use first row data as column names in r

This should be such an easy problem but I have trouble with. I have a dirty dataset that I could not read it with header=T. After I read and clean it, I would like to use the now first row data as the column name. I tried multiple methods on stackoverflow without success. What could be the problem? The dataset t1 should look like this after clean up:

      V1    V2  V3  V4  V5
1   col1    col2    col3    col4
2   row1    2   4   5   56
3   row2    74  74  3   534
4   row3    865 768 8   7
5   row4    68  86  65  87

I tried: colnames(t1)=t1[1,]. Nothing happens.

I tried: names(t1)=ti[1,], Nothing happens.

I tried: lapply(t1, function(x) {names(x)<-x[1,]; x})

it returns an error message:

Error in [.default(x, 1, ) : incorrect number of dimensions

Could anyone help?

header.true <- function(df) {
  names(df) <- as.character(unlist(df[1,]))
  df[-1,]
}

Test

df1 <- data.frame(c("a", 1,2,3), c("b", 4,5,6))
header.true(df1)
  a b
2 1 4
3 2 5
4 3 6

Take a step back, when you read your data use skip=1 in read.table to miss out the first line entirely. This should make life a bit easier when you're cleaning data, particularly for data type. This is key as your problem stems from your data being encoded as factor.

You can then read in your column names separately with nrows=1 in read.table.

Probably, the data type of the data frame columns are factors. That is why the code you tried didn't work, you can check it using str(df):

First option

Use the argument stringsAsFactors = FALSEwhen you import your data:

df <- read.table(text =  "V1    V2  V3  V4  V5
                        col1    col2    col3    col4 col5
                        row1    2   4   5   56
                        row2    74  74  3   534
                        row3    865 768 8   7
                        row4    68  86  65  87", header = TRUE, 
                        stringsAsFactors = FALSE )

Then you can use your first attempt, then remove your first row if you'd like:

colnames(df) <- df[1,]
df <- df[-1, ]

Second option

It will work if your columns are factors or characters:

names(df) <- lapply(df[1, ], as.character)
df <- df[-1,]

Output:

  col1 col2 col3 col4 col5
2 row1    2    4    5   56
3 row2   74   74    3  534
4 row3  865  768    8    7
5 row4   68   86   65   87

How about:

my.names <- t1[1,]

colnames(t1) <- my.names

i.e. specifically naming the row as a variable?

with the following code:

namex <-c("col1","col2","col3","col4")
row1 <- c(2, 4, 5, 56)
row2 <- c(74, 73, 3, 534)
row3 <- c(865, 768, 8, 7)
row4 <- c(68, 58, 65, 87)

t1 <- data.frame(namex, row1, row2, row3, row4)
t1 <- t(t1)

my.names <- t1[1,]

colnames(t1) <- my.names

It seems to work, but maybe I'm missing something?

Using data.table:

library(data.table)

namex <-c("col1","col2","col3","col4")
row1 <- c(2, 4, 5, 56)
row2 <- c(74, 73, 3, 534)
row3 <- c(865, 768, 8, 7)
row4 <- c(68, 58, 65, 87)

t1 <- data.table(namex, row1, row2, row3, row4)
t1 <- data.table(t(t1))

setnames(t1, as.character(t1[1,]))
t1 <- t1[-1,]

Similar to some of the other answers, here is a dplyr/tidyverse option:

library(tidyverse)

names(df) <- df %>% slice(1) %>% unlist()
df <- df %>% slice(-1)

来源：https://stackoverflow.com/questions/32054368/use-first-row-data-as-column-names-in-r

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