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应用group_by和SUMMARE(sum),但保留包含非相关冲突数据的列?

summarize dplyr tidyverse group-by r

mckisa · 技术社区 · 7 年前

我的问题与 Applying group_by and summarise on data while keeping all the columns' info

Label <- c("203c","203c","204a","204a","204a","204a","204a","204a","204a","204a")
Type <- c("wholefish","flesh","flesh","fleshdelip","formula","formuladelip",
          "formula","formuladelip","wholefish", "wholefishdelip")
Proportion <- c(1,1,0.67714,0.67714,0.32285,0.32285,0.32285, 
                0.32285, 0.67714,0.67714)
N <- (1:10)
C <- (1:10)
Code <- c("c","a","a","b","a","b","c","d","c","d")

df <- data.frame(Label,Type, Proportion, N, C, Code)
df

   Label           Type Proportion  N  C Code
1   203c      wholefish     1.0000  1  1    c
2   203c          flesh     1.0000  2  2    a
3   204a          flesh     0.6771  3  3    a
4   204a     fleshdelip     0.6771  4  4    b
5   204a        formula     0.3228  5  5    a
6   204a   formuladelip     0.3228  6  6    b
7   204a        formula     0.3228  7  7    c
8   204a   formuladelip     0.3228  8  8    d
9   204a      wholefish     0.6771  9  9    c
10  204a wholefishdelip     0.6771 10 10    d

total <- df %>% 
  #where the Label and Code are the same the Proportion, N and C 
  #should be added together respectively
  group_by(Label, Code) %>% 
  #total proportion should add up to 1 
  #my way of checking that the correct task has been completed
  summarise_if(is.numeric, sum)

# A tibble: 6 x 5
# Groups:   Label [?]
   Label   Code Proportion     N     C
  <fctr> <fctr>      <dbl> <int> <int>
1   203c      a    1.00000     2     2
2   203c      c    1.00000     1     1
3   204a      a    0.99999     8     8
4   204a      b    0.99999    10    10
5   204a      c    0.99999    16    16
6   204a      d    0.99999    18    18

直到这里,我得到了我想要的。现在我想包括列类型,尽管它被排除在外,因为值是冲突的。这是我想要的结果

# A tibble: 6 x 5
# Groups:   Label [?]
   Label   Code Proportion     N     C    Type
  <fctr> <fctr>      <dbl> <int> <int>  <fctr>
1   203c      a    1.00000     2     2    wholefish
2   203c      c    1.00000     1     1    flesh
3   204a      a    0.99999     8     8    flesh_formula
4   204a      b    0.99999    10    10    fleshdelip_formuladelip
5   204a      c    0.99999    16    16    wholefish_formula
6   204a      d    0.99999    18    18    wholefishdelip_formuladelip

我试过了 ungroup() 以及 mutate unite 但如果没有任何效果,我们将不胜感激

3 回复 | 直到 7 年前

akuiper 7 年前

以下是另外两个选项:

1) 将列嵌套到一列中,然后通过检查数据类型自定义摘要:

df %>% 
    group_by(Label, Code) %>% nest() %>% 
    mutate(data = map(data, 
        ~ as.tibble(map(.x, ~ if(is.numeric(.x)) sum(.x) else paste(.x, collapse="_")))
          )
    ) %>% unnest()

# A tibble: 6 x 6
#   Label   Code                        Type Proportion     N     C
#  <fctr> <fctr>                       <chr>      <dbl> <int> <int>
#1   203c      c                   wholefish    1.00000     1     1
#2   203c      a                       flesh    1.00000     2     2
#3   204a      a               flesh_formula    0.99999     8     8
#4   204a      b     fleshdelip_formuladelip    0.99999    10    10
#5   204a      c           formula_wholefish    0.99999    16    16
#6   204a      d formuladelip_wholefishdelip    0.99999    18    18

2) 单独总结,然后加入结果:

numeric <- df %>% 
    group_by(Label, Code) %>% 
    summarise_if(is.numeric, sum)

character <- df %>% 
    group_by(Label, Code) %>% 
    summarise_if(~ is.character(.) || is.factor(.), ~ paste(., collapse="_"))

inner_join(numeric, character, by = c("Label", "Code"))
# A tibble: 6 x 6
# Groups:   Label [?]
#   Label   Code Proportion     N     C                        Type
#  <fctr> <fctr>      <dbl> <int> <int>                       <chr>
#1   203c      a    1.00000     2     2                       flesh
#2   203c      c    1.00000     1     1                   wholefish
#3   204a      a    0.99999     8     8               flesh_formula
#4   204a      b    0.99999    10    10     fleshdelip_formuladelip
#5   204a      c    0.99999    16    16           formula_wholefish
#6   204a      d    0.99999    18    18 formuladelip_wholefishdelip

Jake Thompson 7 年前

tidyverse公司 解决方案 group_by 声明相同。关键是使用 mutate_if 对于每个变量类型,首先(即数字、字符),然后获得不同的行。

library(tidyverse)
#> Loading tidyverse: ggplot2
#> Loading tidyverse: tibble
#> Loading tidyverse: tidyr
#> Loading tidyverse: readr
#> Loading tidyverse: purrr
#> Loading tidyverse: dplyr
#> Conflicts with tidy packages ----------------------------------------------
#> filter(): dplyr, stats
#> lag():    dplyr, stats

Label <- c("203c", "203c", "204a", "204a", "204a", "204a", "204a", "204a",
  "204a", "204a")
Type <- c("wholefish", "flesh", "flesh", "fleshdelip", "formula", "formuladelip",
  "formula", "formuladelip", "wholefish", "wholefishdelip")
Proportion <- c(1, 1, 0.67714, 0.67714, 0.32285, 0.32285, 0.32285, 0.32285,
  0.67714, 0.67714)
N <- (1:10)
C <- (1:10)
Code <- c("c", "a", "a", "b", "a", "b", "c", "d", "c", "d")

df <- data_frame(Label, Type, Proportion, N, C, Code)
df
#> # A tibble: 10 x 6
#>    Label           Type Proportion     N     C  Code
#>    <chr>          <chr>      <dbl> <int> <int> <chr>
#>  1  203c      wholefish    1.00000     1     1     c
#>  2  203c          flesh    1.00000     2     2     a
#>  3  204a          flesh    0.67714     3     3     a
#>  4  204a     fleshdelip    0.67714     4     4     b
#>  5  204a        formula    0.32285     5     5     a
#>  6  204a   formuladelip    0.32285     6     6     b
#>  7  204a        formula    0.32285     7     7     c
#>  8  204a   formuladelip    0.32285     8     8     d
#>  9  204a      wholefish    0.67714     9     9     c
#> 10  204a wholefishdelip    0.67714    10    10     d

df %>%
  group_by(Label, Code) %>%
  mutate_if(is.numeric, sum) %>%
  mutate_if(is.character, funs(paste(unique(.), collapse = "_"))) %>%
  distinct()
#> # A tibble: 6 x 6
#> # Groups:   Label, Code [6]
#>   Label                        Type Proportion     N     C  Code
#>   <chr>                       <chr>      <dbl> <int> <int> <chr>
#> 1  203c                   wholefish    1.00000     1     1     c
#> 2  203c                       flesh    1.00000     2     2     a
#> 3  204a               flesh_formula    0.99999     8     8     a
#> 4  204a     fleshdelip_formuladelip    0.99999    10    10     b
#> 5  204a           formula_wholefish    0.99999    16    16     c
#> 6  204a formuladelip_wholefishdelip    0.99999    18    18     d

Mako212 7 年前

这是 data.table 解决方案,我假设你想要 mean() 比例,因为这些分组比例可能不是相加的。

setDT(df)

df[, .(Type =paste(Type,collapse="_"), 
  Proportion=mean(Proportion),N= sum(N),C=sum(C)), by=.(Label,Code)]
  [order(Label)]

   Label Code                        Type Proportion  N  C
1:  203c    c                   wholefish   1.000000  1  1
2:  203c    a                       flesh   1.000000  2  2
3:  204a    a               flesh_formula   0.499995  8  8
4:  204a    b     fleshdelip_formuladelip   0.499995 10 10
5:  204a    c           formula_wholefish   0.499995 16 16
6:  204a    d formuladelip_wholefishdelip   0.499995 18 18

dplyr 解决方案,但它有效:

df %>% group_by(Label, Code) %>% 
  mutate(Type = paste(Type,collapse="_")) %>% 
  group_by(Label,Type,Code) %>% 
  summarise(N=sum(N),C=sum(C),Proportion=mean(Proportion))

注意,这里的关键是在创建组合后重新分组 Type

   Label                        Type   Code     N     C Proportion
  <fctr>                       <chr> <fctr> <int> <int>      <dbl>
1   203c                       flesh      a     2     2   1.000000
2   203c                   wholefish      c     1     1   1.000000
3   204a               flesh_formula      a     8     8   0.499995
4   204a     fleshdelip_formuladelip      b    10    10   0.499995
5   204a           formula_wholefish      c    16    16   0.499995
6   204a formuladelip_wholefishdelip      d    18    18   0.499995