合并两列的属性并对重复行的值求和

Question

本题由this one稍作修改。

我有一个长表格式的数据框，如下所示：

df1 <- data.frame(ID=c(1,1,1,1,1,1,2,2),
                  name=c("a","c","a","c","a","c","a","c"),
                  value=c("broad",50,"mangrove",50,"mangrove",50,"coniferous",50))

ID name       value
1    a        broad
1    c           50
1    a     mangrove
1    c           50
1    a     mangrove
1    c           50
2    a   coniferous
2    c           50

关于数据：第二行的值50对应第一行的值broad。同样，第四行的值 50 对应于第三行的值 ma​​ngrove 依此类推。简单来说，名称 c 的值strong> 与名称 a 相关。

我想以这样一种方式组合值，以便我可以获得每个名称的对应值，这也会聚合具有相似名称的值：

df2 <- data.frame(ID=c(1,1,2),
                  name=c("c_broad","c_mangrove","c_coniferous"),
                  value=c(50,100,50))

应该是这样的：

ID         name    value
1       c_broad       50
1    c_mangrove      100
2  c_coniferous       50

Answer 1

使用reshape2：

library(reshape2)

df1$grp = cumsum(df1$name == "a")
df2 = dcast(df1, ID + grp ~ name)
df2$c = as.numeric(df2$c)

aggregate(c ~ ID + a, df2, sum)

  ID          a   c
1  1      broad  50
2  2 coniferous  50
3  1   mangrove 100

如果需要，可以更改列名，也可以使用粘贴将“c_”添加到名称中。

Answer 2

使用 tidyverse：

value_a <- df1 %>% dplyr::filter(name=="a") %>% dplyr::pull(value) 
df1 %>%
  dplyr::filter(name=="c") %>% #Modify into a sensible data frame from here
  dplyr::mutate(a = value_a,
         name = stringr::str_c(name, "_" ,a)) %>%
  dplyr::select(-a) %>% # to here
  dplyr::group_by(ID, name) %>%
  dplyr::summarise(value=sum(as.numeric(value)))

# A tibble: 3 x 3
# Groups:   ID [2]
     ID name         value
  <dbl> <chr>        <dbl>
1     1 c_broad         50
2     1 c_mangrove     100
3     2 c_coniferous    50

您在数据框中发现的主要问题是单个列包含名称和值，这是您应该解决的第一件事。我的建议是始终将原始数据框修改为整洁的格式 (https://tidyr.tidyverse.org/articles/tidy-data.html)，然后利用所有 tidyverse 功能、data.table 或您选择的框架。

请注意，时间变量 value_a 可以直接包含在管道中，为了清楚起见，我没有这样做。主要思想是分离不同列中的值和种类，管道中的前三个调用，然后应用通常的 tidyverse 操作。

Answer 3

可能不是最优雅的，但它确实有效：

df1 <- data.frame(ID=c(1,1,1,1,1,1,2,2),
                  name=c("a","c","a","c","a","c","a","c"),
                  value=c("broad",50,"mangrove",50,"mangrove",50,"coniferous",50)
)

df1 %>% group_by( 1+floor((1:n()-1)/2) ) %>%
    summarize(
        ID = ID[1],
        name = paste0( name[2], "_", value[1] ),
        value = as.numeric(value[2])
    ) %>% ungroup %>% select( -1 ) %>% group_by(name) %>%
    mutate( value = sum(value) ) %>%
    unique

这里有一些改进，实际上是人类可读的：

i <- seq( 1, nrow(df1), 2 )
df1 %>% summarise(
            ID = ID[i],
            name = paste0( name[i+1], "_", value[i] ),
            value = as.numeric(value[i+1])
        ) %>% group_by(name) %>%
    summarize(
        ID=ID[1], value = sum( value )
    ) %>% arrange(ID)

Answer 4

基础 R 解决方案：

# Nullify numeric values belonging to a grouping category: grps => character vector
grps <- gsub("\\d+", NA, df1$value)

# Interpolate NA values using prior string value: a => character vector
df1$a <- na.omit(grps)[cumsum(!(is.na(grps)))]

# Split-Apply-Combine aggregation: data.frame => stdout(console)
data.frame(do.call(rbind, lapply(with(df1, split(df1, a)), function(x){
      y <- transform(subset(x, !grepl("\\D+", value)), value = as.numeric(value))
      setNames(
        aggregate(value ~ ID + a, y, FUN = function(z){sum(z, na.rm = TRUE)}),
        c("ID", "a", "c")
        )
      }
    )
  ), 
row.names = NULL
)

Answer 5

附加选项

df1 <- data.frame(ID=c(1,1,1,1,1,1,2,2),
                  name=c("a","c","a","c","a","c","a","c"),
                  value=c("broad",50,"mangrove",50,"mangrove",50,"coniferous",50))

library(tidyverse)
df1 %>% 
  pivot_wider(ID, names_from = name, values_from = value) %>% 
  unnest(c("a", "c")) %>% 
  group_by(ID, name = a) %>% 
  summarise(value = sum(as.numeric(c), na.rm = T), .groups = "drop") 

#> # A tibble: 3 x 3
#>      ID name       value
#>   <dbl> <chr>      <dbl>
#> 1     1 broad         50
#> 2     1 mangrove     100
#> 3     2 coniferous    50

^{由reprex package (v2.0.0) 于 2021-04-12 创建}