How to change the first row to be the header in R?

Question 1

If you don't want to re-read the data into R (which it seems like you don't from the comments), you can do the following. I had to add some zeros to get your data to read completely, so disregard those.

dat
##       V2        V3        V4        V5           V6  V7           V8    V9    V10
## 17   Zip CuCurrent PaCurrent PoCurrent      Contact Ext          Fax email Status
## 18 74136         0         1         0 918-491-6998   0 918-491-6659     0      1
## 19 30329         1         0         0 404-321-5711   0            0     0      1
## 20 74136         1         0         0 918-523-2516   0 918-523-2522     0      1
## 21 80203         0         1         0 303-864-1919   0            0     0      1
## 22 80120         1         0         0 345-098-8890 456            0     0      1

First take the first row as the column names. Next remove the first row. Finish it off by converting the columns to their appropriate types.

names(dat) <- as.matrix(dat[1, ])
dat <- dat[-1, ]
dat[] <- lapply(dat, function(x) type.convert(as.character(x)))
dat
##     Zip CuCurrent PaCurrent PoCurrent      Contact Ext          Fax email Status
## 1 74136         0         1         0 918-491-6998   0 918-491-6659     0      1
## 2 30329         1         0         0 404-321-5711   0            0     0      1
## 3 74136         1         0         0 918-523-2516   0 918-523-2522     0      1
## 4 80203         0         1         0 303-864-1919   0            0     0      1
## 5 80120         1         0         0 345-098-8890 456            0     0      1

Question 2

This could be in a simple way :

step 1: Copy 1st row to header:

names(dat) <- dat[1,]

step 2: Delete 1st row :

dat <- dat[-1,]

Question 3

The cleanest way to do this is a simple function already designed for this purpose. You need janitor package.

janitor::row_to_names(dat)

If you want the nth row to be used for column names the second parameter of the function is the row number to be used. The default is 1.

Question 4

If you get it from a csv file, use the argument 'header' from read.csv

dat=read.csv("gas.csv", header=TRUE)

if you already have your data and don't want to / or cannot get it in a clean way, you can alwasy do

dat=structure(list(X.5 = structure(c(26L, 14L, 6L, 14L, 17L, 16L), .Label = c("", "1104", "1234 I don't know Ave.", "139.98", "300 Morgan St.", "30329", "312.95", "4101 S. 4th Street, Traff", "500 Highway 89 North", "644.04", "656.73", "72160", "72336-7000", "74136", "75501", "80120", "80203", "877.87", "Address1", "BZip", "General Svcs Admin (WPY)", "InvFileName2", "LDC_Org_Cost", "N/A", "NULL", "Zip"), class = "factor"), X.6 = structure(c(7L, 2L, 3L, 3L, 2L, 3L), .Label = c("", "0", "1", "301 7th St. SW", "800-688-6160", "Address2", "CuCurrent", "Emergency", "LDC_Cost_Adj", "Mtelemetry", "N/A", "NULL", "Suite 1402"), class = "factor"), X.7 = structure(c(8L, 3L, 2L, 2L, 3L, 2L), .Label = c("", "0", "1", "Address3", "Cucustomer", "LDC_Misc_Fee", "NULL", "PaCurrent", "Room 7512"), class = "factor"), X.8 = structure(c(14L, 2L, 2L, 2L, 2L, 2L), .Label = c("", "0", "100.98", "237.02", "242.33", "335.04", "50.6", "City", "Durham", "LDC_FinalVolume", "Leavenwoth", "Pacustomer", "Petersburg", "PoCurrent", "Prescott", "Washington"), class = "factor"), X.9 = structure(c(18L, 16L, 10L, 17L, 7L, 9L), .Label = c("", "0", "1", "139.98", "20024", "27701", "303-864-1919", "312.95", "345-098-8890", "404-321-5711", "644.04", "656.73", "66048", "86313", "877.87", "918-491-6998", "918-523-2516", "Contact", "LDC_FinalCost", "PoCustomer", "Zip"), class = "factor"), X.10 = structure(c(14L, 2L, 1L, 2L, 2L, 9L), .Label = c("", "0", "2.620194604", "2.710064788", "2.717239052", "2.766403162", "202-708-4995", "3.09912854", "456", "804-504-7200", "913-682-2000", "919-956-5541", "928-717-7472", "Ext", "InvoicesNeeded", "LDC_UnitPrice", "NULL", "Phone"), class = "factor"), X.11 = structure(c(7L, 4L, 1L, 5L, 1L, 1L), .Label = c("", " ", "1067", "918-491-6659", "918-523-2522", "Ext", "Fax", "InvoiceMonths", "LDC_UnitPrice_Original", "NULL", "x2951"), class = "factor"), X.12 = structure(c(13L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "0", "100.98", "202-401-3722", "237.02", "242.33", "335.04", "50.6", "716- 344-3303", "804-504-7227", "913- 758-4230", "919- 956-7152", "email", "Fax", "GSA", "Supp_Vol"), class = "factor"), X.13 = structure(c(10L, 2L, 2L, 2L, 2L, 2L), .Label = c("", "1", "15", "202-497-6164", "3", "804-504-7200", "Emergency", "MajorTypeId", "NULL", "Status", "Supp_Vol_Adj"), class = "factor")), .Names = c("X.5", "X.6", "X.7", "X.8", "X.9", "X.10", "X.11", "X.12", "X.13"), row.names = 17:22, class = "data.frame")
dat2 = dat[2:6,]   
colnames(dat2) = dat[1,] 
dat2

Question 5

If you are able to re-read the data into R from the file, you could also just add the "skip" argument to read.csv to skip over the first 16 lines and use line 17 as the header:

dat=read.csv("contacts.csv", skip=16, nrows=5, header=TRUE)

Question 6

Please use header=TRUE when importing data into R!

Question 7

Shalini Baranwal's answer is the best, so I will upvote it. However, future readers may get an error message when running that solution. My error was:

"Error in setnames(x, value) : Passed a vector of type 'list'. Needs to be type 'character'."

To get around this, my modified solution was to add an as.character() wrapper to the first step. The full solution below:

step 1: Copy 1st row to header:

dat <- mtcars
names(dat) <- as.character(dat[1,])

step 2: Delete 1st row :

dat <- dat[-1,]