正如@Roland指出,你有matrix
,不是data.frame
,而這些具有不同的默認print
方法。堅持matrix
,您可以在print
中明確地設置quote = FALSE
,或者您可以使用noquote
。
這是一個基本的例子:
## 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
而且,當你在你自己的,想通了轉換您matrix
爲data.frame
使得報價消失。如果每列是不同類型的數據(數字,字符,因子等),則data.frame
是一個更適合於保存數據的結構。但是,將matrix
轉換爲data.frame
不會自動爲您轉換列。相反,你可以利用type.convert
(使用read.table
和家人創造一個data.frame
時也使用):
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
-1自稱有data.frame時,它顯然是一個矩陣。 – Roland