Error: No Such Column using SQLDF

问题

Below are the scripts

> library(sqldf)
> turnover = read.csv("turnover.csv")
> names(turnover)
[1] "Report.Date"       "PersID"            "Status"            "DOB"
[5] "Age"               "Tenure"            "Current.Hire.Date" "Term.Date"
[9] "Gender"            "Function"          "Grade"             "Job.Category"
[13] "City"              "State"             "Retiree"           "Race"
> turnover_hiredate = sqldf("select Status, Current.Hire.Date from turnover")

I get an error msg: no such column: Current.Hire.Date. But this variable is listed as the 7th variable.

What did I do wrong?

回答1:

sqldf(...) does not like . (period) in column names, so you need to change it to something else. Try this:

library(sqldf)
turnover = read.csv("turnover.csv")
colnames(turnover) <- gsub("\\.","_",colnames(turnover))
turnover_hiredate = sqldf("select Status, Current_Hire_Date from turnover")

The reason is that the period is used in SQL to indicate a table column, e.g. turnover.Status.

回答2:

There is no need to change column names.

Starting with RSQLite 1.0.0 and sqldf 0.4-9 dots in column names are no longer translated to underscores. https://code.google.com/p/sqldf/

We only need to write the SQL statement between single quotes, and the column names including dots between double quotes or backticks/backquotes interchangeably.

Two examples:

require(sqldf)
# 1
turnover <- data.frame(Status = c("A", "B", "C"), 
                       Current.Hire.Date = c("4/10/10", "13/11/10", "1/7/13"))
sqldf('select Status, "Current.Hire.Date" from turnover') 

#2. Double quotes and backticks interchangeably    
sqldf('select Species, avg("Sepal.Length") `Sepal.Length`, 
   avg("Sepal.Width") `Sepal.Width` from iris group by Species')

One more way to achieve the solution

#3 Using square brackets 
sqldf('select Species, avg([Sepal.Length]) `Sepal.Length`, 
   avg([Sepal.Width])  `Sepal.Width` from iris group by Species')

回答3:

You can also use

library(sqldf)
turnover <- read.csv("turnover.csv", header=TRUE, check.names=FALSE) #Leave Names Alone

names(turnover) <- gsub(x = names(turnover),
                    pattern = " ",
                    replacement = "_")  # Replace Var Name blank with _

来源：https://stackoverflow.com/questions/23451139/error-no-such-column-using-sqldf

标签

sqldf