我正在尝试使用rbind合并两个数据帧。以下是两个dfs
ab1
structure(list(Product = c("Black Menthol", "Gold ", "Green ",
"Red "), `Apr 2017` = structure(c(`Black Menthol` = 2L, `Gold ` = 3L,
`Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02", "0.07",
"0.09", "Apr 2017"), class = "factor"), `May 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02",
"0.07", "0.08", "May 2017"), class = "factor"), `Jun 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 4L, `Green ` = 1L, `Red ` = 3L), .Label = c("0", "0.02",
"0.07", "0.08", "Jun 2017"), class = "factor"), `Jul 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02",
"0.07", "0.09", "Jul 2017"), class = "factor"), `Aug 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02",
"0.07", "0.09", "Aug 2017"), class = "factor"), `Sep 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02",
"0.07", "0.09", "Sep 2017"), class = "factor"), `Oct 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02",
"0.07", "0.08", "Oct 2017"), class = "factor"), `Nov 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.02",
"0.07", "0.09", "Nov 2017"), class = "factor"), `Dec 2017` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 4L), .Label = c("0", "0.03",
"0.07", "0.08", "Dec 2017"), class = "factor"), `Jan 2018` = structure(c(`Black Menthol` = 2L,
`Gold ` = 3L, `Green ` = 1L, `Red ` = 3L), .Label = c("0", "0.03",
"0.06", "Jan 2018"), class = "factor")), row.names = c(NA, -4L
), class = "data.frame")
ab2
structure(list(Product = c("Black Menthol", "Black Non-Menthol",
"Green ", "Red "), `Apr 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.07", "0.17", "0.23", "Apr 2017"), class = "factor"), `May 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.06", "0.17", "0.24", "May 2017"), class = "factor"), `Jun 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.07", "0.18", "0.25", "Jun 2017"), class = "factor"), `Jul 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.06", "0.17", "0.24", "Jul 2017"), class = "factor"), `Aug 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.06", "0.16", "0.23", "Aug 2017"), class = "factor"), `Sep 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.06", "0.16", "0.23", "Sep 2017"), class = "factor"), `Oct 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.07", "0.14", "0.22", "Oct 2017"), class = "factor"), `Nov 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.02",
"0.08", "0.15", "0.22", "Nov 2017"), class = "factor"), `Dec 2017` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.03",
"0.09", "0.16", "0.20", "Dec 2017"), class = "factor"), `Jan 2018` = structure(c(`Black Menthol` = 1L,
`Black Non-Menthol` = 3L, `Green ` = 2L, `Red ` = 4L), .Label = c("0.03",
"0.08", "0.16", "0.22", "Jan 2018"), class = "factor")), row.names = c(NA,
-4L), class = "data.frame")
下面是我用来组合它们的代码：
library(shiny)
library(dplyr)
ab3 <- reactive({rbind.data.frame(ab1,ab2)})
ab4<- reactive({ ab3 %>% group_by(`Product`) %>%summarize_all(funs(sum))})
执行此操作时，出现以下错误：Evaluation error: ‘sum’ not meaningful for factors
我想得到的输出是：
Product Apr.2017 May.2017 Jun.2017 Jul.2017 Aug.2017 Sep.2017 Oct.2017
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Black … 0.04 0.04 0.04 0.04 0.04 0.04 0.04
#2 Black … 0.17 0.17 0.18 0.17 0.16 0.16 0.14
#3 Gold 0.07 0.07 0.08 0.07 0.07 0.07 0.07
#4 Green 0.07 0.06 0.07 0.06 0.06 0.06 0.07
#5 Red 0.32 0.32 0.32 0.330 0.32 0.32 0.3
答案 0 :(得分：2)
这是你的追求吗？
library(tidyverse)
bind_rows(ab1, ab2) %>%
group_by(Product) %>%
summarise_all(sum)
## A tibble: 5 x 9
# Product Apr.2017 May.2017 Jun.2017 Jul.2017 Aug.2017 Sep.2017 Oct.2017
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Black … 0.04 0.04 0.04 0.04 0.04 0.04 0.04
#2 Black … 0.17 0.17 0.18 0.17 0.16 0.16 0.14
#3 Gold 0.07 0.07 0.08 0.07 0.07 0.07 0.07
#4 Green 0.07 0.06 0.07 0.06 0.06 0.06 0.07
#5 Red 0.32 0.32 0.32 0.330 0.32 0.32 0.3
## ... with 1 more variable: Nov.2017 <dbl>
这假设ab1和ab2具有相同的列结构。
ab1 <- read.table(text =
" Product 'Apr 2017' 'May 2017' 'Jun 2017' 'Jul 2017' 'Aug 2017' 'Sep 2017' 'Oct 2017' 'Nov 2017'
1 'Black Menthol' 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02
2 Gold 0.07 0.07 0.08 0.07 0.07 0.07 0.07 0.07
3 Green 0 0 0 0 0 0 0 0
4 Red 0.09 0.08 0.07 0.09 0.09 0.09 0.08 0.09
", header = T)
ab2 <- read.table(text =
" Product 'Apr 2017' 'May 2017' 'Jun 2017' 'Jul 2017' 'Aug 2017' 'Sep 2017' 'Oct 2017' 'Nov 2017'
1 'Black Menthol' 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02
2 'Black Non-Menthol' 0.17 0.17 0.18 0.17 0.16 0.16 0.14 0.15
3 Green 0.07 0.06 0.07 0.06 0.06 0.06 0.07 0.08
4 Red 0.23 0.24 0.25 0.24 0.23 0.23 0.22 0.22
", header = T)
要解决factor问题：在绑定行之前，将Product以外的所有列转换为numeric。
bind_rows(
ab1 %>% mutate_at(vars(-Product), function(x) as.numeric(as.character(x))),
ab2 %>% mutate_at(vars(-Product), function(x) as.numeric(as.character(x)))) %>%
group_by(Product) %>%
summarise_all(sum)
## A tibble: 5 x 11
# Product `Apr 2017` `May 2017` `Jun 2017` `Jul 2017` `Aug 2017` `Sep 2017`
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Black … 0.04 0.04 0.04 0.04 0.04 0.04
#2 Black … 0.17 0.17 0.18 0.17 0.16 0.16
#3 "Gold " 0.07 0.07 0.08 0.07 0.07 0.07
#4 "Green… 0.07 0.06 0.07 0.06 0.06 0.06
#5 "Red " 0.32 0.32 0.32 0.330 0.32 0.32
## ... with 4 more variables: `Oct 2017` <dbl>, `Nov 2017` <dbl>, `Dec
## 2017` <dbl>, `Jan 2018` <dbl>