我有下表：
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
怎样才能使第一排＆＃39;拉链，黄瓜，太平洋......＆＃39;列标题？
谢谢，
下面的是dput(dat)
dput（DAT） 结构（列表（X.5 =结构（c（26L，14L，6L，14L，17L，16L），。Label = c（＆＃34;＆＃34;， ＆＃34; 1104＆＃34;，＆＃34; 1234我不认识Ave.＆＃34;，＆＃34; 139.98＆＃34;，＆＃34; 300 Morgan St.＆＃34; ， ＆＃34; 30329＆＃34;，＆＃34; 312.95＆＃34;，＆＃34; 4101 S. 4th Street，Traff＆＃34;，＆＃34; 500 Highway 89 North＆＃34;， ＆＃34; 644.04＆＃34;，＆＃34; 656.73＆＃34;，＆＃34; 72160＆＃34;，＆＃34; 72336-7000＆＃34;，＆＃34; 74136＆＃34;，＆ ＃34; 75501＆＃34 ;, ＆＃34; 80120＆＃34;，＆＃34; 80203＆＃34;，＆＃34; 877.87＆＃34;，＆＃34;地址1＆＃34;，＆＃34; BZip＆＃34;，＆＃34;一般Svcs管理员（WPY）＆＃34;， ＆＃34; InvFileName2＆＃34;，＆＃34; LDC_Org_Cost＆＃34;，＆＃34; N / A＆＃34;，＆＃34; NULL＆＃34;，＆＃34; Zip＆＃34;），class = ＆＃34;因子＆＃34）， X.6 =结构（c（7L，2L，3L，3L，2L，3L），。Label = c（＆＃34;＆＃34;， ＆＃34; 0＆＃34;，＆＃34; 1＆＃34;，＆＃34; 301 7th St. SW＆＃34;，＆＃34; 800-688-6160＆＃34;，＆＃34; Address2＆＃ 34;，＆＃34; CuCurrent＆＃34;， ＆＃34;紧急＆＃34;，＆＃34; LDC_Cost_Adj＆＃34;，＆＃34; Mtelemetry＆＃34;，＆＃34; N / A＆＃34;，＆＃34; NULL＆＃34;， ＆＃34; Suite 1402＆＃34;），class =＆＃34; factor＆＃34;），X.7 = structure（c（8L，3L， 2L，2L，3L，2L）,.标签= c（＆＃34;＆＃34;，＆＃34; 0＆＃34;，＆＃34; 1＆＃34;，＆＃34;地址3＆＃34;， ＆＃34; Cucustomer＆＃34 ;, ＆＃34; LDC_Misc_Fee＆＃34;，＆＃34; NULL＆＃34;，＆＃34; PaCurrent＆＃34;，＆＃34; Room 7512＆＃34;），class =＆＃34; factor＆＃34;）， X.8 =结构（c（14L，2L，2L，2L，2L，2L），。Label = c（＆＃34;＆＃34;， ＆＃34; 0＆＃34;，＆＃34; 100.98＆＃34;，＆＃34; 237.02＆＃34;，＆＃34; 242.33＆＃34;，＆＃34; 335.04＆＃34;，＆ ＃34; 50.6＆＃34;，＆＃34; City＆＃34;， ＆＃34; Durham＆＃34;，＆＃34; LDC_FinalVolume＆＃34;，＆＃34; Leavenwoth＆＃34;，＆＃34; Pacustomer＆＃34;， ＆＃34;彼得堡＆＃34;，＆＃34; PoCurrent＆＃34;，＆＃34; Prescott＆＃34;，＆＃34; Washington＆＃34;），class =＆＃34; factor＆＃34;）， X.9 =结构（c（18L，16L，10L，17L，7L，9L），。Label = c（＆＃34;＆＃34;， ＆＃34; 0＆＃34;，＆＃34; 1＆＃34;，＆＃34; 139.98＆＃34;，＆＃34; 20024＆＃34;，＆＃34; 27701＆＃34;，＆＃34; 303-864-1919＆＃34;，＆＃34; 312.95＆＃34;， ＆＃34; 345-098-8890＆＃34;，＆＃34; 404-321-5711＆＃34;，＆＃34; 644.04＆＃34;，＆＃34; 656.73＆＃34;，＆＃34; 66048＆＃34 ;, ＆＃34; 86313＆＃34;，＆＃34; 877.87＆＃34;，＆＃34; 918-491-6998＆＃34;，＆＃34; 918-523-2516＆＃34;，＆＃34;联系＆ ＃34 ;, ＆＃34; LDC_FinalCost＆＃34;，＆＃34; PoCustomer＆＃34;，＆＃34; Zip＆＃34;），class =＆＃34; factor＆＃34;）， X.10 =结构（c（14L，2L，1L，2L，2L，9L），。Label = c（＆＃34;＆＃34;， ＆＃34; 0＆＃34;，＆＃34; 2.620194604＆＃34;，＆＃34; 2.710064788＆＃34;，＆＃34; 2.717239052＆＃34;，＆＃34; 2.766403162＆＃34;， ＆＃34; 202-708-4995＆＃34;，＆＃34; 3.09912854＆＃34;，＆＃34; 456＆＃34;，＆＃34; 804-504-7200＆＃34;，＆＃34; 913 -682-2000＆＃34 ;, ＆＃34; 919-956-5541＆＃34;，＆＃34; 928-717-7472＆＃34;，＆＃34; Ext＆＃34;，＆＃34; InvoicesNeeded＆＃34;， ＆＃34; LDC_UnitPrice＆＃34;，＆＃34; NULL＆＃34;，＆＃34; Phone＆＃34;），class =＆＃34; factor＆＃34;），X.11 = structure（c（7L， 4L，1L，5L，1L，1L），。标签= c（＆＃34;＆＃34;，＆＃34;＆＃34;，＆＃34; 1067＆＃34;，＆＃34; 918-491 -6659＆＃34 ;, ＆＃34; 918-523-2522＆＃34;，＆＃34; Ext＆＃34;，＆＃34; Fax＆＃34;，＆＃34; InvoiceMonths＆＃34;，＆＃34; LDC_UnitPrice_Original＆＃34;， ＆＃34; NULL＆＃34;，＆＃34; x2951＆＃34;），class =＆＃34; factor＆＃34;），X.12 =结构（c（13L， 1L，1L，1L，1L，1L），。标签= c（＆＃34;＆＃34;，＆＃34; 0＆＃34;，＆＃34; 100.98＆＃34;，＆＃34; 202- 401-3722＆＃34 ;, ＆＃34; 237.02＆＃34;，＆＃34; 242.33＆＃34;，＆＃34; 335.04＆＃34;，＆＃34; 50.6＆＃34;，＆＃34; 716-344-3303＆＃ 34;，＆＃34; 804-504-7227＆＃34;， ＆＃34; 913-758-4230＆＃34;，＆＃34; 919-956-7152＆＃34;，＆＃34;电子邮件＆＃34;，＆＃34;传真＆＃34;，＆＃34; GSA＆＃ 34 ;, ＆＃34; Supp_Vol＆＃34;），class =＆＃34; factor＆＃34;），X.13 =结构（c（10L，2L， 2L，2L，2L，2L）,. Label = c（＆＃34;＆＃34;，＆＃34; 1＆＃34;，＆＃34; 15＆＃34;，＆＃34; 202-497-6164＆ ＃34 ;, ＆＃34; 3＆＃34;，＆＃34; 804-504-7200＆＃34;，＆＃34;紧急＆＃34;，＆＃34; MajorTypeId＆＃34;，＆＃34; NULL＆＃34;， ＆＃34;状态＆＃34;，＆＃34; Supp_Vol_Adj＆＃34;），类=＆＃34;因素＆＃34;））,. Name = c（＆＃34; X.5＆＃34;， ＆＃34; X.6＆＃34;，＆＃34; X.7＆＃34;，＆＃34; X.8＆＃34;，＆＃34; X.9＆＃34;，＆＃34; X. 10＆＃34;，＆＃34; X.11＆＃34;，＆＃34; X.12＆＃34;，＆＃34; X.13＆＃34;），row.names = 17：22，class =＆ ＃34; data.frame＆＃34）
答案 0 :(得分：11)
如果您不想将数据重新读入R（您似乎没有从评论中删除），则可以执行以下操作。我不得不添加一些零来让你的数据完全读取，所以忽略这些。
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
首先将第一行作为列名。接下来删除第一行。通过将列转换为适当的类型来完成它。
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
答案 1 :(得分：4)
如果从csv文件中获取它，请使用read.csv
中的参数'header'dat=read.csv("gas.csv", header=TRUE)
如果您已经拥有自己的数据并且不想/或无法以干净的方式获取数据，那么您可以这样做
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
答案 2 :(得分：2)
最干净的方法是为此目的设计一个简单的函数。 您需要管理员软件包。
janitor::row_to_names(dat)
如果要将第n行用作列名，则该函数的第二个参数是要使用的行号。默认值为1。
答案 3 :(得分：1)
将数据导入R！
时请使用header = TRUE答案 4 :(得分：1)
这可以用一种简单的方式：
步骤1：将第一行复制到标题：
names(dat) <- dat[1,]
第2步：删除第1行：
dat <- dat[-1,]