I have a dataset, In which one column has a values in the format of [A-Z][A-Z][0-1][0-9][0-1][0-1][0-1][0-9][0-9] ie, AC1200019
Now I want to convert this format to [A-Z][A-Z][-][0-1][0-9][-][0-1][0-1][0-1][-][0-9][0-9] ie, AC-12-000-19
([A-Z][A-Z])([0-1][0-9])([0-1][0-1][0-1])([0-9][0-9])
Try this.Replace by $1-$2-$3-$4 or \\1-\\2-\\3-\\4.See demo.
https://regex101.com/r/uK9cD8/5
Try
gsub('^([A-Z]{2})([0-1][0-9])([0-1]{3})([0-9]{2})', '\\1-\\2-\\3-\\4', str1)
#[1] "AC-12-000-19"
data
str1 <- 'AC1200019'
Assuming the entire column has the same number of characters, here a simple version.
library(stringr)
x <- data.frame(X1 = c("AC1510018", "AC1200019", "BT1801007"))
paste(str_sub(x$X1,1,2), str_sub(x$X1,3,4),
str_sub(x$X1,5,7), str_sub(x$X1,8,9) , sep= "-")
I like the dplyr suite so here a version using dplyr and tidyr:
library(dplyr)
library(tidyr)
x %>%
separate(X1, into = c("X2", "X3", "X4", "X5"), sep = c(2,4,7)) %>%
unite("X1", X2, X3, X4, X5, sep="-")
or
x %>%
transmute(X2 = paste(str_sub(X1,1,2), str_sub(X1,3,4),
str_sub(X1,5,7), str_sub(X1,8,9) , sep= "-"))
Related
I am looking for a shorter and more pretty solution (possibly in tidyverse) to the following problem. I have a data.frame "data":
id string
1 A 1.001 xxx 123.123
2 B 23,45 lorem ipsum
3 C donald trump
4 D ssss 134, 1,45
What I wanted to do is to extract all numbers (no matter if the delimiter is "." or "," -> in this case I assume that string "134, 1,45" can be extracted into two numbers: 134 and 1.45) and create a data.frame "output" looking similar to this:
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
I managed to do this (code below) but the solution is pretty ugly for me also not so efficient (two for-loops). Could someone suggest a better way to do do this (preferably using dplyr)
# data
data <- data.frame(id = c("A", "B", "C", "D"),
string = c("1.001 xxx 123.123",
"23,45 lorem ipsum",
"donald trump",
"ssss 134, 1,45"),
stringsAsFactors = FALSE)
# creating empty data.frame
len <- length(unlist(sapply(data$string, function(x) gregexpr("[0-9]+[,|.]?[0-9]*", x))))
output <- data.frame(id = rep(NA, len), string = rep(NA, len))
# main solution
start = 0
for(i in 1:dim(data)[1]){
tmp_len <- length(unlist(gregexpr("[0-9]+[,|.]?[0-9]*", data$string[i])))
for(j in (start+1):(start+tmp_len)){
output[j,1] <- data$id[i]
output[j,2] <- regmatches(data$string[i], gregexpr("[0-9]+[,|.]?[0-9]*", data$string[i]))[[1]][j-start]
}
start = start + tmp_len
}
# further modifications
output$string <- gsub(",", ".", output$string)
output$string <- as.numeric(ifelse(substring(output$string, nchar(output$string), nchar(output$string)) == ".",
substring(output$string, 1, nchar(output$string) - 1),
output$string))
output
1) Base R This uses relatively simple regular expressions and no packages.
In the first 2 lines of code replace any comma followed by a space with a
space and then replace all remaining commas with a dot. After these two lines s will be: c("1.001 xxx 123.123", "23.45 lorem ipsum", "donald trump", "ssss 134 1.45")
In the next 4 lines of code trim whitespace from beginning and end of each string field and split the string field on whitespace producing a
list. grep out those elements consisting only of digits and dots. (The regular expression ^[0-9.]*$ matches the start of a word followed by zero or more digits or dots followed by the end of the word so only words containing only those characters are matched.) Replace any zero length components with NA. Finally add data$id as the names. After these 4 lines are run the list L will be list(A = c("1.001", "123.123"), B = "23.45", C = NA, D = c("134", "1.45")) .
In the last line of code convert the list L to a data frame with the appropriate names.
s <- gsub(", ", " ", data$string)
s <- gsub(",", ".", s)
L <- strsplit(trimws(s), "\\s+")
L <- lapply(L, grep, pattern = "^[0-9.]*$", value = TRUE)
L <- ifelse(lengths(L), L, NA)
names(L) <- data$id
with(stack(L), data.frame(id = ind, string = values))
giving:
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
2) magrittr This variation of (1) writes it as a magrittr pipeline.
library(magrittr)
data %>%
transform(string = gsub(", ", " ", string)) %>%
transform(string = gsub(",", ".", string)) %>%
transform(string = trimws(string)) %>%
with(setNames(strsplit(string, "\\s+"), id)) %>%
lapply(grep, pattern = "^[0-9.]*$", value = TRUE) %>%
replace(lengths(.) == 0, NA) %>%
stack() %>%
with(data.frame(id = ind, string = values))
3) dplyr/tidyr This is an alternate pipeline solution using dplyr and tidyr. unnest converts to long form, id is made factor so that we can later use complete to recover id's that are removed by subsequent filtering, the filter removes junk rows and complete inserts NA rows for each id that would otherwise not appear.
library(dplyr)
library(tidyr)
data %>%
mutate(string = gsub(", ", " ", string)) %>%
mutate(string = gsub(",", ".", string)) %>%
mutate(string = trimws(string)) %>%
mutate(string = strsplit(string, "\\s+")) %>%
unnest() %>%
mutate(id = factor(id))
filter(grepl("^[0-9.]*$", string)) %>%
complete(id)
4) data.table
library(data.table)
DT <- as.data.table(data)
DT[, string := gsub(", ", " ", string)][,
string := gsub(",", ".", string)][,
string := trimws(string)][,
string := setNames(strsplit(string, "\\s+"), id)][,
list(string = list(grep("^[0-9.]*$", unlist(string), value = TRUE))), by = id][,
list(string = if (length(unlist(string))) unlist(string) else NA_character_), by = id]
DT
Update Removed assumption that junk words do not have digit or dot. Also added (2), (3) and (4) and some improvements.
We can replace the , in between the numbers with . (using gsub), extract the numbers with str_extract_all (from stringr into a list), replace the list elements that have length equal to 0 with NA, set the names of the list with 'id' column, stack to convert the list to data.frame and rename the columns.
library(stringr)
setNames(stack(setNames(lapply(str_extract_all(gsub("(?<=[0-9]),(?=[0-9])", ".",
data$string, perl = TRUE), "[0-9.]+"), function(x)
if(length(x)==0) NA else as.numeric(x)), data$id))[2:1], c("id", "string"))
# id string
#1 A 1.001
#2 A 123.123
#3 B 23.45
#4 C NA
#5 D 134
#6 D 1.45
Same idea as Gabor's. I had hoped to use R's built-in parsing of strings (type.convert, used in read.table) rather than writing custom regex substitutions:
sp = setNames(strsplit(data$string, " "), data$id)
spc = lapply(sp, function(x) {
x = x[grep("[^0-9.,]$", x, invert=TRUE)]
if (!length(x))
NA_real_
else
mapply(type.convert, x, dec=gsub("[^.,]", "", x), USE.NAMES=FALSE)
})
setNames(rev(stack(spc)), names(data))
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
Unfortunately, type.convert is not robust enough to consider both decimal delimiters at once, so we need this mapply malarkey instead of type.convert(x, dec = "[.,]").
I have a file with multiple columns. I am showing two columns in which I am interested two columns
Probe.Set.ID Entrez.Gene
A01157cds_s_at 50682
A03913cds_s_at 29366
A04674cds_s_at 24860 /// 100909612
A07543cds_s_at 24867
A09811cds_s_at 25662
---- ----
A16585cds_s_at 25616
I need to replace /// with "\t"(tab) and the output should be like
A01157cds_s_at;50682
A03913cds_s_at;29366
A04674cds_s_at;24860 100909612
Also, I need to avoid the ones with "---"
Here is slightly more different approach using dplyr:
data <- data.frame(Probe.Set.ID = c("A01157cds_s_at",
"A03913cds_s_at",
"A04674cds_s_at",
"A07543cds_s_at",
"A09811cds_s_at",
"----",
"A16585cds_s_at"),
Entrez.Gene = c("50682",
"29366",
"24860 /// 100909612",
"24867",
"25662",
"----",
"25616")
)
if(!require(dplyr)) install.packages("dplyr")
library(dplyr)
data %>%
filter(Entrez.Gene != "----") %>%
mutate(new_column = paste(Probe.Set.ID,
gsub("///", "\t", Entrez.Gene),
sep = ";"
)
) %>% select(new_column)
Looks like you will want to subset the data, then paste the two columns together, then use gsub to make the replace the '///'. Here is what I came up with, with dat being the dataframe containing the two columns.
dat = dat[dat$Probe.Set.ID != "----",] # removes the rows with "---"
dat = paste0(dat$Probe.Set.ID, ";", dat$Entrez.Gene) # pastes the columns together and adds the ";"
dat = gsub("///","\t",dat) # replaces the "///" with a tab
Also, use cat() to view the tab as opposed to "\t". I got that from here: How to replace specific characters of a string with tab in R. This will output a list as opposed to a data.frame. You can convert back with data.frame(), but then you cannot use cat() to view.
We can use dplyr and tidyr here.
library(dplyr)
library(tidyr)
> df <- data.frame(
col1 = c('A01157cds_s_at', 'A03913cds_s_at', 'A04674cds_s_at', 'A07543cds_s_at', '----'),
col2 = c('50682', '29366', '24860 /// 100909612', '24867', '----'))
> df %>% filter(col1 != '----') %>%
separate(col2, c('col2_first', 'col2_second'), '///', remove = T) %>%
unite(col1_new, c(col1, col2_first), sep = ';', remove = T)
> df
## col1_new col2_second
## 1 A01157cds_s_at;50682 <NA>
## 2 A03913cds_s_at;29366 <NA>
## 3 A04674cds_s_at;24860 100909612
## 4 A07543cds_s_at;24867 <NA>
filter removes the observations with col1 == '----'.
separate splits col2 into two columns, namely col2_first and col2_second
unite concatenates col1 and col2_first with ; as separator.
Trying to convert the following durations into seconds
x <- "1005d 16h 09m 57s"
x1 <- "16h 09m 57s"
x2 <- "06h 09m 57s"
x3 <- "09m 57s"
x4 <- "57s"
I've modified the answer from Jthorpe in this post Convert factor of format Hh Mm Ss to time duration.
days <- as.numeric(gsub('^*([0-9]+)d.*$','\\1',x3))
hours <- as.numeric(gsub('^.*([0-9][0-9])h.*$','\\1',x3))
minutes <- as.numeric(gsub('^.*([0-9][0-9])m.*$','\\1',x4))
seconds <- as.numeric(gsub('^.*([0-9][0-9])s.*$','\\1',x4))
duration_seconds <- seconds + 60*minutes + 60*60*hours + 24*60*60*days
However, this is only working with x, but not x1-x4. Now, I know I can probably use if logic to get around the issue, but is there a better way?
Thanks in advance.
We can change the space character (\\s+) with + using gsub, then we can replace 'd', 'h', 'm', 's' with gsubfn and loop through the output and evaluate the string.
library(gsubfn)
v2 <- gsubfn("[a-z]", list(d="*24*60*60", h = "*60*60", m = "*60",
s="*1"), gsub("\\s+", "+", v1))
unname(sapply(v2, function(x) eval(parse(text=x))))
#[1] 86890197 58197 22197 597 57
data
v1 <- c(x, x1, x2, x3, x4)
Use:
ifelse(is.na(your_exp),0)
So that whenever na is the output of your expression it becomes 0.
Eg:
days <- ifelse(is.na(as.numeric(gsub('^*([0-9]+)d.*$','\\1',x1))),0)
hours <- ifelse(is.na(as.numeric(gsub('^.*([0-9][0-9])h.*$','\\1',x1))),0)
minutes <- ifelse(is.na(as.numeric(gsub('^.*([0-9][0-9])m.*$','\\1',x1))),0)
seconds <- ifelse(is.na(as.numeric(gsub('^.*([0-9][0-9])s.*$','\\1',x1))),0)
Output:(after duration_seconds <- seconds + 60*minutes + 60*60*hours + 24*60*60*days)
> duration_seconds
[1] 58197
I have a file of baby names that I am reading in and then trying to get the last character in the baby name. For example, the file looks like..
Name Sex
Anna F
Michael M
David M
Sarah F
I read this in using
sourcenames = read.csv("babynames.txt", header=F, sep=",")
I ultimately want to end up with my result looking like..
Name Last Initial Sex
Michael l M
Sarah h F
I've managed to split the name into separate characters..
sourceout = strsplit(as.character(sourcenames$Name),'')
But now where I'm stuck is how to get the last letter, so in the case of Michael, how to get 'l'. I thought tail() might work but its returning the last few records, not the last character in each Name element.
Any help or advice is greatly appreciated.
Thanks :)
For your strsplit method to work, you can use tail with sapply
df$LastInit <- sapply(strsplit(as.character(df$Name), ""), tail, 1)
df
# Name Sex LastInit
# 1 Anna F a
# 2 Michael M l
# 3 David M d
# 4 Sarah F h
Alternatively, you can use substring
with(df, substring(Name, nchar(Name)))
# [1] "a" "l" "d" "h"
Try this function from stringi package:
require(stringi)
x <- c("Ala", "Sarah","Meg")
stri_sub(x, from = -1, to = -1)
This function extracts substrings between from and to index. If indexes are negative, then it counts characters from the end of a string. So if from=-1 and to=-1 it means that we want substring from last to last character :)
Why use stringi? Just look at this benchmarks :)
require(microbenchmark)
x <- sample(x,1000,T)
microbenchmark(stri_sub(x,-1), str_extract(x, "[a-z]{1}$"), gsub(".*(.)$", "\\1", x),
sapply(strsplit(as.character(x), ""), tail, 1), substring(x, nchar(x)))
Unit: microseconds
expr min lq median uq max neval
stri_sub(x, -1) 56.378 63.4295 80.6325 85.4170 139.158 100
str_extract(x, "[a-z]{1}$") 718.579 764.4660 821.6320 863.5485 1128.715 100
gsub(".*(.)$", "\\\\1", x) 478.676 493.4250 509.9275 533.8135 673.233 100
sapply(strsplit(as.character(x), ""), tail, 1) 12165.470 13188.6430 14215.1970 14771.4800 21723.832 100
substring(x, nchar(x)) 133.857 135.9355 141.2770 147.1830 283.153 100
Here is another option using data.table (for relatively clean syntax) and stringr (easier grammar).
library(data.table); library(stringr)
df = read.table(text="Name Sex
Anna F
Michael M
David M
Sarah F", header=T)
setDT(df) # convert to data.table
df[, "Last Initial" := str_extract(Name, "[a-z]{1}$") ][]
Name Sex Last Initial
1: Anna F a
2: Michael M l
3: David M d
4: Sarah F h
One liner:
x <- c("abc","123","Male")
regmatches(x,regexpr(".$", x))
## [1] "c" "3" "e"
You can do it with a Regular Expression and gsub:
sourcenames$last.letter = gsub(".*(.)$", "\\1", sourcenames$Name)
sourcenames
Name Sex last.letter
1 Anna F a
2 Michael M l
3 David M d
4 Sarah F h
you can try this one... str_sub() function in stringr package would help you.
library(dplyr)
library(stringr)
library(babynames)
babynames %>%
select(name,sex) %>%
mutate(last_letter = str_sub(name,-1,-1)) %>%
head()
dplyr approach:
sourcenames %>% rowwise() %>% mutate("Last Initial" = strsplit(as.character(Name),'') %>% unlist() %>% .[length(.)])
I have a CSV file like
Market,CampaignName,Identity
Wells Fargo,Gary IN MetroChicago IL Metro,56
EMC,Los Angeles CA MetroBoston MA Metro,78
Apple,Cupertino CA Metro,68
Desired Output to a CSV file with the first row as the headers
Market,City,State,Identity
Wells Fargo,Gary,IN,56
Wells Fargo,Chicago,IL,56
EMC,Los Angeles,CA,78
EMC,Boston,MA,78
Apple,Cupertino,CA,68
res <-
gsub('(.*) ([A-Z]{2})*Metro (.*) ([A-Z]{2}) .*','\\1,\\2:\\3,\\4',
xx$Market)
How to modify the above regular expressions to get the result in R?
New to R, any help is appreciated.
library(stringr)
xx.to.split <- with(xx, setNames(gsub("Metro", "", as.character(CampaignName)), Market))
do.call(rbind, str_match_all(xx.to.split, "(.+?) ([A-Z]{2}) ?"))[, -1]
Produces:
[,1] [,2]
Wells Fargo "Gary" "IN"
Wells Fargo "Chicago" "IL"
EMC "Los Angeles" "CA"
EMC "Boston" "MA"
Apple "Cupertino" "CA"
This should work even if you have different number of Compaign Names in each market. Unfortunately I think base options are annoying to implement because frustratingly there isn't a gregexec, although I'd be curious if someone comes up with something comparably compact in base.
Here is a solution using base R. Split the CampaignName column on the string Metro adding sequential numbers as names. stack turns it into a data frame with columns ind and values which we massage into DF1. Merge that with xx by the sequence numbers of DF1 and the row numbers of xx. Move Market to the front of DF2 and remove ind and CampaignName. Finally write it out.
xx <- read.csv("Campaign.csv", as.is = TRUE)
s <- strsplit(xx$CampaignName, " Metro")
names(s) <- seq_along(s)
ss <- stack(s)
DF1 <- with(ss, data.frame(ind,
City = sub(" ..$", "", values),
State = sub(".* ", "", values)))
DF2 <- merge(DF1, xx, by.x = "ind", by.y = 0)
DF <- DF2[ c("Market", setdiff(names(DF2), c("ind", "Market", "CampaignName"))) ]
write.csv(DF, file = "myfile.csv", row.names = FALSE, quote = FALSE)
REVISED to handle extra columns after poster modified the question to include such. Minor improvements.