popFprojMed)
data(popMprojMed)

#combine past and future female population
df0 <- popF %>% 
  left_join(popFprojMed) %>%
  mutate(gender = "female")

#combine past and future male population, add on female population
df1 <- popM %>% 
  left_join(popMprojMed) %>%
  mutate(gender = "male") %>%
  bind_rows(df0) %>%
  mutate(age = factor(age, levels = unique(age)))

#stack data for ggplot, filter World population and required years
df2 <- df1 %>%
  gather(key = year, value = pop, -country, -country_code, -age, -gender) %>%
  mutate(pop = pop/1e03) %>%
  filter(country == "World", year %in% c(1950, 2000, 2050, 2100))

#add extra rows and numeric age variable for geom_step used for 2100
df2 <- df2 %>% 
  mutate(ageno = as.numeric(age) - 0.5)

df2 <- df2 %>%
  bind_rows(df2 %>% filter(year==2100, age=="100+") %>% mutate(ageno = ageno + 1)) 

df2 
# Source: local data frame [170 x 7]
# 
#    country country_code    age gender  year       pop ageno
#     (fctr)        (int) (fctr)  (chr) (chr)     (dbl) (dbl)
# 1    World          900    0-4   male  1950 171.85124   0.5
# 2    World          900    5-9   male  1950 137.99242   1.5
# 3    World          900  10-14   male  1950 133.27428   2.5
# 4    World          900  15-19   male  1950 121.69274   3.5
# 5    World          900  20-24   male  1950 112.39438   4.5
# 6    World          900  25-29   male  1950  96.59408   5.5
# 7    World          900  30-34   male  1950  83.38595   6.5
# 8    World          900  35-39   male  1950  80.59671   7.5
# 9    World          900  40-44   male  1950  73.08263   8.5
# 10   World          900  45-49   male  1950  63.13648   9.5
# ..     ...          ...    ...    ...   ...       ...   ...

使用标准ggplot函数，您可以获得类似的内容，并根据答案here进行调整：

library("ggplot2")
ggplot(data = df2, aes(x = age, y = pop, fill = year)) +
  #bars for all but 2100
  geom_bar(data = df2 %>% filter(gender == "female", year != 2100) %>% arrange(rev(year)),
           stat = "identity",
           position = "identity") +
  geom_bar(data = df2 %>% filter(gender == "male", year != 2100) %>% arrange(rev(year)),
           stat = "identity",
           position = "identity",
           mapping = aes(y = -pop)) +
  #steps for 2100
  geom_step(data =  df2 %>% filter(gender == "female", year == 2100), 
            aes(x = ageno)) +
  geom_step(data =  df2 %>% filter(gender == "male", year == 2100), 
            aes(x = ageno, y = -pop)) +
  coord_flip() +
  scale_y_continuous(labels = abs)

注意：您需要执行arrange(rev(year))，因为条形图是叠加层。

使用ggthemes套餐，您可以非常接近原始的经济学家情节。

library("ggthemes") 
ggplot(data = df2, aes(x = age, y = pop, fill = year)) +
  #bars for all but 2100
  geom_bar(data = df2 %>% filter(gender == "female", year != 2100) %>% arrange(rev(year)),
           stat = "identity",
           position = "identity") +
  geom_bar(data = df2 %>% filter(gender == "male", year != 2100) %>% arrange(rev(year)),
           stat = "identity",
           position = "identity",
           mapping = aes(y = -pop)) +
  #steps for 2100
  geom_step(data =  df2 %>% filter(gender == "female", year == 2100), 
        aes(x = ageno), size = 1) +
  geom_step(data =  df2 %>% filter(gender == "male", year == 2100), 
        aes(x = ageno, y = -pop), size = 1) +
  coord_flip() +
  #extra style shazzaz
  scale_y_continuous(labels = abs, limits = c(-400, 400), breaks = seq(-400, 400, 100)) +
  geom_hline(yintercept = 0) +
  theme_economist(horizontal = FALSE) +
  scale_fill_economist() +
  labs(fill = "", x = "", y = "")

（我相信你可以更接近，但我现在已经停在这里了。）

Answer 2

你可以使用this question的答案轻松烹饪（这里我使用了@ timriffle的答案以及我的答案）。
首先是一些数据（来自您提供的链接）：

wp <- structure(list(M.1990 = c(325814, 295272, 269351, 265163, 249651, 220027, 196523, 178295, 141789, 115097, 106579, 91763, 77150, 56845, 38053, 25716, 19442), M.2000 = c(319675, 317296, 317072, 290827, 262992, 256378, 241401, 212924, 188905, 169133, 131813, 103162, 90921, 72231, 53449, 32707, 25868), M.2010 = c(328759, 315119, 311456, 312831, 311077, 284258, 255596, 248575, 232217, 202633, 176241, 153494, 114194, 83129, 65266, 43761, 39223), F.1990 = c(308121, 281322, 257432, 254065, 238856, 211943, 188433, 170937, 138358, 112931, 106510, 93425, 82667, 67057, 47679, 37435, 36724), F.2000 = c(298455, 297012, 299757, 277706, 252924, 248127, 233583, 207518, 183646, 165444, 132307, 105429, 96681, 80227, 64956, 45832, 46413), F.2010 = c(307079, 293664, 290598, 293313, 295739, 273379, 247383, 241938, 226914, 201142, 176440, 156283, 121200, 92071, 77990, 56895, 66029)), .Names = c("M.1990", "M.2000", "M.2010", "F.1990", "F.2000", "F.2010"), row.names = c("0-4", "5-9", "10-14", "15-19", "20-24", "25-29", "30-34", "35-39", "40-44", "45-49", "50-54", "55-59", "60-64", "65-69", "70-74", "75-79", "80+"), class = "data.frame")

wp
      M.1990 M.2000 M.2010 F.1990 F.2000 F.2010
0-4   325814 319675 328759 308121 298455 307079
5-9   295272 317296 315119 281322 297012 293664
10-14 269351 317072 311456 257432 299757 290598
15-19 265163 290827 312831 254065 277706 293313
20-24 249651 262992 311077 238856 252924 295739
25-29 220027 256378 284258 211943 248127 273379
30-34 196523 241401 255596 188433 233583 247383
35-39 178295 212924 248575 170937 207518 241938
40-44 141789 188905 232217 138358 183646 226914
45-49 115097 169133 202633 112931 165444 201142
50-54 106579 131813 176241 106510 132307 176440
55-59  91763 103162 153494  93425 105429 156283
60-64  77150  90921 114194  82667  96681 121200
65-69  56845  72231  83129  67057  80227  92071
70-74  38053  53449  65266  47679  64956  77990
75-79  25716  32707  43761  37435  45832  56895
80+    19442  25868  39223  36724  46413  66029

xrange <- range(c(0,wp))
yrange <- range(c(0,nrow(wp)))

然后是绘图部分（在两个面板中）：

par(mfcol=c(1,2))
par(mar=c(5,4,4,0))
plot(NA,type="n", main="Men", xlab="", ylab="", xaxs="i", 
     xlim=rev(xrange), ylim=yrange, axes=FALSE, yaxs="i")
rect(xrange[1],yrange[1],xrange[2],yrange[2], col="cadetblue")
abline(v=seq(0,xrange[2],by=1e5), col="white")
# All years with bars you want to represent filled 
# should be entered in reverse order
polygon(c(0,rep(wp$M.2000,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)),
        col="lightblue",border="lightblue")
polygon(c(0,rep(wp$M.1990,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)), 
        col="darkblue",border="darkblue")
# And those you want with just a border, afterwards:
polygon(c(0,rep(wp$M.2010,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)), 
        col=NA,border="darkred",lwd=2)
axis(1, at=c(0,1e5,2e5,3e5), labels=format(c(0,1e5,2e5,3e5),scientific=FALSE))
axis(2, at=1:nrow(wp)-0.5,labels=row.names(wp),las=2)
box()

par(mar=c(5,0,4,4))
plot(NA,type="n", main="Women", xlab="", ylab="", xaxs="i", 
     xlim=xrange, ylim=yrange, axes=FALSE, yaxs="i")
rect(xrange[1],yrange[1],xrange[2],yrange[2], col="cadetblue")
abline(v=seq(0,xrange[2],by=1e5), col="white")
polygon(c(0,rep(wp$F.2000,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)),
        col="lightblue",border="lightblue")
polygon(c(0,rep(wp$F.1990,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)), 
        col="darkblue",border="darkblue")
polygon(c(0,rep(wp$F.2010,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)), 
        col=NA,border="darkred",lwd=2)
axis(1, at=c(0,1e5,2e5,3e5), labels=format(c(0,1e5,2e5,3e5),scientific=FALSE))
axis(4, at=1:nrow(wp)-0.5,labels=row.names(wp),las=2)
box()

enter image description here

为了避免@Spacedman在评论中强调的问题，你可以在某些年份使用alpha。

library(scales)
[...]
polygon(c(0,rep(wp$M.1990,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)), 
        col=alpha("darkblue",0.4),border="darkblue")
[...]
polygon(c(0,rep(wp$F.1990,each=2), 0), c(0,0,rep(1:nrow(wp),each=2)), 
        col=alpha("darkblue",0.4),border="darkblue")
[...]

enter image description here

R中的人口金字塔w投影

2 个答案: