如何使用dplyr在行组之间划分变量而不列出它们?

时间:2018-02-13 15:19:34

标签: r dplyr

关注此问题How to divide between groups of rows using dplyr?

如果我有这个数据框:

id = c("a","a","b","b","c","c")
condition = c(0,1,0,1,0,1)
gene1 = sample(1:100,6)
gene2 = sample(1:100,6)
#...
geneN = sample(1:100,6)

df = data.frame(id,condition,gene1,gene2,geneN)

我希望按ID进行分组,并将条件== 0的行的值与条件== 1的行除以得到:

df[condition == 0,3:5]/ df[condition == 1,3:5]
#
      gene1     gene2     geneN
1 0.2187500 0.4946237 0.3750000
3 0.4700000 0.6382979 0.5444444
5 0.7674419 0.5471698 2.3750000

我可以按如下方式使用dplyr:

df %>% 
    group_by(id) %>%
    summarise(gene1 = gene1[condition == 0] / gene1[condition == 1],
              gene2 = gene2[condition == 0] / gene2[condition == 1],
              geneN = geneN[condition == 0] / geneN[condition == 1])

但我有例如100个变量如下。如何在不必列出所有基因的情况下做到这一点。

id = c("a","a","b","b","c","c")
condition = c(0,1,0,1,0,1)
genes = matrix(1:600,ncol = 100)
df = data.frame(id,condition,genes)

3 个答案:

答案 0 :(得分:3)

我们可以使用summarise_at将相同的操作应用于多列。

library(dplyr)

df2 <- df %>%
  group_by(id) %>%
  arrange(condition) %>%
  summarise_at(vars(-condition), funs(first(.)/last(.))) %>%
  ungroup()
df2
# # A tibble: 3 x 4
#   id    gene1 gene2 geneN
#   <fct> <dbl> <dbl> <dbl>
# 1 a     0.524 2.28  0.654
# 2 b     1.65  0.616 1.38 
# 3 c     0.578 2.00  2.17 

答案 1 :(得分:1)

你可以尝试

df %>% 
  gather(k,v, -id, -condition) %>% 
  spread(condition, v) %>% 
  mutate(ratio=`0`/`1`) %>% 
  select(id, k, ratio) %>% 
  spread(k, ratio)
  id      gene1     gene2    geneN
1  a  0.3670886 0.5955056 1.192982
2  b  0.4767442 1.2222222 0.125000
3  c 18.2000000 2.0909091 6.000000

将您的数据用于set.seed(123)

答案 2 :(得分:0)

如果您的数据集已排序且没有异常,则可以使用purr::map_dfr

执行此操作
df[paste0("gene",c(1,2,"N"))] %>% map_dfr(~.x[c(F,T)]/.x[c(T,F)])
# # A tibble: 3 x 3
#       gene1    gene2      geneN
#       <dbl>    <dbl>      <dbl>
# 1 0.1764706 1.323944 38.5000000
# 2 0.4895833 0.531250  0.3478261
# 3 0.3278689 2.705882  1.2424242

或其基本R等价物:

as.data.frame(lapply(df[paste0("gene",c(1,2,"N"))],function(x) x[c(F,T)]/x[c(T,F)]))

您可能需要绑定observations,我跳过此步骤,因为它不在预期的输出中。