如果我想按主题和每个日期获取小计，并根据主题总数排列主题，我该怎么办？
最终输出应该如下所示（蓝色部分是我们需要添加的部分，也是总 ELA (23) 可以使用代码构建示例表：df <- structure(list(Subject = c("Math", "Math", "Math", "Math", "ELA",
"ELA", "ELA"), date = c(1, 7, 14, 21, 1, 7, 21), A = c(1, 2,
0, 9, 2, 6, 0), B = c(3, 5, 5, 1, 0, 5, 0), C = c(2, 1, 0, 8,
0, 0, 0), D = c(0, 0, 2, 8, 0, 8, 2)), row.names = c(NA, -7L), class = c("tbl_df",
"tbl", "data.frame"))
【问题讨论】：
标签： r
使用包Janitor快速完成
df %>% mutate(date = as.character(date)) %>%
group_split(Subject) %>%
map_df(., janitor::adorn_totals, fill = "All Dates", name = "Sub-Total") %>%
adorn_totals(where = "col")
Subject date A B C D Total
ELA 1 2 0 0 0 2
ELA 7 6 5 0 8 19
ELA 21 0 0 0 2 2
Sub-Total All Dates 8 5 0 10 23
Math 1 1 3 2 0 6
Math 7 2 5 1 0 8
Math 14 0 5 0 2 7
Math 21 9 1 8 8 26
Sub-Total All Dates 12 14 11 10 47
如果你不将列date改成一个字符，它也会被总计
【讨论】：
这是一个基本的 R 解决方案。主要功能有
by 和 addmargins，计算 Subject 的每组总数和行总数；lapply) 将列总计作为第一行。其余代码将所有内容放在一起。
res <- by(df[-1], df[1], FUN = function(x){
x <- as.matrix(x)
rownames(x) <- x[, 1]
addmargins(x[, -1], margin = 1:2)
})
res <- lapply(seq_along(res), function(i){
x <- as.data.frame(res[[i]])
row.names(x)[row.names(x) == "Sum"] <- "All dates"
y <- cbind.data.frame(Subject = names(res)[i], date = row.names(x), x)
names(y)[ncol(y)] <- "Total"
y[order(y[["Total"]], decreasing = TRUE), ]
})
i <- sapply(res, '[', 1, "Total")
res <- do.call(rbind.data.frame, res[order(i, decreasing = TRUE)])
row.names(res) <- NULL
res
# Subject date A B C D Total
#1 Math All dates 12 14 11 10 47
#2 Math 21 9 1 8 8 26
#3 Math 7 2 5 1 0 8
#4 Math 14 0 5 0 2 7
#5 Math 1 1 3 2 0 6
#6 ELA All dates 8 5 0 10 23
#7 ELA 7 6 5 0 8 19
#8 ELA 1 2 0 0 0 2
#9 ELA 21 0 0 0 2 2
【讨论】：
addmargins。
这是一种 dplyr 方式：
library(dplyr)
c_order <- c('All dates', 1, 7, 14, 21)
df %>%
group_by(Subject) %>%
summarise(across(A:D, sum)) %>%
mutate(date = 'All dates', .after = 'Subject') %>%
bind_rows(df %>% mutate(date = as.character(date))) %>%
arrange(Subject, match(date, c_order)) %>%
mutate(Total = rowSums(select(., A:D)))
# Subject date A B C D Total
# <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 ELA All dates 8 5 0 10 23
#2 ELA 1 2 0 0 0 2
#3 ELA 7 6 5 0 8 19
#4 ELA 21 0 0 0 2 2
#5 Math All dates 12 14 11 10 47
#6 Math 1 1 3 2 0 6
#7 Math 7 2 5 1 0 8
#8 Math 14 0 5 0 2 7
#9 Math 21 9 1 8 8 26
首先为每个Subject sum 列A:D 添加一个值为“所有日期”的列“date”。将此绑定到原始数据框并根据需要的顺序排列数据并执行逐行求和。
【讨论】：
这行得通吗：
library(dplyr)
library(tidyr)
df %>% rowwise() %>% mutate(Total = sum(c_across(A:D))) %>%
bind_rows(df %>% rowwise() %>% mutate(Total = sum(c_across(A:D))) %>% group_by(Subject) %>% summarise_at(vars(A:Total), sum)) %>%
mutate(date = replace_na(date, 'All Dates')) %>% arrange(Subject, desc(Total))
# A tibble: 9 x 7
# Rowwise:
Subject date A B C D Total
<chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 ELA All Dates 8 5 0 10 23
2 ELA 7 6 5 0 8 19
3 ELA 1 2 0 0 0 2
4 ELA 21 0 0 0 2 2
5 Math All Dates 12 14 11 10 47
6 Math 21 9 1 8 8 26
7 Math 7 2 5 1 0 8
8 Math 14 0 5 0 2 7
9 Math 1 1 3 2 0 6
【讨论】：