在 r 中将 dataframe 从宽转换为长

Question

I have a dataframe which is like below:

structure(list(A = c(6.53920197406645, 6.12380136266864, 8.01553257692446, 
4.62636832157394, 7.58222133679378), B = c(6.56200038984423, 
6.09642510342734, 7.73715705458708, 4.64560570976, 7.23920390575521
), C = c(6.80800376627205, 7.92368949337286, 8.01633247802198, 
4.87384339969836, 6.83446360366941), D = c(6.69117551163928, 
5.93689715688807, 8.40247900956586, 4.20000164335469, 6.78643597456963
), E = c(6.30449572859692, 5.99369984659008, 7.86273536430256, 
4.11510456695528, 7.11972911832181), eID = c("hsa:5982", "hsa:3310", 
"hsa:7849", "hsa:2978", "hsa:7318")), row.names = c("X1053_at", 
"X117_at", "X121_at", "X1255_g_at", "X1294_at"), class = "data.frame")

I would like to extend this dataframe as in each row I would have:

A X1053_at 6.539202 hsa:5982
B X1053_at 6.562000 hsa:5982

and so on.

How I can do so?

Answer 1

You may try using reshape2::melt

library(dplyr)
library(tibble)

df %>%
  rownames_to_column("something") %>%
  reshape2::melt(id.cols = c(something, eID), variable.name = "a", value.name = "b") %>%
  select(a, something, b, eID)

   a  something        b      eID
1  A   X1053_at 6.539202 hsa:5982
2  A    X117_at 6.123801 hsa:3310
3  A    X121_at 8.015533 hsa:7849
4  A X1255_g_at 4.626368 hsa:2978
5  A   X1294_at 7.582221 hsa:7318
6  B   X1053_at 6.562000 hsa:5982
7  B    X117_at 6.096425 hsa:3310
8  B    X121_at 7.737157 hsa:7849
9  B X1255_g_at 4.645606 hsa:2978
10 B   X1294_at 7.239204 hsa:7318
11 C   X1053_at 6.808004 hsa:5982
12 C    X117_at 7.923689 hsa:3310
13 C    X121_at 8.016332 hsa:7849
14 C X1255_g_at 4.873843 hsa:2978
15 C   X1294_at 6.834464 hsa:7318
16 D   X1053_at 6.691176 hsa:5982
17 D    X117_at 5.936897 hsa:3310
18 D    X121_at 8.402479 hsa:7849
19 D X1255_g_at 4.200002 hsa:2978
20 D   X1294_at 6.786436 hsa:7318
21 E   X1053_at 6.304496 hsa:5982
22 E    X117_at 5.993700 hsa:3310
23 E    X121_at 7.862735 hsa:7849
24 E X1255_g_at 4.115105 hsa:2978
25 E   X1294_at 7.119729 hsa:7318