如何使用 tidycensus 和 tidyverse 准确聚合子组误差值？

如何解决如何使用 tidycensus 和 tidyverse 准确聚合子组误差值？

我正在尝试使用 R 中的美国社区调查来计算明尼苏达州每个县按种族划分的 20 岁以下人口。使用 Tidycensus 我知道这可以使用 R 中每个种族和年龄组的 B01001H 变量来完成。但是我需要汇总每个种族群体 20 岁以下人群的所有变量。根据此网页 (https://www.census.gov/content/dam/Census/library/publications/2018/acs/acs_general_handbook_2018_ch08.pdf)，虽然汇总估计值只是每个子组值的总和，但汇总误差幅度需要我计算以下公式：

MOE = sqrt(moe_1^2 + moe_2^2 + ... + moe_n^2)

对于子组内的每个 MOE。那么我究竟如何使用 tidyverse 来准确计算这个聚合的 MOE 值？

到目前为止，我的代码如下所示：

## age race 
age_vars_male = c(w1="B01001H_003",w2="B01001H_004",w3="B01001H_005",w4="B01001H_006",b1="B01001B_003",b2="B01001B_004",b3="B01001B_005",b4="B01001B_006",AN1="B01001C_003",AN2="B01001C_004",AN3="B01001C_005",AN4="B01001C_006",AS1="B01001D_003",AS2="B01001D_004",AS3="B01001D_005",AS4="B01001D_006",H1="B01001I_003",H2="B01001I_004",H3="B01001I_005",H4="B01001I_006")

## obtaining variables listed above for MN counties
pop_un20 <- get_acs(geography = "county",variables = age_vars_male,state = "MN",geometry=T)

pop_un20 = pop_un20 %>% mutate(Race = case_when(variable %in% c("w1","w2","w3","w4") ~ "White",variable %in% c("b1","b2","b3","b4") ~ "Black",variable %in% c("AN1","AN2","AN3","AN4") ~"AI/AN",variable %in% c("AS1","AS2","AS3","AS4") ~"Asian",variable %in% c("H1","H2","H3","H4") ~"Hispanic/Latino"),moe_sqrd = moe^2) %>% select(-variable)


moe_aggregate = pop_un20 %>% group_by(NAME,Race) %>% summarise(moe_aggregate = sqrt(sum(moe_sqrd,na.rm = T))) %>% st_set_geometry(NULL)
est_aggregate = pop_un20 %>% group_by(NAME,Race) %>% summarise(estimate_aggregate = sum(estimate,na.rm = T)) %>% st_set_geometry(NULL)


pop_under20  = pop_un20 %>% right_join(moe_aggregate,by = c("NAME","Race")) %>% right_join(est_aggregate,"Race")) %>%
  select(-estimate,-moe,moe_sqrd)

我已经计算了我的要求，首先为 moe squared 创建一个列，然后取每个组和种族的总和的平方根。但是，有没有办法一次性做到这一点？

解决方法

您可以使用 summarise 直接在数据中添加新列，而不是 mutate 和 join。

library(dplyr)

pop_under20 <- pop_un20 %>% 
  group_by(NAME,Race) %>% 
  mutate(moe_aggregate = sqrt(sum(moe_sqrd,na.rm = TRUE)),estimate_aggregate = sum(estimate,na.rm = TRUE))

,

tidycensus 有一个函数，moe_sum()，that does this for you。调整您的代码：

library(tidycensus)
library(tidyverse)

age_vars_male = c(w1="B01001H_003",w2="B01001H_004",w3="B01001H_005",w4="B01001H_006",b1="B01001B_003",b2="B01001B_004",b3="B01001B_005",b4="B01001B_006",AN1="B01001C_003",AN2="B01001C_004",AN3="B01001C_005",AN4="B01001C_006",AS1="B01001D_003",AS2="B01001D_004",AS3="B01001D_005",AS4="B01001D_006",H1="B01001I_003",H2="B01001I_004",H3="B01001I_005",H4="B01001I_006")

## obtaining variables listed above for MN counties
pop_un20 <- get_acs(geography = "county",variables = age_vars_male,state = "MN")

pop_un20_grouped <- pop_un20 %>% 
  mutate(Race = case_when(variable %in% c("w1","w2","w3","w4") ~ "White",variable %in% c("b1","b2","b3","b4") ~ "Black",variable %in% c("AN1","AN2","AN3","AN4") ~"AI/AN",variable %in% c("AS1","AS2","AS3","AS4") ~"Asian",variable %in% c("H1","H2","H3","H4") ~"Hispanic/Latino")) %>%
           group_by(NAME,Race) %>%
           summarize(group_estimate = sum(estimate,na.rm = TRUE),group_moe = moe_sum(moe = moe,estimate = estimate,na.rm = TRUE))

pop_un20_grouped

# A tibble: 435 x 4
# Groups:   NAME [87]
   NAME                     Race            group_estimate group_moe
   <chr>                    <chr>                    <dbl>     <dbl>
 1 Aitkin County,Minnesota AI/AN                       70      24.5
 2 Aitkin County,Minnesota Asian                        3      14.9
 3 Aitkin County,Minnesota Black                        5      15.1
 4 Aitkin County,Minnesota Hispanic/Latino             71      22.6
 5 Aitkin County,Minnesota White                     1223      59.9
 6 Anoka County,Minnesota  AI/AN                      322      97.9
 7 Anoka County,Minnesota  Asian                     1983     219. 
 8 Anoka County,Minnesota  Black                     4015     303. 
 9 Anoka County,Minnesota  Hispanic/Latino           3176     219. 
10 Anoka County,Minnesota  White                    30557     478. 
# … with 425 more rows