Question

当我执行以下操作时：

data_control %>%
  group_by(politics, partner_politics) %>%
  summarize(pd_sent_amount = mean(as.numeric(pd_sent_amount)),
            n = n(),
            pd_sent_amount_sd =  sd(as.numeric(pd_sent_amount), na.rm = T)
           )

我得到当前输出：

# A tibble: 4 x 5
# Groups:   politics [?]
  politics   partner_politics pd_sent_amount     n pd_sent_amount_sd
  <fct>      <fct>                     <dbl> <int>             <dbl>
1 Democrat   Democrat                  0.598    76                NA
2 Democrat   Republican                0.479    34                NA
3 Republican Democrat                  0.404    34                NA
4 Republican Republican                0.404    70                NA

不确定为什么要进行标准偏差计算，因为我可以按组手动进行计算，例如：

test = subset(data_control, politics == "Democrat" & partner_politics == "Democrat")
with(test, sd(pd_sent_amount) / sqrt(nrow(test)))

> with(test, sd(pd_sent_amount) / sqrt(nrow(test)))
[1] 0.05008275

以下是数据样本的副本：

structure(list(politics = structure(c(1L, 2L, 1L, 2L, 1L, 1L, 
2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 
1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L), .Label = c("Democrat", "Republican"
), class = "factor"), partner_politics = structure(c(2L, 1L, 
2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 1L), .Label = c("Democrat", 
"Republican"), class = "factor"), pd_sent_amount = c(0.2, 0, 
0.75, 0, 0, 0, 0, 0, 0.5, 0, 1, 0, 1, 0.5, 1, 1, 1, 0.5, 1, 0.5, 
1, 1, 0.25, 0, 0, 0.25, 0, 0, 0.5, 1)), row.names = 5:34, class = "data.frame")

Answer 1

对sd()的调用引用了pd_sent_amount的就地突变（摘要）。为汇总列重新命名。

data_control %>%
  group_by(politics, partner_politics) %>%
  summarize(pd_sent_amount_mean = mean(as.numeric(pd_sent_amount)),
            n = n(),
            pd_sent_amount_sd =  sd(as.numeric(pd_sent_amount), na.rm = T)
           )

dplyr网站上给here的第四个示例提到“新创建的摘要会立即覆盖现有变量”，并且该示例与您的情况相同，并依次调用mean()和{{ 1}}。

标准偏差在dplyr链中显示为NA

1 个答案: