I have a function that takes input as the column name of dataframe as columnname~1.
The dataframe consists of about 50 columns in which I want to repeat the process, I can
use a for loop to generate column name as a character which unfortunately the function does
not recognize. The difference is just M1~1 (works) vs "M1"~1
Any suggestions are welcome
Try:
as.formula(paste("M1","1",sep="~"))
Use the "[[ function?
datafrm[[M1]] ~ 1
That function will interpret the character value and convert it to a language value. Or you could use the do.call function. Best answers will be forthcoming if we have specifics of the dataframe and the function.
Or as an alternative to paste you can use sprintf:
variablename = "M1"
as.formula(sprintf("%s ~ 1", variablename))
You can also use this:
substitute(columnname~1,list(columnname=as.name("M1")))
EDIT
It will also work when you substitute "M1" with variable containing string (M1 <- "M1").
Related
I am currently in a loop in my R code and I want to use the following code:
distinct.values <- unique(SQL_Table$column.names2[for.num])
column.names2 looks like this
column.names2
[1] "plan" "gender" "marital_status" "acceleration" "extension"
[6] "inflation"
depending on the for.num I want a different variable after the $.
Example:
when for.num = 1.
I want distinct.values to be set as unique(SQL_Table$plan)
when for.num = 2.
I want distinct.values to be set as unique(SQL_Table$gender)
and so on...
How can I do this?
To expand my comment on the original question, there a few different ways to access columns of data frames.
my_df$column_name
When using the $ operator, "column_name" is specified as a literal token in the R script. Note that because the column name is unquoted, this method does not allow variable substitution.
my_df[["column_name"]]
When using the [[ ]] operator, a string (or vector of strings) is expected. In this case, variable substitution is allowed, so the following is valid:
my_col <- "column_name"
my_df[[my_col]]
This would be equivalent to my_df$column_name.
So your code could be modified to read:
unique(SQL_Table[[column.names2[for.num]]])
This is what my text file looks like:
1241105.41129.97Y317052.03
2282165.61187.63N364051.40
2251175.87190.72Y366447.49
2243125.88150.81N276045.45
328192.89117.68Y295050.51
2211140.81165.77N346053.11
1291125.61160.61Y335048.3
3273127.73148.76Y320048.04
2191132.22156.94N336051.38
3221118.73161.03Y349349.5
2341189.01200.31Y360048.02
1253144.45180.96N305051.51
2251125.19152.75N305052.72
2192137.82172.25N240046.96
3351140.96174.85N394048.09
1233135.08173.36Y265049.82
1201112.59140.75N380051.25
2202128.19159.73N307048.29
2192132.82172.25Y240046.96
3351148.96174.85Y394048.09
1233132.08173.36N265049.82
1231114.59140.75Y380051.25
3442128.19159.73Y307048.29
2323179.18191.27N321041.12
All these values are continuous and each character indicates something. I am unable to figure out how to separate each value into columns and specify a heading for all these new columns which will be created.
I used this code, however it does not seem to work.
birthweight <- read.table("birthweighthw1.txt", sep="", col.names=c("ethnic","age","smoke","preweight","delweight","breastfed","brthwght","brthlngth”))
Any help would be appreciated.
Assuming that you have a clear definition for every column, you can use regular expressions to solve this in no time.
From your column names and example data, I guess that the regular expression that matches each field is:
ethnic: \d{1}
age: \d{1,2}
smoke: \d{1}
preweight: \d{3}\.\d{2}
delweight: \d{3}\.\d{2}
breastfed: Y|N
brthwght: \d{3}
brthlngth: \d{3}\.\d{1,2}
We can put all this together in a regular expression that captures each of these fields
reg.expression <- "(\\d{1})(\\d{1,2})(\\d{1})(\\d{3}\\.\\d{2})(\\d{3}\\.\\d{2})(Y|N)(\\d{3})(\\d{3}\\.\\d{1,2})"
Note: In R, we need to scape "\" that's why we write \d instead of \d.
That said, here comes the code to solve the problem.
First, you need to read your strings
lines <- readLines("birthweighthw1.txt")
Now, we define our regular expression and use the function str_match from the package stringr to get your data into character matrix.
require(stringr)
reg.expression <- "(\\d{1})(\\d{1,2})(\\d{1})(\\d{3}\\.\\d{2})(\\d{3}\\.\\d{2})(Y|N)(\\d{3})(\\d{3}\\.\\d{1,2})"
captured <- str_match(string= lines, pattern= reg.expression)
You can check that the first column in the matrix contains the text matched, and the following columns the data captured. So, we can get rid of the first column
captured <- captured[,-1]
and transform it into a data.frame with appropriate column names
result <- as.data.frame(captured,stringsAsFactors = FALSE)
names(result) <- c("ethnic","age","smoke","preweight","delweight","breastfed","brthwght","brthlngth")
Now, every column in result is of type character, you can transform each of them into other types. For example:
require(dplyr)
result <- result %>% mutate(ethnic=as.factor(ethnic),
age=as.integer(age),
smoke=as.factor(smoke),
preweight=as.numeric(preweight),
delweight=as.numeric(delweight),
breastfed=as.factor(breastfed),
brthwght=as.integer(brthwght),
brthlngth=as.numeric(brthlngth)
)
I am working with a long list of data frames.
Here is a simple hypothetical example of a data frame:
DFrame<-data.frame(c(1,0),c("Yes","No"))
colnames(DFrame)<-c("ColOne","ColTwo")
I am trying to retrieve a specified column of the data frame using paste function.
get(paste("DFrame","$","ColTwo",sep=""))
The get function returns the following error, when trying to retrieve a specified column:
Error in get(paste("DFrame", "$", "ColTwo", sep = "")) :object 'DFrame$ColTwo' not found
When I enter the constructed name of the data frame DFrame$ColTwo it returns the desired output of the second column.
If I reconstruct an example without the '$' sign then I get the desired answer from the get function. For example the code yields 2:
enter code here
Ans <- 2
get(paste("An","s",sep=""))
[1] 2
I am looking for the same desired outcome, but struggling to get past the error that the object could not be found.
I also attempted using the following format, but the quotation in the column name breaks the paste function:
paste("DFrame","[,"ColTwo"]",sep="")
Thank you very much for the input,
Kind regards
You can do that using the following syntax:
get("DFrame")[,"ColTwo"]
You can use paste() in both of these strings, for example:
get(paste("D", "Frame", sep=""))[,paste("Col", "Two", sep="")]
Edit: Despite someone downvoting this answer without leaving a comment, this does exactly what the original poster asked for. If you feel that it does not or is in some way dangerous, I would encourage you to leave a comment.
Stop trying to use paste and get entirely.
The whole point of having a list (of data frames, say) is that you can reference them using names:
DFrame<-data.frame(c(1,0),c("Yes","No"))
colnames(DFrame)<-c("ColOne","ColTwo")
#A list of data frames
l <- list(DFrame,DFrame)
#The data frames in the list can have names
names(l) <- c("DF1",'DF2')
# Now you just use `[[`
> l[["DF1"]][["ColOne"]]
[1] 1 0
> l[["DF1"]][["ColTwo"]]
[1] Yes No
Levels: No Yes
If you have to, you can use paste to construct the indices passed inside [[.
I want to get a cell value after dynamically passing its address. So I am trying paste command to join the address of the cell like following:
paste0("DT1$", eval(cols[1]),"[1]")
where DT1 is datatable, cols[1] is refering to 1 column and [1] is first row of that column. While running this I am getting the string(address of the cell):
> paste0("DT1$", eval(cols[1]),"[1]")
[1] "DT1$BCC1[1]"
But I want the value of the cell like if I run:
> DT1$BCC1[1]
[1] 0
So how to run call the result of the paste expression to get value of cell like "0" in previous example. I tried eval() and do.call(), but nothing seems to be working. I am sorry for this basic question as I am new to R. Any help is really appreciated.
You can use eval(), but you have to parse the string "DT1$BCC1[1]" first:
str <-paste0("DT1$", eval(cols[1]),"[1]")
eval(parse(text = str))
The $ dollar is suitable for console use(partial name matching). You should Use the subsetting [ operator.
For example you can call it like this :
DT1[1,cols[1]]
Ore more general :
x= 1
y = "BCC1"
DT1[x,y]
Note that DT1 that here is a data.frame not a data.table. You can do the same thing with a data.table:
DT1[x,y,with=F]
I have a dataset and unfortunately some of the column labels in my dataframe contain signs (- or +). This doesn't seem to bother the dataframe, but when I try to plot this with qplot it throws me an error:
x <- 1:5
y <- x
names <- c("1+", "2-")
mydf <- data.frame(x, y)
colnames(mydf) <- names
mydf
qplot(1+, 2-, data = mydf)
and if I enclose the column names in quotes it will just give me a category (or something to that effect, it'll give me a plot of "1+" vs. "2-" with one point in the middle).
Is it possible to do this easily? I looked at aes_string but didn't quite understand it (at least not enough to get it to work).
Thanks in advance.
P.S. I have searched for a solution online but can't quite find anything that helps me with this (it could be due to some aspect I don't understand), so I reason it might be because this is a completely retarded naming scheme I have :p.
Since you have non-standard column names, you need to to use backticks (`)in your column references.
For example:
mydf$`1+`
[1] 1 2 3 4 5
So, your qplot() call should look like this:
qplot(`1+`, `2-`, data = mydf)
You can find more information in ?Quotes and ?names
As said in the other answer you have a problem because you you don't have standard names. When solution is to avoid backticks notation is to convert colnames to a standard form. Another motivation to convert names to regular ones is , you can't use backticks in a lattice plot for example. Using gsub you can do this:
gsub('(^[0-9]+)[+|-]+|[+|-]+','a\\1',c("1+", "2-","a--"))
[1] "a1" "a2" "aa"
Hence, applying this to your example :
colnames(mydf) <- gsub('(^[0-9]+)[+|-]+|[+|-]+','a\\1',colnames(mydf))
qplot(a1,a2,data = mydf)
EIDT
you can use make.names with option unique =T
make.names(c("10+", "20-", "10-", "a30++"),unique=T)
[1] "X10." "X20." "X10..1" "a30.."
If you don't like R naming rules, here a custom version with using gsubfn
library(gsubfn)
gsubfn("[+|-]|^[0-9]+",
function(x) switch(x,'+'= 'a','-' ='b',paste('x',x,sep='')),
c("10+", "20-", "10-", "a30++"))
"x10a" "x20b" "x10b" "a30aa" ## note x10b looks better than X10..1