Convert data frame row into new data frame column R

0
votes

I have a data frame, cul7_expr, with one row and 22 columns. I want to create a new data frame containing one column with all the numerical values from cul7_expr and another column with all the column names (TCGA.BC...). The new data frame would have 2 columns and 22 rows. However, when I try to take the row of cul7_expr, a warning pops up and the data frame is empty.

Sample data (cul7_expr): 

df <- structure(list(TCGA.BC.A10Q.11A=819.3685,TCGA.BC.A10Q.01A=2757.486, 
TCGA.DD.A1EB.11A=698.5818,TCGA.DD.A1EB.01A=1625.094,TCGA.DD.A1EG.11A=409.9332,
TCGA.DD.A1EG.01A=2221.012,TCGA.DD.A1EH.11A=391.0916,TCGA.DD.A1EH.01A=2122.782, 
TCGA.DD.A1EI.11A=717.2073,TCGA.DD.A1EI.01A=768.7468,TCGA.DD.A3A6.11A=464.6395,
TCGA.DD.A3A6.01A=1175.928,TCGA.DD.A3A8.11A=934.9738,TCGA.DD.A3A8.01A=931.8955,
TCGA.ES.A2HT.11A=599.736,TCGA.ES.A2HT.01A=894.8324,TCGA.FV.A23B.11A=970.1805,
TCGA.FV.A23B.01A=3018.075,TCGA.FV.A3I0.11A=337.222,TCGA.FV.A3I0.01A=3895.477,
TCGA.FV.A3R2.11A=912.8499,TCGA.FV.A3R2.01A=2226.921), 
.Names=c("TCGA.BC.A10Q.11A","TCGA.BC.A10Q.01A","TCGA.DD.A1EB.11A",
"TCGA.DD.A1EB.01A","TCGA.DD.A1EG.11A","TCGA.DD.A1EG.01A","TCGA.DD.A1EH.11A",
"TCGA.DD.A1EH.01A","TCGA.DD.A1EI.11A","TCGA.DD.A1EI.01A","TCGA.DD.A3A6.11A", 
"TCGA.DD.A3A6.01A","TCGA.DD.A3A8.11A","TCGA.DD.A3A8.01A","TCGA.ES.A2HT.11A", 
"TCGA.ES.A2HT.01A","TCGA.FV.A23B.11A","TCGA.FV.A23B.01A","TCGA.FV.A3I0.11A",
"TCGA.FV.A3I0.01A","TCGA.FV.A3R2.11A","TCGA.FV.A3R2.01A"),row.names = c(NA, -1L), 
class = c("data.table","data.frame"))
2
rdataframe
How are you trying to "take the row"? - Elin
@Elin I am trying to convert row 1 of cul7_expr, which is basically all the numerics, and make them into a column of the new data frame - merryberry
So how are you trying to convert that row? Show us some code. Like this it is hard to figure out what you mean and we can't know where the problem is - KenHBS
The idea is to encourage the OP to provide a minimal reproducable example, which this is not. - Elin

2 Answers

1
votes

Try the melt function. To help in the future, this is called changing data from wide to long format.

require(data.table)
melt(df,measure.vars=1:22)

Output:

            variable     value
 1: TCGA.BC.A10Q.11A  819.3685
 2: TCGA.BC.A10Q.01A 2757.4860
 3: TCGA.DD.A1EB.11A  698.5818
 4: TCGA.DD.A1EB.01A 1625.0940
 5: TCGA.DD.A1EG.11A  409.9332
 6: TCGA.DD.A1EG.01A 2221.0120
 7: TCGA.DD.A1EH.11A  391.0916
 8: TCGA.DD.A1EH.01A 2122.7820
 9: TCGA.DD.A1EI.11A  717.2073
10: TCGA.DD.A1EI.01A  768.7468
11: TCGA.DD.A3A6.11A  464.6395
12: TCGA.DD.A3A6.01A 1175.9280
13: TCGA.DD.A3A8.11A  934.9738
14: TCGA.DD.A3A8.01A  931.8955
15: TCGA.ES.A2HT.11A  599.7360
16: TCGA.ES.A2HT.01A  894.8324
17: TCGA.FV.A23B.11A  970.1805
18: TCGA.FV.A23B.01A 3018.0750
19: TCGA.FV.A3I0.11A  337.2220
20: TCGA.FV.A3I0.01A 3895.4770
21: TCGA.FV.A3R2.11A  912.8499
22: TCGA.FV.A3R2.01A 2226.9210
0
votes

If I understand what you want correctly you need to get the transpose and the column names and then make a data frame from them. 

# set up data
x <- data.frame(1, 2, 3, 4)
names <- c("A", "B",  "C", "D")
colnames(x) <- names

#convert
names <- colnames(x) 
data.frame(t(x), names)