remove all quotation marks from a data frame

Question

I have a data frame rep that looks like this:

> head(rep)
     position chrom  value label  
[1,] "17408"  "chr1" "0"   "miRNA"
[2,] "17409"  "chr1" "0"   "miRNA"
[3,] "17410"  "chr1" "0"   "miRNA"
[4,] "17411"  "chr1" "0"   "miRNA"
[5,] "17412"  "chr1" "0"   "miRNA"
[6,] "17413"  "chr1" "0"   "miRNA"

How can I remove the quotation marks from all elements?

Note: rep$position and rep$value should be numeric type, rep$chrom and rep$label should be character type.

Answer 1

As indicated by @Roland, you have a matrix, not a data.frame, and these have different default print methods. Sticking with a matrix, you can set quote = FALSE explicitly in print or you can use noquote.

Here is a basic example:

## Sample data
x <- matrix(c(17, "chr1", 0, "miRNA", 18, "chr1", 0, "miRNA"), nrow = 2, 
            byrow = TRUE, dimnames = list(
              NULL, c("position", "chrom", "value", "label")))

## Default printing
x
#      position chrom  value label  
# [1,] "17"     "chr1" "0"   "miRNA"
# [2,] "18"     "chr1" "0"   "miRNA"

## Two options to make the quotes disappear
print(x, quote = FALSE)
#      position chrom value label
# [1,] 17       chr1  0     miRNA
# [2,] 18       chr1  0     miRNA
noquote(x)
#      position chrom value label
# [1,] 17       chr1  0     miRNA
# [2,] 18       chr1  0     miRNA

---

Also, as you figured out on your own, converting your matrix to a data.frame makes the quotes disappear. A data.frame is a more appropriate structure to hold your data if each column is a different type of data (numeric, character, factor, and so on). However, converting a matrix to a data.frame does not take care of the conversion of columns for you automatically. Instead, you can make use of type.convert (which is also used when creating a data.frame using read.table and family):

y <- data.frame(x, stringsAsFactors = FALSE)
str(y)
# 'data.frame':  2 obs. of  4 variables:
#  $ position: chr  "17" "18"
#  $ chrom   : chr  "chr1" "chr1"
#  $ value   : chr  "0" "0"
#  $ label   : chr  "miRNA" "miRNA"
y[] <- lapply(y, type.convert)
str(y)
# 'data.frame':  2 obs. of  4 variables:
#  $ position: int  17 18
#  $ chrom   : Factor w/ 1 level "chr1": 1 1
#  $ value   : int  0 0
#  $ label   : Factor w/ 1 level "miRNA": 1 1
y
#   position chrom value label
# 1       17  chr1     0 miRNA
# 2       18  chr1     0 miRNA

Answer 2

Two steps: 1) get rid of quotation marks, 2) convert columns accordingly:

The data

x <- read.table(text='
position chrom  value label  
"\\"17408\\""  "\\"chr1\\"" "\\"0\\""   "\\"miRNA\\""
"\\"17409\\""  "\\"chr1\\"" "\\"0\\""   "\\"miRNA\\""'
, header=T)

1) get rid of quotation marks

library(stringr)
library(plyr)

del <- colwise(function(x) str_replace_all(x, '\"', ""))
x <- del(x)

2) convert columns accordingly

num <- colwise(as.numeric)    
x[c(1,3)] <- num(x[c(1,3)])
x

  position chrom value label
1    17408  chr1     0 miRNA
2    17409  chr1     0 miRNA

Answer 3

I think I found the answer. What I have is not a data.frame but a matrix. Converting it to a data.frame got rid of the quotation marks. I am still wondering why though....

rep <- data.frame(rep)
> head(rep)
  position chrom value label
1    17408  chr1     0 miRNA
2    17409  chr1     0 miRNA
3    17410  chr1     0 miRNA
4    17411  chr1     0 miRNA
5    17412  chr1     0 miRNA
6    17413  chr1     0 miRNA