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 = "[.,]").
Related
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$Entrez.Gene) # pastes the columns together and adds the ";"
dat = gsub("///","\t",dat) # replaces the "///" with a tab
Also, use cat() to view the tab as opposed to "\t". I got that from here: How to replace specific characters of a string with tab in R. This will output a list as opposed to a data.frame. You can convert back with data.frame(), but then you cannot use cat() to view.
We can use dplyr and tidyr here.
library(dplyr)
library(tidyr)
> df <- data.frame(
col1 = c('A01157cds_s_at', 'A03913cds_s_at', 'A04674cds_s_at', 'A07543cds_s_at', '----'),
col2 = c('50682', '29366', '24860 /// 100909612', '24867', '----'))
> df %>% filter(col1 != '----') %>%
separate(col2, c('col2_first', 'col2_second'), '///', remove = T) %>%
unite(col1_new, c(col1, col2_first), sep = ';', remove = T)
> df
## col1_new col2_second
## 1 A01157cds_s_at;50682 <NA>
## 2 A03913cds_s_at;29366 <NA>
## 3 A04674cds_s_at;24860 100909612
## 4 A07543cds_s_at;24867 <NA>
filter removes the observations with col1 == '----'.
separate splits col2 into two columns, namely col2_first and col2_second
unite concatenates col1 and col2_first with ; as separator.
I'm formating a data set so each entry has the adegenet format for codominant markers, such as:
Loci1
###/###
208/210
200/204
198/208
where the # represents any digit (the number is a allele size in basepairs). My data has some homozygous entries (all 3 digit integers with no separator) that have the the form of:
Loci1
###
208
198
I intend to paste the 3 digit string to itself with sep='/' to produce the first format. I've tried to use grep to subset these homozygous entries by finding all non ###/### and negating the match using the table matching such as:
a <- grep('\\b\\d{3}?[/]\\d{3}', score$Loci1, value =T ) # Subset all ###/###/
score[!(a %in% 1:nrow(score$Loci1)), ] # works but only on vectors...
After the subset I could paste. The problem arises when I apply this to a data frame. grep seems to treat the data frame as a list (which in part it is) and returns columns that have a match.
So in short how can I go from ### to ###/### in a data frame
self contained example of data:
score2 <- NULL
set.seed(9)
Loci1 <- NULL
Loci2 <- NULL
Loci3 <- NULL
for (i in 1:5) Loci1 <- append(Loci1, paste(sample(seq(from = 230, to=330, by=3), 2, replace = F), collapse = '/'))
for (i in 1:5) Loci2 <- append(Loci2, paste(sample(seq(from = 230, to=330, by=3), 2, replace = F), collapse = '/'))
for (i in 1:5) Loci3 <- append(Loci3, paste(sample(seq(from = 230, to=330, by=3), 2, replace = F), collapse = '/'))
score2 <- data.frame(Loci1, Loci2, Loci3, stringsAsFactors = F)
score2[2,3] <- strsplit(score2[2,3], split = '/')[1]
score2[5,2] <- strsplit(score2[3,3], split = '/')[1]
score2[1,1] <- strsplit(score2[1,1], split = '/')[1]
score2[c(1, 4),c(2,3)] <- NA
score2
You could just replace the 3 digit items with the separator and a copy:
sub("^(...)$", "\\1/\\1", Loci1)
Use lapply with an anonymized function:
data.frame( lapply(score2, function(x) sub("^(...)$", "\\1/\\1", x) ) )
Loci1 Loci2 Loci3
1 251/251 <NA> <NA>
2 251/329 320/257 260/260
3 275/242 278/329 281/320
4 269/266 <NA> <NA>
5 296/326 281/281 326/314
(Not sure what the "paste-part" was supposed to refer to, but I think this was the intent of your question)
If the numeric values could have a varying number of digits then use a pattern argument like "^([0-9]{1,9})$"
An option using grep/paste,
m1 <- as.matrix(score2)
indx <- grep('^...$', m1)
m1[indx] <- paste(m1[indx], m1[indx], sep="/")
as.data.frame(m1)
# Loci1 Loci2 Loci3
#1 251/251 <NA> <NA>
#2 251/329 320/257 260/260
#3 275/242 278/329 281/320
#4 269/266 <NA> <NA>
#5 296/326 281/281 326/314
Or without converting to matrix, this can be done using lapply
score2[] <- lapply(score2, function(x) ifelse(grepl('^...$', x),
paste(x, x, sep="/"),x))
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(" ..$", "", values),
State = sub(".* ", "", values)))
DF2 <- merge(DF1, xx, by.x = "ind", by.y = 0)
DF <- DF2[ c("Market", setdiff(names(DF2), c("ind", "Market", "CampaignName"))) ]
write.csv(DF, file = "myfile.csv", row.names = FALSE, quote = FALSE)
REVISED to handle extra columns after poster modified the question to include such. Minor improvements.
I have a dataframe with 2 columns GL and GLDESC and want to add a 3rd column called KIND based on some data that is inside of column GLDESC.
The dataframe is as follows:
GL GLDESC
1 515100 Payroll-Indir Salary Labor
2 515900 Payroll-Indir Compensated Absences
3 532300 Bulk Gas
4 539991 Area Charge In
5 551000 Repairs & Maint-Spare Parts
6 551100 Supplies-Operating
7 551300 Consumables
For each row of the data table:
If GLDESC contains the word Payroll anywhere in the string then I want KIND to be Payroll
If GLDESC contains the word Gas anywhere in the string then I want KIND to be Materials
In all other cases I want KIND to be Other
I looked for similar examples on stackoverflow but could not find any, also looked in R for dummies on switch, grep, apply and regular expressions to try and match only part of the GLDESC column and then fill the KIND column with the kind of account but was unable to make it work.
Since you have only two conditions, you can use a nested ifelse:
#random data; it wasn't easy to copy-paste yours
DF <- data.frame(GL = sample(10), GLDESC = paste(sample(letters, 10),
c("gas", "payroll12", "GaSer", "asdf", "qweaa", "PayROll-12",
"asdfg", "GAS--2", "fghfgh", "qweee"), sample(letters, 10), sep = " "))
DF$KIND <- ifelse(grepl("gas", DF$GLDESC, ignore.case = T), "Materials",
ifelse(grepl("payroll", DF$GLDESC, ignore.case = T), "Payroll", "Other"))
DF
# GL GLDESC KIND
#1 8 e gas l Materials
#2 1 c payroll12 y Payroll
#3 10 m GaSer v Materials
#4 6 t asdf n Other
#5 2 w qweaa t Other
#6 4 r PayROll-12 q Payroll
#7 9 n asdfg a Other
#8 5 d GAS--2 w Materials
#9 7 s fghfgh e Other
#10 3 g qweee k Other
EDIT 10/3/2016 (..after receiving more attention than expected)
A possible solution to deal with more patterns could be to iterate over all patterns and, whenever there is match, progressively reduce the amount of comparisons:
ff = function(x, patterns, replacements = patterns, fill = NA, ...)
{
stopifnot(length(patterns) == length(replacements))
ans = rep_len(as.character(fill), length(x))
empty = seq_along(x)
for(i in seq_along(patterns)) {
greps = grepl(patterns[[i]], x[empty], ...)
ans[empty[greps]] = replacements[[i]]
empty = empty[!greps]
}
return(ans)
}
ff(DF$GLDESC, c("gas", "payroll"), c("Materials", "Payroll"), "Other", ignore.case = TRUE)
# [1] "Materials" "Payroll" "Materials" "Other" "Other" "Payroll" "Other" "Materials" "Other" "Other"
ff(c("pat1a pat2", "pat1a pat1b", "pat3", "pat4"),
c("pat1a|pat1b", "pat2", "pat3"),
c("1", "2", "3"), fill = "empty")
#[1] "1" "1" "3" "empty"
ff(c("pat1a pat2", "pat1a pat1b", "pat3", "pat4"),
c("pat2", "pat1a|pat1b", "pat3"),
c("2", "1", "3"), fill = "empty")
#[1] "2" "1" "3" "empty"
I personally like matching by index. You can loop grep over your new labels, in order to get the indices of your partial matches, then use this with a lookup table to simply reassign the values.
If you wanna create new labels, use a named vector.
DF <- data.frame(GL = sample(10), GLDESC = paste(sample(letters, 10),
c(
"gas", "payroll12", "GaSer", "asdf", "qweaa", "PayROll-12",
"asdfg", "GAS--2", "fghfgh", "qweee"
), sample(letters, 10),
sep = " "
))
lu <- stack(sapply(c(Material = "gas", Payroll = "payroll"), grep, x = DF$GLDESC, ignore.case = TRUE))
DF$KIND <- DF$GLDESC
DF$KIND[lu$values] <- as.character(lu$ind)
DF$KIND[-lu$values] <- "Other"
DF
#> GL GLDESC KIND
#> 1 6 x gas f Material
#> 2 3 t payroll12 q Payroll
#> 3 5 a GaSer h Material
#> 4 4 s asdf x Other
#> 5 1 m qweaa y Other
#> 6 10 y PayROll-12 r Payroll
#> 7 7 g asdfg a Other
#> 8 2 k GAS--2 i Material
#> 9 9 e fghfgh j Other
#> 10 8 l qweee p Other
Created on 2021-11-13 by the reprex package (v2.0.1)
I have a large text file with a variable number of fields in each row. The first entry in each row corresponds to a biological pathway, and each subsequent entry corresponds to a gene in that pathway. The first few lines might look like this
path1 gene1 gene2
path2 gene3 gene4 gene5 gene6
path3 gene7 gene8 gene9
I need to read this file into R as a list, with each element being a character vector, and the name of each element in the list being the first element on the line, for example:
> pathways <- list(
+ path1=c("gene1","gene2"),
+ path2=c("gene3","gene4","gene5","gene6"),
+ path3=c("gene7","gene8","gene9")
+ )
>
> str(pathways)
List of 3
$ path1: chr [1:2] "gene1" "gene2"
$ path2: chr [1:4] "gene3" "gene4" "gene5" "gene6"
$ path3: chr [1:3] "gene7" "gene8" "gene9"
>
> str(pathways$path1)
chr [1:2] "gene1" "gene2"
>
> print(pathways)
$path1
[1] "gene1" "gene2"
$path2
[1] "gene3" "gene4" "gene5" "gene6"
$path3
[1] "gene7" "gene8" "gene9"
...but I need to do this automatically for thousands of lines. I saw a similar question posted here previously, but I couldn't figure out how to do this from that thread.
Thanks in advance.
Here's one way to do it:
# Read in the data
x <- scan("data.txt", what="", sep="\n")
# Separate elements by one or more whitepace
y <- strsplit(x, "[[:space:]]+")
# Extract the first vector element and set it as the list element name
names(y) <- sapply(y, `[[`, 1)
#names(y) <- sapply(y, function(x) x[[1]]) # same as above
# Remove the first vector element from each list element
y <- lapply(y, `[`, -1)
#y <- lapply(y, function(x) x[-1]) # same as above
One solution is to read the data in via read.table(), but use the fill = TRUE argument to pad the rows with fewer "entries", convert the resulting data frame to a list and then clean up the "empty" elements.
First, read your snippet of data in:
con <- textConnection("path1 gene1 gene2
path2 gene3 gene4 gene5 gene6
path3 gene7 gene8 gene9
")
dat <- read.table(con, fill = TRUE, stringsAsFactors = FALSE)
close(con)
Next we drop the first column, first saving it for the names of the list later
nams <- dat[, 1]
dat <- dat[, -1]
Convert the data frame to a list. Here I just split the data frame on the indices 1,2,...,n where n is the number of rows:
ldat <- split(dat, seq_len(nrow(dat)))
Clean up the empty cells:
ldat <- lapply(ldat, function(x) x[x != ""])
Finally, apply the names
names(ldat) <- nams
Giving:
> ldat
$path1
[1] "gene1" "gene2"
$path2
[1] "gene3" "gene4" "gene5" "gene6"
$path3
[1] "gene7" "gene8" "gene9"
A quick solution based on the linked page...
inlist <- strsplit(readLines("file.txt"), "[[:space:]]+")
pathways <- lapply(inlist, tail, n = -1)
names(pathways) <- lapply(inlist, head, n = 1)
One more solution:
sl <- c("path1 gene1 gene2", "path2 gene1 gene2 gene3") # created by readLines
f <- function(l, s) {
v <- strsplit(s, " ")[[1]]
l[[v[1]]] <- v[2:length(v)]
return(l)
}
res <- Reduce(f, sl, list())