I am creating a simple data frame like this:
qcCtrl <- data.frame("2D6"="DNS00012345", "3A4"="DNS000013579")
My understanding is that the column names should be "2D6" and "3A4", but they are actually "X2D6" and "X3A4". Why are the X's being added and how do I make that stop?
I do not recommend working with column names starting with numbers, but if you insist, use the check.names=FALSE argument of data.frame:
qcCtrl <- data.frame("2D6"="DNS00012345", "3A4"="DNS000013579",
check.names=FALSE)
qcCtrl
2D6 3A4
1 DNS00012345 DNS000013579
One of the reasons I caution against this, is that the $ operator becomes more tricky to work with. For example, the following fails with an error:
> qcCtrl$2D6
Error: unexpected numeric constant in "qcCtrl$2"
To get round this, you have to enclose your column name in back-ticks whenever you work with it:
> qcCtrl$`2D6`
[1] DNS00012345
Levels: DNS00012345
The X is being added because R does not like having a number as the first character of a column name. To turn this off, use as.character() to tell R that the column name of your data frame is a character vector.
Related
please see the the column name "if" in the second column,the deifference is :when check.name=F,"." beside "if" disappear
Sorry for the code,because I try to type some codes to generate this data.frame like in the picture,but i failed due to the "if".We know that "if" is a reserved word in R(like else,for, while ,function).And here, i deliberately use the "if" as the column name (the 2nd column),and see whether R will generate some novel things.
So using another way, I type the "if" in the excel and save as the format of csv in order to use read.csv.
Question is:
Why "if." changes to "if"?(After i use check.names=FALSE)
enter image description here
?read.csv describes check.names= in a similar fashion:
check.names: logical. If 'TRUE' then the names of the variables in the
data frame are checked to ensure that they are syntactically
valid variable names. If necessary they are adjusted (by
'make.names') so that they are, and also to ensure that there
are no duplicates.
The default action is to allow you to do something like dat$<column-name>, but unfortunately dat$if will fail with Error: unexpected 'if' in "dat$if", ergo check.names=TRUE changing it to something that the parser will not trip over. Note, though, that dat[["if"]] will work even when dat$if will not.
If you are wondering if check.names=FALSE is ever a bad thing, then imagine this:
dat <- read.csv(text = "a,a\n2,3")
dat
# a a.1
# 1 2 3
dat <- read.csv(text = "a,a\n2,3", check.names = FALSE)
dat
# a a
# 1 2 3
In the second case, how does one access the second column by-name? dat$a returns 2 only. However, if you don't want to use $ or [[, and instead can rely on positional indexing for columns, then dat[,colnames(dat) == "a"] does return both of them.
I am trying to add 0s into character strings, but only under certain conditions.
I have a vector of file names like such:
my.fl <- c("res_P1_R1.rds", "res_P2_R1.rds",
"res_P1_R19.rds", "res_P2_R2.rds",
"res_P10_R1.rds", "res_P10_R19.rds")
I want to sort(my.fl) so that the file names are ordered by the numbers following the P and R, but as it stands sorting results in this:
"res_P1_R1.rds" "res_P1_R19.rds" "res_P10_R1.rds" "res_P10_R19.rds" "res_P2_R1.rds" "res_P2_R2.rds"
To fix this I need to add 0s after P and R, but only when the following number ranges from 1-9, if the following number is > 9 I want to do nothing.
The result should be as follows:
"res_P01_R01.rds" "res_P01_R19.rds" "res_P10_R01.rds" "res_P10_R19.rds" "res_P02_R01.rds" "res_P02_R02.rds"
and if I sort it, it is ordered as expected e.g.:
"res_P01_R01.rds" "res_P01_R19.rds" "res_P02_R01.rds" "res_P02_R02.rds" "res_P10_R01.rds" "res_P10_R19.rds"
I can add 0s based on position, but since the required position changes my solution only works on a subset of the file names. I think this would be a common problem but I haven't managed to find an answer on SO (or anywhere), any help much appreciated.
You should be able to just use mixedsort from the gtools package which removes the need to insert zeroes.
my.fl <- c("res_P1_R1.rds", "res_P2_R1.rds",
"res_P1_R19.rds", "res_P2_R2.rds",
"res_P10_R1.rds", "res_P10_R19.rds")
library(gtools)
mixedsort(my.fl)
[1] "res_P1_R1.rds" "res_P1_R19.rds" "res_P2_R1.rds" "res_P2_R2.rds" "res_P10_R1.rds" "res_P10_R19.rds"
But if you do want to insert the zeroes you could use something like:
sort(gsub("(?<=\\D)(\\d{1})(?=\\D)", "0\\1", my.fl, perl = TRUE))
[1] "res_P01_R01.rds" "res_P01_R19.rds" "res_P02_R01.rds" "res_P02_R02.rds" "res_P10_R01.rds" "res_P10_R19.rds"
I have a list, e.g. mylist=c("A","B","C"), and I wish to use list elements to extract factors of a data frame in R.
If MyDataFrame has a column name "A", I can extract the column/factor as MyDataFrame$A. However,
MyDataFrame$mylist[1]
fails. At first I thought that this was because mycolumn[3] is "A" whereas I need $A without the quotes. However, using
MyDataFrame$as.name(mylist[1])
fails as well, presumably because R looks for the string as.name(mylist[1]) as a factor name rather than processing the function (the rror it gives is "attempt to apply non-function". Setting x=as.name(mylist[1]) and then using MyDataFrame$x runs into the same problem of x not being treated as a variable.
Is there a straightforward way to do this, as I need to loop over a long list of column names in order to call the factors of interest.
Try this : rather than $
MyDataFrame[,mylist[1]]
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 [[.