I need help conditionally adding leading or trailing zeros.
I have a dataframe with one column containing icd9 diagnoses. as a vector, the column looks like:
"33.27" "38.45" "9.25" "4.15" "38.45" "39.9" "84.1" "41.5" "50.3"
I need all the values to have the length of 5, including the period in the middle (not counting ""). If the value has one digit before the period, it need to have a leading zero. If value has one digit after the period, it need to have zero at the end. So the result should look like this:
"33.27" "38.45" "09.25" "04.15" "38.45" "39.90" "84.10" "41.50" "50.30"
Here is the vector for R:
icd9 <- c("33.27", "38.45", "9.25", "4.15", "38.45", "39.9", "84.1", "41.5", "50.3" )
This does it in one line
formatC(as.numeric(icd9),width=5,format='f',digits=2,flag='0')
ICD-9 codes have some formatting quirks which can lead to misinterpretation with simple string processing. The icd package on CRAN takes care of all the corner cases when doing ICD processing, and has been battle-tested over about six years of use by many R users.
Using this function called change that accepts the argument of the max number of characters, i think it can help
change<-function(x, n=max(nchar(x))) gsub(" ", "0", formatC(x, width=n))
icd92<-gsub(" ","",paste(change(icd9,5)))
You can also use sprintf after converting the vector into numeric.
sprintf("%05.2f", as.numeric(icd9))
[1] "33.27" "38.45" "09.25" "04.15" "38.45" "39.90" "84.10" "41.50" "50.30"
Notes
The examples in ?sprint to get work out the proper format.
There is some risk of introducing errors due to numerical precision here, though it works well in the example.
Related
I have a dataframe called "Camera_data" and a column called "Numeric_time"
My "Numeric_time" column is in character format and includes hours, minutes and seconds, it looks like this: 08:40:01
I need to remove the numbers that pertain to seconds and replace the semicolons with periods to make a decimal number for my time. I need it to look like this: 08.40 in order to turn my time into radians for an analysis I'm running.
I've looked for a few solutions in stringr, but so far can't work out how to consistently take off the last three characters. I think once I have removed the seconds and replaced the : with a . I can just use as.numeric to turn the character column into a numerical column, but would really appreciate any help!
We could do
Camera_data$Numeric_time <- as.numeric(chartr(":", ".",
sub(":\\d{2}$", "", Camera_data$Numeric_time )))
Or use substr
Camera_data$Numeric_time <- substr(Camera_data$Numeric_time, 1, nchar(Camera_data$Numeric_time)-3)
Using gsub with two capture groups.
as.numeric(gsub('(\\d+):(\\d+).*', '\\1.\\2', x))
# [1] 8.40 18.41 0.00
Data:
x <- c('08:40:01', '18:41:01', '00:00:01')
I have created a list (Based on items in a column) in order to subset my dataset into smaller datasets relating to a particular variable. This list contains strings with hyphens in them -.
dim.list <- c('Age_CareContactDate-Gender', 'Age_CareContactDate-Group',
'Age_ServiceReferralReceivedDate-Gender',
'Age_ServiceReferralReceivedDate-Gender-0-18',
'Age_ServiceReferralReceivedDate-Group',
'Age_ServiceReferralReceivedDate-Group-ReferralReason')
I have then written some code to loop through each item in this list subsetting my main data.
for (i in dim.list) {assign(paste("df1.",i,sep=""),df[df$Dimension==i,])}
This works fine, however when I come to aggregate this in order to get some summary statistics I can't reference the dataset as R stops reading after the hyphen (I assume that the hyphen is some special character)
If I use a different list without hyphens e.g.
dim.list.abr <- c('ACCD_Gen','ACCD_Grp',
'ASRRD_Gen',
'ASRRD_Gen_0_18',
'ASRRD_Grp',
'ASRRD_Grp_RefRsn')
When my for loop above executes I get 6 data.frames with no observations.
Why is this happening?
Comment to answer:
Hyphens aren't allowed in standard variable names. Think of a simple example: a-b. Is it a variable name with a hyphen or is it a minus b? The R interpreter assumes a minus b, because it doesn't require spaces for binary operations. You can force non-standard names to work using backticks, e.g.,
# terribly confusing names:
`a-b` <- 5
`x+y` <- 10
`mean(x^2)` <- "this is awful"
but you're better off following the rules and using standard names without special characters like + - * / % $ # # ! & | ^ ( [ ' " in them. At ?quotes there is a section on Names and Identifiers:
Identifiers consist of a sequence of letters, digits, the period (.) and the underscore. They must not start with a digit nor underscore, nor with a period followed by a digit. Reserved words are not valid identifiers.
So that's why you're getting an error, but what you're doing isn't good practice. I completely agree with Axeman's comments. Use split to divide up your data frame into a list. And keep it in a list rather than use assign, it will be much easier to loop over or use lapply with that way. You might want to read my answer at How to make a list of data frames for a lot of discussion and examples.
Regarding your comment "dim.list is not the complete set of unique entries in the Dimensions column", that just means you need to subset before you split:
nice_list = df[df$Dimension %in% dim.list, ]
nice_list = split(nice_list, nice_list$Dimension)
If you load the pracma package into the r console and type
gammainc(2,2)
you get
lowinc uppinc reginc
0.5939942 0.4060058 0.5939942
This looks like some kind of a named tuple or something.
But, I can't work out how to extract the number below the lowinc, namely 0.5939942. The code (gammainc(2,2))[1] doesn't work, we just get
lowinc
0.5939942
which isn't a number.
How is this done?
As can be checked with str(gammainc(2,2)[1]) and class(gammainc(2,2)[1]), the output mentioned in the OP is in fact a number. It is just a named number. The names used as attributes of the vector are supposed to make the output easier to understand.
The function unname() can be used to obtain the numerical vector without names:
unname(gammainc(2,2))
#[1] 0.5939942 0.4060058 0.5939942
To select the first entry, one can use:
unname(gammainc(2,2))[1]
#[1] 0.5939942
In this specific case, a clearer version of the same might be:
unname(gammainc(2,2)["lowinc"])
Double brackets will strip the dimension names
gammainc(2,2)[[1]]
gammainc(2,2)[["lowinc"]]
I don't claim it to be intuitive, or obvious, but it is mentioned in the manual:
For vectors and matrices the [[ forms are rarely used, although they
have some slight semantic differences from the [ form (e.g. it drops
any names or dimnames attribute, and that partial matching is used for
character indices).
The partial matching can be employed like this
gammainc(2, 2)[["low", exact=FALSE]]
In R vectors may have names() attribute. This is an example:
vector <- c(1, 2, 3)
names(vector) <- c("first", "second", "third")
If you display vector, you should probably get desired output:
vector
> vector
first second third
1 2 3
To ensure what type of output you get after the function you can use:
class(your_function())
I hope this helps.
I have a data set in which I want to pad zeroes in front of a set of dates that don't have six characters. For example, I have a date that reads 91003 (October 3rd, 2009) and I want it to read 091003, as well as any other date that is missing a zero in front. When I use the sprintf function, the code is:
Data1$entrydate <- sprintf("%06d", data1$entrydate)
But what it spits out is something like 000127, or some other other random number for all the other dates in the problem. I don't understand what's going on, and I would appreciate some help on the issue. Thanks.
PS. I am sometimes also getting a error message that sprintf is only for character values, I don't know if there is any code for numerical values.
I guess you got different results than expected because the column class was factor. You can convert the column to numeric either by as.numeric(as.character(datacolumn)) or as.numeric(levels(datacolumn)). According to ?factor
To transform a factor ‘f’ to approximately its
original numeric values, ‘as.numeric(levels(f))[f]’ is recommended
and slightly more efficient than ‘as.numeric(as.character(f))’.
So, you can use
levels(data1$entrydate) <- sprintf('%06d', as.numeric(levels(data1$entrydate)))
Example
Here is an example that shows the problem
v1 <- factor(c(91003, 91104,90103))
sprintf('%06d', v1)
#[1] "000002" "000003" "000001"
Or, it is equivalent to
sprintf('%06d', as.numeric(v1)) #the formatted numbers are
# the numeric index of factor levels.
#[1] "000002" "000003" "000001"
When you convert it back to numeric, works as expected
sprintf('%06d', as.numeric(levels(v1)))
#[1] "090103" "091003" "091104"
Is it possible to add or retain one or more leading zeros to a number without the result being converted to character? Every solution I have found for adding leading zeros returns a character string, including: paste, formatC, format, and sprintf.
For example, can x be 0123 or 00123, etc., instead of 123 and still be numeric?
x <- 0123
EDIT
It is not essential. I was just playing around with the following code and the last two lines gave the wrong answer. I just thought maybe if I could have leading zeros with numeric format obtaining the correct answer would be easier.
a7 = c(1,1,1,0); b7=c(0,1,1,1); # 4
a77 = '1110' ; b77='0111' ; # 4
a777 = 1110 ; b777=0111 ; # 4
length(b7[(b7 %in% intersect(a7,b7))])
R - count matches between characters of one string and another, no replacement
keyword <- unlist(strsplit(a77, ''))
text <- unlist(strsplit(b77, ''))
sum(!is.na(pmatch(keyword, text)))
ab7 <- read.fwf(file = textConnection(as.character(rbind(a777, b777))), widths = c(1,1,1,1), colClasses = rep("character", 2))
length(ab7[2,][(ab7[2,] %in% intersect(ab7[1,],ab7[2,]))])
You are not thinking correctly about what a "number" is. Programming languages store an internal representation which retains full precision to the machine limit. You are apparently concerned with what gets printed to your screen or console. By definition, those number characters are string elements, which is to say, a couple bytes are processed by the ASCII decoder (or equivalent) to determine what to draw on the screen. What x "is," to draw happily on Presidential Testimony, depends on your definition of what "is" is.
You could always create your own class of objects that has one slot for the value of the number (but if it is stored as numeric then what we see as 123 will actually be stored as as a binary value, something like 01111011 (though probably with more leading 0's)) and another slot or attribute for either the number of leading 0's or the number of significant digits. Then you can write methods for what to do with the number (and what effect that will have on the leading 0's, sig digits, etc.).
The print method could then make sure to print it with the leading zeros while keeping the internal value as a number.
But this seems a bit overkill in most cases (though I know that some fields make a big deal about indicating number of significant digits so that leading 0's could be important). It may be simpler to use the conversion to character methods that you already know about, but just do the printing in a way that does not look obviously like a number, see the cat and print functions for the options.