通过dplyr计算中位数与R中的聚集数

Question

你好：当我通过dplyr/tidyr与聚合生成数据时，我看到的数据集看起来像下面创建的数据集略有不同。任何人都可以解释这个区别吗谢谢！因为aggregate正在下降整个行，其中的任何值是NA，即使在该行中的一些变量包含数据

#dataset 
out2<-structure(list(d3 = structure(c(1L, 2L, NA, NA, 1L, 1L, NA, 
2L,NA,3L,1L, NA, NA, 1L, 3L, NA, 1L, 2L, 3L, 2L, 1L, 3L, 2L, 3L, 1L), .Label 
=  c("Professional journalist", "Elected politician", "Online blogger"), 
class = "factor"), Accessible = c(3, 5, 2,NA, 1, 2, NA, 3, NA, 4, 2, 5, NA, 
3, 4, NA, 2, NA, 3, 4, 4, 4,2, 2, 2), Information = c(1, 2, 1, NA, 4, 1, NA, 
2, NA, 2, 1, 1, NA, 4, 1, NA, 1, 1, 1, 3, 1, 3, 3, 4, 1), Responsive = c(5, 
4, 6, NA, 2, 3, NA, 1, NA, 5, 4, 4, NA, 6, 3, NA, 4, NA, 2, 2, 6, 2, 1, 1, 
3), Debate = c(6, 3, 4, NA, 3, 4, NA, 5, NA, 6, 5,6, NA, 1, 5, NA, 5, 2, NA, 
1, 5, 6, 5, 5, 7), Officials = c(2,1, 5, NA, 5, 5, NA, 6, NA, 3, 6, 2, NA, 2, 
2, NA, 6, 3, NA, 5,2, 5, 4, 6, 5), Social = c(7, 6, 7, NA, 7, 7, NA, 4, NA, 
7, 7,                                                         
7, NA, 7, 7, NA, 7, NA, NA, 7, 7, 1, 6, 7, 6), `Trade-Offs` = c(4, 
7, 3, NA, 6, 6, NA, 7, NA, 1, 3, 3, NA, 5, 6, NA, 3, NA, NA, 
6, 3, 7, 7, 3, 4)), .Names = c("d3", "Accessible", "Information",  
"Responsive", "Debate", "Officials", "Social", "Trade-Offs"), row.names = 
c(171L, 126L, 742L, 379L, 635L, 3L, 303L, 419L, 324L, 97L, 758L, 136L, 
770L, 405L, 101L, 674L, 386L, 631L, 168L, 590L, 731L, 387L, 673L, 208L, 
728L), class = "data.frame") 

#Find Medians via tidyR and dplyr 
test<-out2 %>% 
gather(variable, value, -1) %>% 
filter(is.na(d3)==FALSE)%>% 
group_by(d3, variable) %>% 
summarise(value=median(value, na.rm=TRUE)) 

#dataframe 
test<-data.frame(test) 

#find Medians via aggregate 
test2<-aggregate(.~d3, data=out2, FUN=median, na.rm=TRUE) 

#Gather for plotting 
test2<-test2 %>% 
gather(variable, value, -d3) 

#Plot Medians via tidyr 
ggplot(test, aes(x=d3, y=value,  
group=d3))+facet_wrap(~variable)+ 
geom_bar(stat='identity')+labs(title='Medians via TidyR') 

#Plot Medians Via aggregate 
ggplot(test2, aes(x=d3, y=value,  
group=d3))+facet_wrap(~variable)+geom_bar(stat='identity')+ 
labs(title='Medians via Aggregate') 

#Compare Debate, Information and Responsive 

Answer 1

aggregate通过产生的结果是不同的。

您可以通过指定na.action参数的值来解决此问题，如this accepted answer中所述。这将是：

test2<-aggregate(.~d3, data=out2, FUN=median, na.rm = TRUE, na.action=NULL) 
test2<-test2 %>% 
    gather(variable, value, -d3) 

确认的结果是相同的：

identical(as.data.frame(test %>% arrange(d3, variable, value)), 
      as.data.frame(test2 %>% arrange(d3, variable, value))) 
[1] TRUE