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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")
290
Click anywhere in the table. On the Home tab on the ribbon, click the down arrow next to Table and select Toggle Header Row. Click the Table Design tab > Style Options > Header Row.
Data Visualization using R Programming First of all, create a data frame. Convert the row into column names by using colnames function and subsetting the data frame.
colnames() function in R is used to set headers or names to columns of a dataframe or matrix. Syntax: colnames(dataframe) <- c(“col_name-1”, “col_name-2”, “col_name-3”, “col_name-4”,…..)
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.
17
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
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)
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