Question

我有以下几点：

df <- structure(list(treatment = c("control", "control", "control", 
"control", "control", "control", "treated", "treated", "treated", 
"treated", "treated", "treated"), `0610005C13Rik` = c(5L, 2L, 
2L, 5L, 1L, 0L, 6L, 1L, 0L, 5L, 1L, 2L), `0610007P14Rik` = c(300L, 
249L, 166L, 104L, 248L, 136L, 164L, 121L, 191L, 187L, 289L, 169L
), `0610009B22Rik` = c(251L, 158L, 92L, 82L, 239L, 107L, 147L, 
96L, 153L, 200L, 211L, 80L), `0610009L18Rik` = c(42L, 17L, 16L, 
17L, 10L, 6L, 18L, 1L, 15L, 8L, 19L, 13L), `0610009O20Rik` = c(187L, 
77L, 86L, 37L, 81L, 24L, 83L, 57L, 98L, 83L, 113L, 48L), `0610010B08Rik` = c(16L, 
3L, 6L, 3L, 2L, 3L, 3L, 2L, 3L, 2L, 3L, 1L)), .Names = c("treatment", 
"0610005C13Rik", "0610007P14Rik", "0610009B22Rik", "0610009L18Rik", 
"0610009O20Rik", "0610010B08Rik"), row.names = c(NA, -12L), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), vars = "treatment", drop = TRUE, indices = list(
    0:5, 6:11), group_sizes = c(6L, 6L), biggest_group_size = 6L, labels = structure(list(
    treatment = c("control", "treated")), row.names = c(NA, -2L
), class = "data.frame", vars = "treatment", drop = TRUE, .Names = "treatment"))

看起来像这样：

Source: local data frame [12 x 7]
Groups: treatment [2]

   treatment `0610005C13Rik` `0610007P14Rik` `0610009B22Rik` `0610009L18Rik` `0610009O20Rik` `0610010B08Rik`
       <chr>           <int>           <int>           <int>           <int>           <int>           <int>
1    control               5             300             251              42             187              16
2    control               2             249             158              17              77               3
3    control               2             166              92              16              86               6
4    control               5             104              82              17              37               3
5    control               1             248             239              10              81               2
6    control               0             136             107               6              24               3
7    treated               6             164             147              18              83               3
8    treated               1             121              96               1              57               2
9    treated               0             191             153              15              98               3
10   treated               5             187             200               8              83               2
11   treated               1             289             211              19             113               3
12   treated               2             169              80              13              48               1

我想要做的是根据分组mean计算treatment和 http://codepen.io/gsound/ 。简历基本上是 ~~mean / sd~~ sd / mean。最终的预期结果如下：

gene_symbol  control.mean    treated.mean control.cv treated.cv
0610005C13Rik      2.5000        2.500000 0.829457    ...
0610007P14Rik    200.5000      186.833333  ...        ...
... etc ...

我怎么能用dplyr做到这一点？

Answer 1

我们可以gather然后获取mean/sd

library(tidyverse)
df %>%
  gather(gene_symbol, Val, -treatment) %>%
  group_by(treatment, gene_symbol) %>% 
  summarise(Mean = mean(Val), cv = sd(Val)/mean(Val)) %>%
  gather(Var1, Val, -treatment,-gene_symbol) %>% 
  unite(new, treatment, Var1) %>%
  spread(new, Val)
# A tibble: 6 × 5
#    gene_symbol control_cv control_Mean treated_cv treated_Mean
#*         <chr>      <dbl>        <dbl>      <dbl>        <dbl>
#1 0610005C13Rik  0.8294577       2.5000  0.9715966     2.500000
#2 0610007P14Rik  0.3809605     200.5000  0.2992429   186.833333
#3 0610009B22Rik  0.4823019     154.8333  0.3582799   147.833333
#4 0610009L18Rik  0.6983225      18.0000  0.5515103    12.333333
#5 0610009O20Rik  0.6996217      82.0000  0.3040676    80.333333
#6 0610010B08Rik  0.9672317       5.5000  0.3499271     2.333333

或另一种选择是将mean，cv与summarise_all匹配，然后重塑为“长”格式并将其重新转换回“广泛”

df %>% 
   group_by(treatment) %>%
   summarise_all(funs(mean = mean(.), cv = sd(.)/mean(.))) %>%
   gather(Var, Val, -treatment) %>% 
   separate(Var, into = c('gene_symbol', 'Var2')) %>% 
   unite(new, treatment, Var2) %>%
   spread(new, Val)
# A tibble: 6 × 5
#    gene_symbol control_cv control_mean treated_cv treated_mean
#*         <chr>      <dbl>        <dbl>      <dbl>        <dbl>
#1 0610005C13Rik  0.8294577       2.5000  0.9715966     2.500000
#2 0610007P14Rik  0.3809605     200.5000  0.2992429   186.833333
#3 0610009B22Rik  0.4823019     154.8333  0.3582799   147.833333
#4 0610009L18Rik  0.6983225      18.0000  0.5515103    12.333333
#5 0610009O20Rik  0.6996217      82.0000  0.3040676    80.333333
#6 0610010B08Rik  0.9672317       5.5000  0.3499271     2.333333

或者我们可以使用melt/dcast

中的data.table执行此操作

library(data.table)
dcast(melt(setDT(df), id.var = "treatment", variable.name = "gene_symbol"
      )[, .(mean = mean(value), cv = sd(value)/mean(value)), .(treatment, gene_symbol)
        ], gene_symbol~treatment, value.var = c('mean', 'cv'))
# gene_symbol mean_control mean_treated cv_control cv_treated
#1: 0610005C13Rik       2.5000     2.500000  0.8294577  0.9715966
#2: 0610007P14Rik     200.5000   186.833333  0.3809605  0.2992429
#3: 0610009B22Rik     154.8333   147.833333  0.4823019  0.3582799
#4: 0610009L18Rik      18.0000    12.333333  0.6983225  0.5515103
#5: 0610009O20Rik      82.0000    80.333333  0.6996217  0.3040676
#6: 0610010B08Rik       5.5000     2.333333  0.9672317  0.3499271

编辑：反映OP公式中的变化

Answer 2

这是一种使用连接的方法

library("tidyverse")

df %>% gather(key = gene_symbol, value = value,-treatment) %>%
  group_by(treatment, gene_symbol) %>%
  summarise(mean = mean(value), cv = mean / sd(value)) %>%
  ungroup() %>% 
  left_join(
    x = filter(., treatment == "control"),
    y = filter(., treatment == "treated"),
    by = "gene_symbol",
    suffix = c(".control", ".treated")
    ) %>% 
  select(-starts_with("treatment"))

如何使用dplyr在分组tibble上应用多个函数

2 个答案: