This question already has answers here:
Split data frame string column into multiple columns
(16 answers)
Closed 6 years ago.
i have a dataframe with 2 columns id, cat_list
id cat_list
1 A
2 A|B
3 E|F|G
4 I
5 P|R|T|Z
i want to achieve the below using R code.
id cat_list1 cat_list2 cat_list3 cat_list4
1 A
2 A B
3 E F G
4 I
5 P R T Z
tidyr::separate is handy:
library(tidyr)
df %>% separate(cat_list, into = paste0('cat_list', 1:4), fill = 'right')
## id cat_list1 cat_list2 cat_list3 cat_list4
## 1 1 A <NA> <NA> <NA>
## 2 2 A B <NA> <NA>
## 3 3 E F G <NA>
## 4 4 I <NA> <NA> <NA>
## 5 5 P R T Z
We can use cSplit. Here, we don't need to worry to about the number of splits as it will automatically detect it.
library(splitstackshape)
cSplit(df1, "cat_list", "|")
# id cat_list_1 cat_list_2 cat_list_3 cat_list_4
#1: 1 A NA NA NA
#2: 2 A B NA NA
#3: 3 E F G NA
#4: 4 I NA NA NA
#5: 5 P R T Z
NOTE: It may be better to fill with NA rather than ''.
Related
This question already has answers here:
MATCH function in r [duplicate]
(1 answer)
New column in dataframe based on match between two columns [duplicate]
(1 answer)
Closed 4 years ago.
What's an elegant way (without additional packages) to "expand" a given data.frame according to one of its columns?
Given:
df <- data.frame(values = 1:5, strings = c("e", "g", "h", "b", "c"))
more.strings <- letters[c(3, 5, 7, 1, 4, 8, 6)]
Desired outcome: A data.frame containing:
5 c
1 e
2 g
NA a
NA d
3 h
NA f
So those values of df$strings appearing in more.strings should be used to fill the new data.frame (otherwise NA).
you can do a join:
In base R you could do:
merge(df, more.strings, by.y="y",by.x="strings", all.y=TRUE)
strings values
1 c 5
2 e 1
3 g 2
4 h 3
5 a NA
6 d NA
7 f NA
or even as given by #thelatemailin the comment section below:
merge(df, list(strings=more.strings),by="strings", all.y=TRUE)
Using library:
library(tidyverse)
right_join(df,data.frame(strings=more.strings),by="strings")
values strings
1 5 c
2 1 e
3 2 g
4 NA a
5 NA d
6 3 h
7 NA f
We can do this without using any library i.e. using only base R
data.frame(value = with(df, match(more.strings, strings)),
strings = more.strings)
# value strings
#1 5 c
#2 1 e
#3 2 g
#4 NA a
#5 NA d
#6 3 h
#7 NA f
Or we can use complete
library(tidyverse)
complete(df, strings = more.strings) %>%
arrange(match(strings, more.strings)) %>%
select(names(df))
# A tibble: 7 x 2
# values strings
# <int> <chr>
#1 5 c
#2 1 e
#3 2 g
#4 NA a
#5 NA d
#6 3 h
#7 NA f
I am working in R with a dataset that is created from mongodb with the use of mongolite.
I am getting a list that looks like so:
_id A B A B A B NA NA
1 a 1 b 2 e 5 NA NA
2 k 4 l 3 c 3 d 4
I would like to merge the datasetto look like this:
_id A B
1 a 1
2 k 4
1 b 2
2 l 3
1 e 5
2 c 3
1 NA NA
2 d 4
The NAs in the last columns are there because the columns are named from the first entry and if a later entry has more columns than that they don't get names assigned to them, (if I get help for this as well it would be awesome but it's not the reason I am here).
Also the number of columns might differ for different subsets of the dataset.
I have tried melt() but since it is a list and not a dataframe it doesn't work as expected, I have tried stack() but it dodn't work because the columns have the same name and some of them don't even have a name.
I know this is a very weird situation and appreciate any help.
Thank you.
using library(magrittr)
data:
df <- fread("
_id A B A B A B NA NA
1 a 1 b 2 e 5 NA NA
2 k 4 l 3 c 3 d 4 ",header=T)
setDF(df)
Code:
df2 <- df[,-1]
odds<- df2 %>% ncol %>% {(1:.)%%2} %>% as.logical
even<- df2 %>% ncol %>% {!(1:.)%%2}
cbind(df[,1,drop=F],
A=unlist(df2[,odds]),
B=unlist(df2[,even]),
row.names=NULL)
result:
# _id A B
# 1 1 a 1
# 2 2 k 4
# 3 1 b 2
# 4 2 l 3
# 5 1 e 5
# 6 2 c 3
# 7 1 <NA> NA
# 8 2 d 4
We can use data.table. Assuming A and B are always following each other. I created an example with 2 sets of NA's in the header. With grep we can find the ones fread has named V8 etc. Using R's recycling of vectors, you can rename multiple headers in one go. If in your case these are named differently change the pattern in the grep command. Then we melt the data in via melt
library(data.table)
df <- fread("
_id A B A B A B NA NA NA NA
1 a 1 b 2 e 5 NA NA NA NA
2 k 4 l 3 c 3 d 4 e 5",
header = TRUE)
df
_id A B A B A B A B A B
1: 1 a 1 b 2 e 5 <NA> NA <NA> NA
2: 2 k 4 l 3 c 3 d 4 e 5
# assuming A B are always following each other. Can be done in 1 statement.
cols <- names(df)
cols[grep(pattern = "^V", x = cols)] <- c("A", "B")
names(df) <- cols
# melt data (if df is a data.frame replace df with setDT(df)
df_melted <- melt(df, id.vars = 1,
measure.vars = patterns(c('A', 'B')),
value.name=c('A', 'B'))
df_melted
_id variable A B
1: 1 1 a 1
2: 2 1 k 4
3: 1 2 b 2
4: 2 2 l 3
5: 1 3 e 5
6: 2 3 c 3
7: 1 4 <NA> NA
8: 2 4 d 4
9: 1 5 <NA> NA
10: 2 5 e 5
Thank you for your help, they were great inspirations.
Even though #Andre Elrico gave a solution that worked in the reproducible example better #phiver gave a solution that worked better on my overall problem.
By using both those I came up with the following.
library(data.table)
#The data were in a list of lists called list for this example
temp <- as.data.table(matrix(t(sapply(list, '[', seq(max(sapply(list, lenth))))),
nrow = m))
# m here is the number of lists in list
cols <- names(temp)
cols[grep(pattern = "^V", x = cols)] <- c("B", "A")
#They need to be the opposite way because the first column is going to be substituted with id, and this way they fall on the correct column after that
cols[1] <- "id"
names(temp) <- cols
l <- melt.data.table(temp, id.vars = 1,
measure.vars = patterns(c("A", "B")),
value.name = c("A", "B"))
That way I can use this also if I have more than 2 columns that I need to manipulate like that.
I am relatively new to R, so bear with me. I have a list of data frames that I need to combine into one data frame. so:
dfList <- list(
df1 = data.frame(x=letters[1:2],y=1:2),
df2 = data.frame(x=letters[3:4],z=3:4)
)
comes out as:
$df1
x y
1 a 1
2 b 2
$df2
x z
1 c 3
2 d 4
and I want them to combine common columns and add anything not already there. the result would be:
final result
x y z
1 a 1
2 b 2
3 c 3
4 d 4
Is this even possible?
Yep, it's pretty easy, actually:
library(dplyr)
df_merged <- bind_rows(dfList)
df_merged
x y z
1 a 1 NA
2 b 2 NA
3 c NA 3
4 d NA 4
And if you don't want NA in the empty cells, you can replace them like this:
df_merged[is.na(df_merged)] <- 0 # or whatever you want to replace NA with
Just using do.call with rbind.fill
do.call(rbind.fill,dfList)
x y z
1 a 1 NA
2 b 2 NA
3 c NA 3
4 d NA 4
You could do that with base function merge():
merge(dfList$df1, dfList$df2, by = "x", all = TRUE)
# x y z
# 1 a 1 NA
# 2 b 2 NA
# 3 c NA 3
# 4 d NA 4
Or with dplyr package with function full_join:
dplyr::full_join(dfList$df1, dfList$df2, by = "x")
# x y z
# 1 a 1 NA
# 2 b 2 NA
# 3 c NA 3
# 4 d NA 4
They both join everything that is in both data.frames.
Hope that works for you.
I have a data.frame with a single column, a vector of strings.
These strings have duplicate values.
I want to find the character strings that have duplicates in this vector and write their index of position in a new column.
So for example consider I have:
DT<- data.frame(string=A,B,C,D,E,F,A,C,F,Z,A)
I want to get:
string match2 match2 match3 matchx....
A 1 7 11
B 2 NA NA
C 3 8 NA
D 4 NA NA
E 5 NA NA
F 6 9 NA
A 1 7 11
C 3 8 NA
F 6 9 NA
Z 10 NA NA
A 1 7 11
The string is ways longer than in this example and I do not know the amount of maximum columns I need.
What will be the most effective way to do this?
I know that there is the duplicate function but I am not exactly sure how to combine it to the result I want to get here.
Many thanks!
Here's one way of doing this. I'm sure a data.table one liner follows.
DT<- data.frame(string=c("A","B","C","D","E","F","A","C","F","Z","A"))
# find matches
rbf <- sapply(DT$string, FUN = function(x, DT) which(DT %in% x), DT = DT$string)
# fill in NAs to have a pretty matrix
out <- sapply(rbf, FUN = function(x, mx) c(x, rep(NA, length.out = mx - length(x))), max(sapply(rbf, length)))
# bind it to the original data
cbind(DT, t(out))
string 1 2 3
1 A 1 7 11
2 B 2 NA NA
3 C 3 8 NA
4 D 4 NA NA
5 E 5 NA NA
6 F 6 9 NA
7 A 1 7 11
8 C 3 8 NA
9 F 6 9 NA
10 Z 10 NA NA
11 A 1 7 11
Here is one option with data.table. After grouping by 'string', get the sequence (seq_len(.N)) and row index (.I), then dcast to 'wide' format and join with the original dataset on the 'string'
library(data.table)
dcast(setDT(DT)[, .(seq_len(.N),.I), string],string ~ paste0("match", V1))[DT, on = "string"]
# string match1 match2 match3
# 1: A 1 7 11
# 2: B 2 NA NA
# 3: C 3 8 NA
# 4: D 4 NA NA
# 5: E 5 NA NA
# 6: F 6 9 NA
# 7: A 1 7 11
# 8: C 3 8 NA
# 9: F 6 9 NA
#10: Z 10 NA NA
#11: A 1 7 11
Or another option would be to split the sequence of rows with 'string', pad the list elements with NA for length that are less, and merge with the original dataset (using base R methods)
lst <- split(seq_len(nrow(DT)), DT$string)
merge(DT, do.call(rbind, lapply(lst, `length<-`, max(lengths(lst)))),
by.x = "string", by.y = "row.names")
data
DT<- data.frame(string=c("A","B","C","D","E","F","A","C",
"F","Z","A"), stringsAsFactors=FALSE)
And here's one that uses tidyverse tools ( not quite a one-liner ;) ):
library( tidyverse )
DT %>% group_by( string ) %>%
do( idx = which(DT$string == unique(.$string)) ) %>%
ungroup %>% unnest %>% group_by( string ) %>%
mutate( m = stringr::str_c( "match", 1:n() ) ) %>%
spread( m, idx )
I'm using 'fill = T' on a file that has single letters separated by commas:
Pred
1 T,T
2 NA
3 D
4 NA
5 NA
6 T
7 P,B
8 NA
9 NA
using the command:
sift <- read.table("/home/pred.txt", header=F, fill=TRUE, sep=',', stringsAsFactors=F)
Which I was hoping the sift will turn out as:
V1 V2
1 T T
2 <NA>
3 D
4 <NA>
5 <NA>
6 T
7 P B
8 <NA>
9 <NA>
However, it comes out like:
V1
1 T
2 <NA>
3 D
4 <NA>
5 <NA>
6 T
7 P
8 <NA>
9 <NA>
This code works when there are multiple sampleIDs (separated by a comma) in each row - but not for single letters. Does 'fill' work for single letters? Stupid question, I know.
So here is a workaround:
url <- "https://dl.dropboxusercontent.com/s/bjb241s16t63ev8/pred.txt?dl=1&token_hash=AAEBzfCGgoeHgNTvhMSVoZK6qRGrdwwuDZB3h8lWTZNtkA"
df.1 <- read.table(url,header=F,sep=",",fill=T,stringsAsFactors=F)
dim(df.1)
# [1] 149792 1 <-- 149,792 rows and ** 1 ** column
df.2 <- read.table(url,header=F,sep=",",fill=T,stringsAsFactors=F,
col.names=c("V1","V2"))
dim(df.2)
# [1] 149633 2 <-- 149,633 rows and ** 2 ** columns
head(df.2[which(nchar(df.2$V2)>0),])
# V1 V2
# 1000 T T
# 2419 T T
# 3507 T T
# 3766 T D
# 4308 T D
# 4545 T D
read.table(...) creates a data frame with number of columns determined by the first 5 rows. Since the first 5 rows in your file have only 1 column, that's what you get. Evidently, by specifying sep="," you force read.table(...) to add the "extra" data as extra rows.
The workaround explicitly sets the number of columns by specifying column names, which could be anything, as long as length(col.names) = 2.