I imported a dataset that unfortunately did not have any separators defined, nor in columns or in rows. I have tried to look for an option to define a specific row separator but could not find one that could be applicable to this situation.
df1 <- data.frame("V1" = "{lat:45.493,lng:-76.4886,alt:22400,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:328,spd:null,postime:2019-01-15 16:10:39},
{lat:45.5049,lng:-76.5285,alt:23425,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:321,spd:null,postime:2019-01-15 16:11:50},
{lat:45.5049,lng:-76.5285,alt:24000,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:321,spd:null,postime:2019-01-15 16:11:50},
{lat:45.5049,lng:-76.5285,alt:24000,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:321,spd:null,postime:2019-01-15 16:11:50}")
df2 <- data.frame("V1" = "{lat:45.493,lng:-76.4886,alt:22400,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:328,spd:null,postime:2019-01-15 16:10:39},
{lat:45.5049,lng:-76.5285,alt:23425,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:321,spd:null,postime:2019-01-15 16:11:50},
{lat:45.5049,lng:-76.5285,alt:24000,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:321,spd:null,postime:2019-01-15 16:11:50},
{lat:45.5049,lng:-76.5285,alt:24000,call:COFPQ,icao:C056P,registration:X-VLMP,sqk:6232,trak:321,spd:null,postime:2019-01-15 16:11:50}")
newdf <- rbind(df1,df2)
This is a model of the data that I am currently struggling with. Ideally, the row separators in this case would have to be defined as "},{" and the column separators as ",". I tried subsetting this pattern to a tab and defining a different separator but this either returned an error (tried with separate_rows from TidyR) or simply did nothing.
Hope you guys can help
This looks like incomplete (incorrect) JSON, so I suggest you bring it up-to-spec and then parse it with known tools. Some problems, easily mitigated:
sqk should have a comma-separator, perhaps a copy/paste issue. This might be generalized as any "number-letter" progression depending on your process. (Edit: your update seems to have resolved this issue, so I'll remove it. If you still need it, I recommend you go with a very literal gsub("([^,])sqk:", "\\1,sql:", s).)
Labels (e.g., lat, alt, sql) should all be double-quoted.
Non-numeric data needs to be quoted, specifically the dates.
Exception to 3: null should remain unquoted.
There are multiple "dicts" that need to be within a "list", i.e., from {...},{...} to [{...},{...}].
Side note with your data: I read them in with stringsAsFactors=FALSE, since we don't need factors.
fixjson <- function(s) {
gsub(",+", ",",
paste(
gsub('"sqk":([^,]+)', '"sqk":"\\1"',
gsub("\\s*\\b([A-Za-z]+)\\s*(?=:)", '"\\1"', # note 2
gsub('(?<=:)"(-?[0-9.]+|null)"', "\\1", # notes 3, 4
gsub("(?<=:)([^,]+)\\b", "\"\\1\"", # quote all data
s, perl = TRUE), perl = TRUE), perl = TRUE)),
collapse = "," )
)
}
fixjson(df1$V1)
# [1] "{\"lat\":45.493,\"lng\":-76.4886,\"alt\":22400,\"call\":\"COFPQ\",\"icao\":\"C056P\",\"registration\":\"X-VLMP\",\"sqk\":\"6232\",\"trak\":328,\"spd\":null,\"postime\":\"2019-01-15 16:10:39\"},\n {\"lat\":45.5049,\"lng\":-76.5285,\"alt\":23425,\"call\":\"COFPQ\",\"icao\":\"C056P\",\"registration\":\"X-VLMP\",\"sqk\":\"6232\",\"trak\":321,\"spd\":null,\"postime\":\"2019-01-15 16:11:50\"},\n {\"lat\":45.5049,\"lng\":-76.5285,\"alt\":24000,\"call\":\"COFPQ\",\"icao\":\"C056P\",\"registration\":\"X-VLMP\",\"sqk\":\"6232\",\"trak\":321,\"spd\":null,\"postime\":\"2019-01-15 16:11:50\"},\n {\"lat\":45.5049,\"lng\":-76.5285,\"alt\":24000,\"call\":\"COFPQ\",\"icao\":\"C056P\",\"registration\":\"X-VLMP\",\"sqk\":\"6232\",\"trak\":321,\"spd\":null,\"postime\":\"2019-01-15 16:11:50\"}"
From here, we use a well-defined json parser (from either jsonlite or RJSONIO, both use similar APIs):
jsonlite::fromJSON(paste("[", fixjson(df1$V1), "]", sep=""))
# lat lng alt call icao registration sqk trak spd postime
# 1 45.4930 -76.4886 22400 COFPQ C056P X-VLMP 6232 328 NA 2019-01-15 16:10:39
# 2 45.5049 -76.5285 23425 COFPQ C056P X-VLMP 6232 321 NA 2019-01-15 16:11:50
# 3 45.5049 -76.5285 24000 COFPQ C056P X-VLMP 6232 321 NA 2019-01-15 16:11:50
# 4 45.5049 -76.5285 24000 COFPQ C056P X-VLMP 6232 321 NA 2019-01-15 16:11:50
From here, rbind as needed. (Note that the null literal was translated into R's NA, which is "as it should be" in my opinion.)
Follow-on suggestion: you can use as.POSIXct directly on your postime column; I hope you are certain all of your data are in the same timezone since the field contains no hint.
Lastly, you mentioned something about non-ASCII characters gumming up the works. My recent edit included a little added robustness for spaces introduced from the use of iconv (e.g., the use of \\s*), so the following might suffice for you:
jsonlite::fromJSON( paste("[", fixjson(iconv(df2$V1, "latin1", "ASCII", sub="")), "]") )
(Use of iconv suggested by https://stackoverflow.com/a/9935242/3358272)
Related
I try to read in a .csv file with, example, such a column:
These values are meant like they are representing thousands of hours, not two or three hours and so on.
When I try to change the reading in options through
read.csv(file, sep = ";, dec = ".") nothing changes. It doesn't matter what I define, dec = "." or dec = "," it will always keep these numbers above.
You can use the following code:
library(readr)
df <- read_csv('data.csv', locale = locale(grouping_mark = "."))
df
Output:
# A tibble: 4 × 1
`X-ray`
<dbl>
1 2771
2 3783
3 1267
4 7798
As you can see, the values are now thousands.
An elegant way (in my opinion) is to create a new class, which you then use in the reading process.
This way, you stay flexible when your data is (really) messed up and the decimal/thousand separator is not equal over all (numeric) columns.
# Define a new class of numbers
setClass("newNumbers")
# Define substitution of dots to nothing
setAs("character", "newNumbers", function(from) as.numeric(gsub("\\.", "", from)))
# Now read
str(data.table::fread( "test \n 1.235 \n 1.265", colClasses = "newNumbers"))
# Classes ‘data.table’ and 'data.frame': 2 obs. of 1 variable:
# $ test: num 1235 1265
Solution proposed by Quinten will work; however, it's worth adding that function which is designed to process numbers with a grouping mark is col_number.
with(asNamespace("readr"),
read_delim(
I("X-ray hours\n---\n2.771\n3.778\n3,21\n"),
delim = ";",
col_names = c("x_ray_hours"),
col_types = cols(x_ray_hours = col_number()),
na = c("---"),
skip = 1
))
There is no need to define specific locale to handle this specific case only. Also locale setting will apply to the whole data and intention in this case to handle only that specific column. From docs:
?readr::parse_number
This drops any non-numeric characters before or after the first number.
Also if the columns use ; as a separator, read_delim is more appropriate.
I downloaded historical prices of an index but all prices are characters, that is are of the form : "24,31" (and i checked the mode).
I tried several things, such as :
as.numeric(as.character(VDAXcsv$Dernier))
which returns only NAs, or :
sapply(VDAXcsv$Dernier, as.numeric) | sapply(VDAXcsv, as.numeric)
Or simply
as.numeric(VDAXcsv)
And still only NAs, besides I tried to put "stringsAsFactors=FALSE" into my "read.zoo" function but it doesn't change anything. as.format doesn't work either.
x <- "24,31"
y <- as.numeric(gsub(",", ".", x))
y
# [1] 24.31
class(y)
# [1] "numeric"
A side note
I think depending on the data file you have, you might even want to prevent this to happen in the first place defining dec. Be careful if your sep is a comma as well though. It still can be an option, so you do not get your values as a character and therefor no need to do any replacements.
fread(file, header = T, sep = ";", dec = ",") # fread is data.table, but I think read.csv and any others support it as well
You can use this function:
library(readr)
parse_number("24,31", locale = locale(decimal_mark = ","))
Is vectorized, so just put VDAXcsv$Dernier as first argument.
I have a file that that has over a hundred million rows and scattered throughout there are extra tab delimiters in fields. I need to read the problematic rows into R whilst ignoring the others due to the large file size involved.
Example txt file with extra delimiters in some rows:
text_file <-"My\tname\tis\tAlpha\nMy\tname\tis\t\t\tBravo\nMy\tname\tis\tCharlie\nMy\tname\tis\t\t\tDelta\nMy\tname\tis\tEcho"
The first thing that I tried was using the 'readLines' function however whilst I can specify the row to stop on it will still read everything else up to that point which could still be too much
readLines(textConnection(text_file), n = 4)
[1] "My\tname\tis\tAlpha" "My\tname\tis\t\t\tBravo" "My\tname\tis\tCharlie" "My\tname\tis\t\t\tDelta"
I then realised that I could also use the other dataset import functions if I specified the delimiter to be something that should probably never appear. The "fread" function from the data.table package would be perfect for this as it is the fastest way to deal with large datasets like mine however when I tried it the data was in a format that I couldnt really work with further:
library(data.table)
library(stringi)
lines <- fread(text_file, sep = NULL, header = FALSE, skip = 1, nrows = 3)
> lines
V1
1: My\tname\tis\t\t\tBravo
2: My\tname\tis\tCharlie
3: My\tname\tis\t\t\tDelta
> invalid_delimiter_rows <- which(stri_count_regex(lines, "\\t") != 3)
Warning message:
In stri_count_regex(lines, "\\t") :
argument is not an atomic vector; coercing
Preferably I shouldnt have to convert this data after importing however when I tried changing this to a character vector or list it was still in a bad format (the concatenation is considered part of the string and not a function). What is the most computing time efficient way that I could approach this issue?
> class(lines)
[1] "data.table" "data.frame"
> as.character(lines)
[1] "c(\"My\\tname\\tis\\t\\t\\tBravo\", \"My\\tname\\tis\\tCharlie\", \"My\\tname\\tis\\t\\t\\tDelta\")"
Let's replicate the process till fread() import:
# your example string
text_file <-"My\tname\tis\tAlpha\nMy\tname\tis\t\t\tBravo\nMy\tname\tis\tCharlie\nMy\tname\tis\t\t\tDelta\nMy\tname\tis\tEcho"
# import
library(data.table)
lines <- fread(text_file, sep = NULL, header = FALSE, skip = 1, nrows = 5)
lines
V1
1: My\tname\tis\t\t\tBravo
2: My\tname\tis\tCharlie
3: My\tname\tis\t\t\tDelta
4: My\tname\tis\tEcho
When you try
as.character(lines)
[1] "c(\"My\\tname\\tis\\t\\t\\tBravo\", \"My\\tname\\tis\\tCharlie\", \"My\\tname\\tis\\t\\t\\tDelta\", \"My\\tname\\tis\\tEcho\")"
it converts all data.table in character, so each column will be a concatenated vector. See below:
as.character(data.table(lines$V1, lines$V1))
[1] "c(\"My\\tname\\tis\\t\\t\\tBravo\", \"My\\tname\\tis\\tCharlie\", \"My\\tname\\tis\\t\\t\\tDelta\", \"My\\tname\\tis\\tEcho\")"
[2] "c(\"My\\tname\\tis\\t\\t\\tBravo\", \"My\\tname\\tis\\tCharlie\", \"My\\tname\\tis\\t\\t\\tDelta\", \"My\\tname\\tis\\tEcho\")"
What you want is extract just lines$V1, which is already a character vector.
lines$V1
[1] "My\tname\tis\t\t\tBravo" "My\tname\tis\tCharlie" "My\tname\tis\t\t\tDelta" "My\tname\tis\tEcho"
I have a .csv file with account codes in the form of 00xxxxx and I need them to stay that way for use with other programs which use the account codes in this format. I was just working on an R script to reconcile account charges on Friday and swore that as.is = T was working for me. Now, it doesn't seem to be. Here's some example data:
test <- data.frame(col1 = c("apple", "banana", "carrot"),
col2 = c(100, 200, 300),
col3 = c("00234", "00345", "00456"))
My write.table strategy:
write.table(test, file = "C:/path/test.csv", quote = T,
sep=",", row.names = F)
Remove the old data.frame and re-read:
rm(test)
test <- read.csv("C:/path/test.csv")
test
col1 col2 col3
1 apple 100 234
2 banana 200 345
3 carrot 300 456
In case it's not clear, it should look like the original data.frame we created:
test
col1 col2 col3
1 apple 100 00234
2 banana 200 00345
3 carrot 300 00456
I also tried the following, after perusing the available read.table options, with the results all the same as above:
test <- read.csv("C:/path/test.csv", quote = '"')
test <- read.csv("C:/path/test.csv", as.is = T)
test <- read.csv("C:/path/test.csv", as.is = T, quote = '"')
StringsAsFactors didn't seem to be relevant in this case (and sounds like as.is will do the same thing.
When I open the file in Emacs, col3 is, indeed, surrounded by quotes, so I'd expect it to be treated like text instead of converted to a number:
Most of the other questions are simply about not treating things like factors, or getting numbers not to be recognized as characters, usually the result of an overlooked character string in that column.
I see I can pursue the colClasses argument from questions like this one, but I'd prefer not to; my "colClasses" are built into the data :) Quoted = character, not quoted = numeric.
After pinging a couple of friends who are R users, they both suggested using colClasses. I was relieved to find that I didn't need to specify each class, since my data is ~25 columns. SO confirmed this (once I knew what I was looking for) in another question.
test <- read.csv("C:/path/test.csv", colClasses = c(col3 = "character"))
test
col1 col2 col3
1 apple 100 00234
2 banana 200 00345
3 carrot 300 00456
As it currently stands, the question is a duplicate of the other with respect to the solution. The difference is that I was looking for ways other than colClasses (since as.is sounds like such a likely candidate), while that question was about how to use colClasses.
I'll reiterate that I don't actually like this solution, even thought it's pretty simple. Quotes denote text fields in a .csv, and they don't seem to be respected in this case. The LibreOffice .csv import has a checkbox for "Treat quoted fields as text," which I'd think is analogous to as.is = T in R. Obviously not! #end_rant
I have this issue too. Of course you can manually specify colClasses, but why is this necessary when data is quoted? I agree with the OP's 'rant' in the answer posted to his own question:
Quotes denote text fields in a .csv, and they don't seem to be
respected in this case.
Anyway, I elected to use data.table's fread() which doesn't have this issue. Still annoying behaviour for read.csv though.
# here's a data frame with chr and int columns
my_df <- data.frame(chars=letters[1:5],
nums=1:5,
txt_nums=sprintf('%02d', 1:5),
stringsAsFactors=F)
# all looks as it should
lapply(my_df, class)
# $chars
# [1] "character"
#
# $nums
# [1] "integer"
#
# $txt_nums
# [1] "character"
But now, write to csv, read it back in, and the third column is coerced to int!
# quote=T redundant since that's the default, but just to be sure
write.csv(my_df, 'my_df.csv', row.names=F, quote=T)
my_df2 <- read.csv('my_df.csv')
lapply(my_df2, class)
# even with as.is=TRUE, same issue
my_df2 <- read.csv('my_df.csv', as.is=T)
lapply(my_df2, class)
# data.table's fread doesn't have this issue, at least
library(data.table)
my_dt <- fread('my_df.csv')
lapply(my_dt, class)
I expect there's a better method, but one option would be to use quote=""
test <- read.csv("C:/path/test.csv", as.is = TRUE, quote = "")
This would make the quotes part of the values, giving you:
test
#col1 col2 col3
#1 apple 100 "00234"
#2 banana 200 "00345"
#3 carrot 300 "00456"
You could then either keep them in that format, or use something like gsub to remove them:
test$col3 <- gsub('"', '', test$col3)
test
#col1 col2 col3
#1 apple 100 00234
#2 banana 200 00345
#3 carrot 300 00456
You can use some kind of apply-type function to do the gsub on the whole data frame at once:
test <- as.data.frame(sapply(test,gsub,pattern='"',replacement=""))
sapply code taken from: R - how to replace parts of variable strings within data frame
Obviously, this method will only be useful to you if you don't need the quotes elsewhere for other reasons.
The popular "readr" package also respects the quotes in .csv files.
test <- read_csv("C:/path/test.csv")
I couldn't agree more that the base R read.csv() behavior is unacceptable.
I've been trying to remove the white space that I have in a data frame (using R). The data frame is large (>1gb) and has multiple columns that contains white space in every data entry.
Is there a quick way to remove the white space from the whole data frame? I've been trying to do this on a subset of the first 10 rows of data using:
gsub( " ", "", mydata)
This didn't seem to work, although R returned an output which I have been unable to interpret.
str_replace( " ", "", mydata)
R returned 47 warnings and did not remove the white space.
erase_all(mydata, " ")
R returned an error saying 'Error: could not find function "erase_all"'
I would really appreciate some help with this as I've spent the last 24hrs trying to tackle this problem.
Thanks!
A lot of the answers are older, so here in 2019 is a simple dplyr solution that will operate only on the character columns to remove trailing and leading whitespace.
library(dplyr)
library(stringr)
data %>%
mutate_if(is.character, str_trim)
## ===== 2020 edit for dplyr (>= 1.0.0) =====
df %>%
mutate(across(where(is.character), str_trim))
You can switch out the str_trim() function for other ones if you want a different flavor of whitespace removal.
# for example, remove all spaces
df %>%
mutate(across(where(is.character), str_remove_all, pattern = fixed(" ")))
If i understood you correctly then you want to remove all the white spaces from entire data frame, i guess the code which you are using is good for removing spaces in the column names.I think you should try this:
apply(myData, 2, function(x)gsub('\\s+', '',x))
Hope this works.
This will return a matrix however, if you want to change it to data frame then do:
as.data.frame(apply(myData, 2, function(x) gsub('\\s+', '', x)))
EDIT In 2020:
Using lapply and trimws function with both=TRUE can remove leading and trailing spaces but not inside it.Since there was no input data provided by OP, I am adding a dummy example to produce the results.
DATA:
df <- data.frame(val = c(" abc", " kl m", "dfsd "),
val1 = c("klm ", "gdfs", "123"),
num = 1:3,
num1 = 2:4,
stringsAsFactors = FALSE)
#situation: 1 (Using Base R), when we want to remove spaces only at the leading and trailing ends NOT inside the string values, we can use trimws
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, cols_to_be_rectified] <- lapply(df[, cols_to_be_rectified], trimws)
# situation: 2 (Using Base R) , when we want to remove spaces at every place in the dataframe in character columns (inside of a string as well as at the leading and trailing ends).
(This was the initial solution proposed using apply, please note a solution using apply seems to work but would be very slow, also the with the question its apparently not very clear if OP really wanted to remove leading/trailing blank or every blank in the data)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, cols_to_be_rectified] <- lapply(df[, cols_to_be_rectified],
function(x) gsub('\\s+', '', x))
## situation: 1 (Using data.table, removing only leading and trailing blanks)
library(data.table)
setDT(df)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, c(cols_to_be_rectified) := lapply(.SD, trimws), .SDcols = cols_to_be_rectified]
Output from situation1:
val val1 num num1
1: abc klm 1 2
2: kl m gdfs 2 3
3: dfsd 123 3 4
## situation: 2 (Using data.table, removing every blank inside as well as leading/trailing blanks)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[, c(cols_to_be_rectified) := lapply(.SD, function(x) gsub('\\s+', '', x)), .SDcols = cols_to_be_rectified]
Output from situation2:
val val1 num num1
1: abc klm 1 2
2: klm gdfs 2 3
3: dfsd 123 3 4
Note the difference between the outputs of both situation, In row number 2: you can see that, with trimws we can remove leading and trailing blanks, but with regex solution we are able to remove every blank(s).
I hope this helps , Thanks
One possibility involving just dplyr could be:
data %>%
mutate_if(is.character, trimws)
Or considering that all variables are of class character:
data %>%
mutate_all(trimws)
Since dplyr 1.0.0 (only strings):
data %>%
mutate(across(where(is.character), trimws))
Or if all columns are strings:
data %>%
mutate(across(everything(), trimws))
Picking up on Fremzy and the comment from Stamper, this is now my handy routine for cleaning up whitespace in data:
df <- data.frame(lapply(df, trimws), stringsAsFactors = FALSE)
As others have noted this changes all types to character. In my work, I first determine the types available in the original and conversions required. After trimming, I re-apply the types needed.
If your original types are OK, apply the solution from MarkusN below https://stackoverflow.com/a/37815274/2200542
Those working with Excel files may wish to explore the readxl package which defaults to trim_ws = TRUE when reading.
Picking up on Fremzy and Mielniczuk, I came to the following solution:
data.frame(lapply(df, function(x) if(class(x)=="character") trimws(x) else(x)), stringsAsFactors=F)
It works for mixed numeric/charactert dataframes manipulates only character-columns.
You could use trimws function in R 3.2 on all the columns.
myData[,c(1)]=trimws(myData[,c(1)])
You can loop this for all the columns in your dataset. It has good performance with large datasets as well.
If you're dealing with large data sets like this, you could really benefit form the speed of data.table.
library(data.table)
setDT(df)
for (j in names(df)) set(df, j = j, value = df[[trimws(j)]])
I would expect this to be the fastest solution. This line of code uses the set operator of data.table, which loops over columns really fast. There is a nice explanation here: Fast looping with set.
R is simply not the right tool for such file size. However have 2 options :
Use ffdply and ff base
Use ff and ffbase packages:
library(ff)
library(ffabse)
x <- read.csv.ffdf(file=your_file,header=TRUE, VERBOSE=TRUE,
first.rows=1e4, next.rows=5e4)
x$split = as.ff(rep(seq(splits),each=nrow(x)/splits))
ffdfdply( x, x$split , BATCHBYTES=0,function(myData)
apply(myData,2,function(x)gsub('\\s+', '',x))
Use sed (my preference)
sed -ir "s/(\S)\s+(/S)/\1\2/g;s/^\s+//;s/\s+$//" your_file
If you want to maintain the variable classes in your data.frame - you should know that using apply will clobber them because it outputs a matrix where all variables are converted to either character or numeric. Building upon the code of Fremzy and Anthony Simon Mielniczuk you can loop through the columns of your data.frame and trim the white space off only columns of class factor or character (and maintain your data classes):
for (i in names(mydata)) {
if(class(mydata[, i]) %in% c("factor", "character")){
mydata[, i] <- trimws(mydata[, i])
}
}
I think that a simple approach with sapply, also works, given a df like:
dat<-data.frame(S=LETTERS[1:10],
M=LETTERS[11:20],
X=c(rep("A:A",3),"?","A:A ",rep("G:G",5)),
Y=c(rep("T:T",4),"T:T ",rep("C:C",5)),
Z=c(rep("T:T",4),"T:T ",rep("C:C",5)),
N=c(1:3,'4 ','5 ',6:10),
stringsAsFactors = FALSE)
You will notice that dat$N is going to become class character due to '4 ' & '5 ' (you can check with class(dat$N))
To get rid of the spaces on the numeic column simply convert to numeric with as.numeric or as.integer.
dat$N<-as.numeric(dat$N)
If you want to remove all the spaces, do:
dat.b<-as.data.frame(sapply(dat,trimws),stringsAsFactors = FALSE)
And again use as.numeric on col N (ause sapply will convert it to character)
dat.b$N<-as.numeric(dat.b$N)