Convert a vector in R to a dataframe by columns

StackOverflow https://stackoverflow.com/questions/23388577

Pergunta

I would like to convert this vector with strings in each row and spaces separating the elements within one string:

> v.input_red 
[1] "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
[2] "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
[3] "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "

to a dataframe with a column for each element. But I'm not quite sure how to extract the elements from the strings. Best way would be to convert the whole thing at once somehow, I guess..

Wanted result-dataframe (created manually):

V1 V2  V3  V4 V5 V6   V7   V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21  V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33 V34 V35
1  pm  0 100 2.1 59 70 15.5 14.8 31 984  32   0  56  55   0   0   0   0   0   0 -60 -260   0   0   6   0   0   0   0   0  20   8 2ff   0 249
2  pm  0 100 2.1 59 70 15.5 14.8 31 984  32   0  56  55   0   0   0   0   0   0 -60 -260   0   0   6   0   0   0   0   0  20   8 2ff   0 249
3  pm  0 100 2.1 59 70 15.5 14.8 31 984  32   0  56  55   0   0   0   0   0   0 -60 -260   0   0   6   0   0   0   0   0  20   8 2ff   0 249

Thanks in advance! Matthias

Foi útil?

Solução 2

Assuming file is a file name that you save on your system:

writeLines(v.input_red, file)
data <- read.table(file)

Outras dicas

For quite some time, read.table and family have had a text argument that lets you read directly from character vectors. There's no need to write the object to a file first.

Your sample data...

v.input_red <- c("pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 ",
"pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 ",
"pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 ")

... directly read in:

read.table(text = v.input_red, header = FALSE)
#   V1 V2  V3  V4 V5 V6   V7   V8 V9 V10 V11 V12 V13 V14 V15 V16 V17
# 1 pm  0 100 2.1 59 70 15.5 14.8 31 984  32   0  56  55   0   0   0
# 2 pm  0 100 2.1 59 70 15.5 14.8 31 984  32   0  56  55   0   0   0
# 3 pm  0 100 2.1 59 70 15.5 14.8 31 984  32   0  56  55   0   0   0
#   V18 V19 V20 V21  V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33
# 1   0   0   0 -60 -260   0   0   6   0   0   0   0   0  20   8 2ff
# 2   0   0   0 -60 -260   0   0   6   0   0   0   0   0  20   8 2ff
# 3   0   0   0 -60 -260   0   0   6   0   0   0   0   0  20   8 2ff
#   V34 V35 V36 V37
# 1   0 249   0   0
# 2   0 249   0   0
# 3   0 249   0   0

Is this solution what you were looking for?

s1 <- "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
s2 <- "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "
s3 <- "pm 0 100 2.1 59 70 15.5 14.8 31 984 32 0 56 55 0 0 0 0 0 0 -60 -260 0 0 6 0 0 0 0 0 20 8 2ff 0 249 0 0 "

df <- t(data.frame(strsplit(s1, " "),strsplit(s2, " "),strsplit(s3, " ")))
row.names(df) <- c("s1", "s2", "s3")

strsplit splits the string at each space char. Concatenated as data.frame gives you a df wih 3 columns. So you have to transpose it with t. I changes row names for better readability.

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