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

I have the following table:

     X.5       X.6       X.7       X.8          X.9 X.10         X.11  X.12   X.13
17   Zip CuCurrent PaCurrent PoCurrent      Contact  Ext          Fax email Status
18  74136         0         1         0 918-491-6998    0 918-491-6659            1
19  30329         1         0         0 404-321-5711                              1
20  74136         1         0         0 918-523-2516    0 918-523-2522            1
21  80203         0         1         0 303-864-1919    0                         1
22  80120         1         0         0 345-098-8890  456                         1

how can make the first row 'zip, cucurrent, pacurrent...' to be the column header?

Thanks,

below is dput(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")

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

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,]

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.

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

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