您好，我有一个带有权重列的数据框，如示例所示：
df <- tibble::tribble(
~id, ~edu, ~q_d1, ~q_d2_1, ~weight,
1L, 1L, 1L, 0L, 1740,
2L, 1L, 1L, 0L, 1428,
3L, 2L, 1L, 2L, 496,
4L, 2L, 1L, 2L, 550,
5L, 3L, 1L, 1L, 1762,
6L, 4L, 1L, 0L, 1004,
7L, 5L, 1L, 0L, 522,
8L, 3L, 2L, 0L, 1099,
9L, 4L, 2L, 2L, 1295
)
我使用 srvyr 包来计算组的汇总统计数据。我的脚本：
sv_design_test <- df %>%
srvyr::as_survey_design(weights = weight)
sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone" ,
q_d2_1 == 0 ~ "No smartphone" ,
TRUE ~ NA_character_)) %>%
group_by(smartphone) %>%
summarize(proportion = srvyr::survey_mean(),
total = srvyr::survey_total(),
total_unweighted = srvyr::unweighted(n())) %>%
select(-proportion_se, -total_se )
输出：
# A tibble: 3 x 4
smartphone proportion total total_unweighted
<chr> <dbl> <dbl> <int>
1 No Internet 0.242 2394 2
2 No smartphone 0.474 4694 4
3 smartphone 0.284 2808 3
但是当我将教育 (edu) 添加到 group_by 时出现错误：
sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone" ,
q_d2_1 == 0 ~ "No smartphone" ,
TRUE ~ NA_character_)) %>%
group_by(edu, smartphone) %>%
summarize(proportion = srvyr::survey_mean(),
total = srvyr::survey_total(),
total_unweighted = srvyr::unweighted(n())) %>%
select(-proportion_se, -total_se )
错误信息是：
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
contrasts can be applied only to factors with 2 or more levels
您的错误消息（有关对比的错误消息）表明您需要使用因素作为分组变量。在原始数据框中，edu是数字，因此您可以在创建调查设计之前将其转换为因子。
library(tidyverse)
library(srvyr)
# ...
sv_design_test <- df %>%
mutate(edu = as.factor(edu)) %>%
srvyr::as_survey_design(weights = weight)
然后在创建 后smartphone，将其也转换为一个因子：
sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone" ,
q_d2_1 == 0 ~ "No smartphone" ,
TRUE ~ NA_character_)) %>%
mutate(smartphone = as.factor(smartphone))
在第二条错误消息（关于长度的错误消息）中，这是因为您的函数中summarise返回不同的行数。您可以通过单独调用这些函数来验证这一点（错误消息显示它是参数 3，意思是n = unweighted(n())问题所在）。
这将返回 15 行：
sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone",
q_d2_1 == 0 ~ "No smartphone",
TRUE ~ NA_character_)) %>%
mutate(smartphone = as.factor(smartphone)) %>%
group_by(edu, smartphone) %>%
summarise(prop = survey_mean(),
total = survey_total())
#> # A tibble: 15 x 6
#> edu smartphone prop prop_se total total_se
#> <fct> <fct> <dbl> <dbl> <dbl> <dbl>
#> 1 1 No Internet 0 0 0 0
#> 2 1 No smartphone 1 0 3168 2108.
#> 3 1 smartphone 0 0 0 0
#> 4 2 No Internet 0 0 0 0
#> 5 2 No smartphone 0 0 0 0
#> 6 2 smartphone 1 0 1046 693.
#> 7 3 No Internet 0.384 0.355 1099 1099.
#> 8 3 No smartphone 0 0 0 0
#> 9 3 smartphone 0.616 0.355 1762 1762.
#> 10 4 No Internet 0.563 0.369 1295 1295.
#> 11 4 No smartphone 0.437 0.369 1004 1004
#> 12 4 smartphone 0 0 0 0
#> 13 5 No Internet 0 0 0 0
#> 14 5 No smartphone 1 0 522 522
#> 15 5 smartphone 0 0 0 0
虽然这仅返回 7，因为edu和只smartphone出现了 7 种组合，因此只计算了 7 个。
sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone",
q_d2_1 == 0 ~ "No smartphone",
TRUE ~ NA_character_)) %>%
mutate(smartphone = as.factor(smartphone)) %>%
group_by(edu, smartphone) %>%
summarise(n = unweighted(n()))
#> # A tibble: 7 x 3
#> edu smartphone n
#> <fct> <fct> <int>
#> 1 1 No smartphone 2
#> 2 2 smartphone 2
#> 3 3 No Internet 1
#> 4 3 smartphone 1
#> 5 4 No Internet 1
#> 6 4 No smartphone 1
#> 7 5 No smartphone 1
.drop = FALSE内部group_by()通过使用函数的参数，您summarize()甚至可以强制生成数据中未出现的因子水平组合的结果。.dropgroup_by()
sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone",
q_d2_1 == 0 ~ "No smartphone",
TRUE ~ NA_character_)) %>%
mutate(smartphone = as.factor(smartphone)) %>%
group_by(edu, smartphone,
.drop = FALSE) %>%
summarize(prop= srvyr::survey_mean(),
total = srvyr::survey_total(),
total_unweighted = srvyr::unweighted(n()))
#> # A tibble: 15 x 7
#> edu smartphone prop prop_se total total_se total_unweighted
#> <fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 1 No Internet 0 0 0 0 0
#> 2 1 No smartphone 1 0 3168 2108. 2
#> 3 1 smartphone 0 0 0 0 0
#> 4 2 No Internet 0 0 0 0 0
#> 5 2 No smartphone 0 0 0 0 0
#> 6 2 smartphone 1 0 1046 693. 2
#> 7 3 No Internet 0.384 0.355 1099 1099. 1
#> 8 3 No smartphone 0 0 0 0 0
#> 9 3 smartphone 0.616 0.355 1762 1762. 1
#> 10 4 No Internet 0.563 0.369 1295 1295. 1
#> 11 4 No smartphone 0.437 0.369 1004 1004 1
#> 12 4 smartphone 0 0 0 0 0
#> 13 5 No Internet 0 0 0 0 0
#> 14 5 No smartphone 1 0 522 522 1
#> 15 5 smartphone 0 0 0 0 0
您可以创建 2 个不同的汇总数据框，然后将它们连接起来。
complete我正在添加对after 的调用n()来填充缺失的级别。制作两个数据框并将它们连接起来会得到以下结果：
props <- sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone",
q_d2_1 == 0 ~ "No smartphone",
TRUE ~ NA_character_)) %>%
mutate(smartphone = as.factor(smartphone)) %>%
group_by(edu, smartphone) %>%
summarise(prop = survey_mean(),
total = survey_total())
counts <- sv_design_test %>%
dplyr::mutate(smartphone = case_when(
q_d1 == 2 ~ "No Internet",
q_d2_1 > 0 ~ "smartphone",
q_d2_1 == 0 ~ "No smartphone",
TRUE ~ NA_character_)) %>%
mutate(smartphone = as.factor(smartphone)) %>%
group_by(edu, smartphone) %>%
summarise(n = unweighted(n())) %>%
complete(edu, smartphone, fill = list(n = 0))
left_join(props, counts, by = c("edu", "smartphone"))
#> # A tibble: 15 x 7
#> edu smartphone prop prop_se total total_se n
#> <fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 1 No Internet 0 0 0 0 0
#> 2 1 No smartphone 1 0 3168 2108. 2
#> 3 1 smartphone 0 0 0 0 0
#> 4 2 No Internet 0 0 0 0 0
#> 5 2 No smartphone 0 0 0 0 0
#> 6 2 smartphone 1 0 1046 693. 2
#> 7 3 No Internet 0.384 0.355 1099 1099. 1
#> 8 3 No smartphone 0 0 0 0 0
#> 9 3 smartphone 0.616 0.355 1762 1762. 1
#> 10 4 No Internet 0.563 0.369 1295 1295. 1
#> 11 4 No smartphone 0.437 0.369 1004 1004 1
#> 12 4 smartphone 0 0 0 0 0
#> 13 5 No Internet 0 0 0 0 0
#> 14 5 No smartphone 1 0 522 522 1
#> 15 5 smartphone 0 0 0 0 0