use*_*890 11 group-by r count dplyr
这是我的例子
mydf<-data.frame('col_1'=c('A','A','B','B'), 'col_2'=c(100,NA, 90,30))
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我想分组col_1 并计算非NA元素col_2
我想这样做dplyr.
以下是我搜索SO后尝试的内容:
mydf %>% group_by(col_1) %>% summarise_each(funs(!is.na(col_2)))
mydf %>% group_by(col_1) %>% mutate(non_na_count = length(col_2, na.rm=TRUE))
mydf %>% group_by(col_1) %>% mutate(non_na_count = count(col_2, na.rm=TRUE))
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没有任何效果.有什么建议?
Ric*_*ord 33
你可以用它
mydf %>% group_by(col_1) %>% summarise(non_na_count = sum(!is.na(col_2)))
# A tibble: 2 x 2
col_1 non_na_count
<fctr> <int>
1 A 1
2 B 2
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我们可以 filter在'col_2'中设置NA元素,然后执行count'col_1'
mydf %>%
filter(!is.na(col_2)) %>%
count(col_1)
# A tibble: 2 x 2
# col_1 n
# <fctr> <int>
#1 A 1
#2 B 2
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或使用 data.table
library(data.table)
setDT(mydf)[, .(non_na_count = sum(!is.na(col_2))), col_1]
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或aggregate从base R
aggregate(cbind(col_2 = !is.na(col_2))~col_1, mydf, sum)
# col_1 col_2
#1 A 1
#2 B 2
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或使用 table
table(mydf$col_1[!is.na(mydf$col_2)])
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library(knitr)
library(dplyr)
mydf <- data.frame("col_1" = c("A", "A", "B", "B"),
"col_2" = c(100, NA, 90, 30))
mydf %>%
group_by(col_1) %>%
select_if(function(x) any(is.na(x))) %>%
summarise_all(funs(sum(is.na(.)))) -> NA_mydf
kable(NA_mydf)
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