This question already has an answer here:
Merging rows with the same ID variable [duplicate]
(1 answer)
Closed 7 years ago.
I have the following dataframe:
st <- data.frame(
se = rep(1:2, 5),
X = rnorm(10, 0, 1),
Y = rnorm(10, 0, 2))
st$xy <- paste(st$X,",",st$Y)
st <- st[c("se","xy")]
but I want it to be the following:
1 2 3 4 5
-1.53697673029089 , 2.10652020463275 -1.02183940974772 , 0.623009466458354 1.33614674072657 , 1.5694345481646 0.270466789820086 , -0.75670874554064 -0.280167896821629 , -1.33313822867893
0.26012874418111 , 2.87972571647846 -1.32317949800031 , -2.92675188421021 0.584199000313255 , 0.565499464846637 -0.555881716346136 , -1.14460518414649 -1.0871665543915 , -3.18687136890236
I mean when the value of se is the same, make a column bind.
Do you have any ideas how to accomplish this?
I had no luck with spread(tidyr), and I guess it's something which involves sapply, cbind and a if statement. Because the real data involves more than 35.000 rows.
It seems as though your eventual goal is to have a data file which has roughly 35000 columns. Are you sure about that? That doesn't sound very tidy.
To do what you want, you are going to need to have a row identifier. In the below, I've called it caseid, and then removed it once it was no longer required. I then transpose the result to get what you asked for.
library(tidyr)
library(dplyr)
st <- data.frame(
se = rep(1:2, 5),
X = rnorm(10, 0, 1),
Y = rnorm(10, 0, 2))
st$xy <- paste(st$X,",",st$Y)
st <- st[c("se","xy")]
st$caseid = rep(1:(nrow(st)/2), each = 2) # temporary
df = spread(st, se, xy) %>%select(-caseid) %>%t()
print(df)
If we need to split the 'xy' column elements into individual units, cSplit from splitstackshape can be used. Then rbind the alternating rows of 'st1' after unlisting`.
library(splitstackshape)
st1 <- cSplit(st, 'xy', ', ', 'wide')
rbind(unlist(st1[c(TRUE,FALSE)][,-1, with=FALSE]),
unlist(st1[c(FALSE, TRUE)][,-1, with=FALSE]))
If we don't need to split the 'xy' column into individual elements, we can use dcast from data.table. It should be fast enough. Convert the 'data.frame' to 'data.table' (setDT(st), create a sequence column ('N') by 'se', and then dcast from 'long' to 'wide'.
library(data.table)
dcast(setDT(st)[, N:= 1:.N, se], se~N, value.var= 'xy')
Related
I would like to create an empty data.table in R with colum names from another existing data.table.
Somehow I could not find a solution for that.
I would like to do something like that:
require(data.table)
dt1 <- data.table(fn = c("A","B","C"), x = c(1,2,3), y = c(2,3,4), a = 1, b = 2, c = 3)
dt2 <- data.table(names=colnames(dt1)) # Gives 6 rows instead of 6 cols
How can this be achieved?
Thanks!
You can also take your old dt1, clear it and keep as dt2
dt2 <- dt1[0,]
dt2
Empty data.table (0 rows and 6 cols): fn,x,y,a,b,c
It isn't precisely what did you want, but it always some solution.
One option could be:
dt2 <- setnames(data.table(matrix(nrow = 0, ncol = length(dt1))), names(dt1))
Empty data.table (0 rows and 6 cols): fn,x,y,a,b,c
This question already has answers here:
How to sum a variable by group
(18 answers)
Closed 3 years ago.
I have a data frame similar to this. I want to sum up values for rows if the text in column "Name" is the same before the - sign.
Remove everything after "-" using sub and then use How to sum a variable by group
df$Name <- sub("-.*", "",df$Name)
aggregate(cbind(val1, val2)~Name, df, sum)
Below is a data.table solution.
Data:
df = data.table(
Name = c('IRON - A', 'IRON - B', 'SABBATH - A', 'SABBATH - B'),
val1 = c(1,2,3,4),
val2 = c(5,6,7,8)
)
Code:
df[, Name := sub("-.*", "", Name)]
mat = df[, .(sum(val1), sum(val2)), by = Name]
> mat
Name V1 V2
1: IRON 3 11
2: SABBATH 7 15
You can rbind your 2 tables (top and bottom) into one data frame and then use dplyr or data.table. The data.table would be much faster for large tables.
data_framme$Name <- sub("-.*", "", data_frame$Name)
library(dplyr)
data_frame %>%
group_by(Name) %>%
summarise_all(sum)
library(data.table)
data.frame <- data.table(data.frame)
data.frame[, lapply(.SD, sum, na.rm=TRUE), by=Name ]
This question already has answers here:
How do I replace NA values with zeros in an R dataframe?
(29 answers)
Fastest way to replace NAs in a large data.table
(10 answers)
Closed 6 years ago.
Quite new to R, I am trying to subselect certain columns in order to set their NA's to 0.
so far I have:
col_names1 <- c('a','b','c')
col_names2 <- c('e','f','g')
col_names <- c(col_names1, col_names2)
data = fread('data.tsv', sep="\t", header= FALSE,na.strings="NA",
stringsAsFactors=TRUE,
colClasses=my_col_Classes
)
setnames(data, col_names)
data[col_names2][is.na(data[col_names2])] <- 0
But I keep getting the error
Error in `[.data.table`(`*tmp*`, column_names2): When i is a data.table (or character vector), x must be keyed (i.e. sorted, and, marked as sorted) so data.table knows which columns to join to and take advantage of x being sorted. Call setkey(x,...) first, see ?setkey.
I believer this error is saying I have the wrong order but I am not sure how I do?
You can do it with data.table assign :=
data <- data.table(a = c(2, NA, 3, 5), b = c(NA,2,3,4), c = c(2,5,NA, 6))
fix_columns <- c('a','b')
fix_fun <- function(x) ifelse(is.na(x), 0 , x)
data[,(fix_columns):=lapply(.SD, fix_fun), .SDcols=fix_columns]
P.S. You cant select columns from data.table like data[col_names2]. If you want select them by character vector, one approach is : data[, col_names2, with = F]
I have a large (6 million row) table of values that I believe needs to be reformatted before it can be used for comparison to my data set. The table has 3 columns that I care about.
The first column contains nucleotide base changes, in the form of C>G, A>C, A>G, etc. I'd like to split these into two separate columns.
The second column has the chromosome and base position, formatted as 10:130448, 2:40483, 5:30821291, etc. I would also like to split this into two columns.
The third column has the allelic fraction in a number of sample populations, formatted like .02/.03/.20. I'd like to extract the third fraction into a new column.
The problem is that the code I have written is currently extremely slow. It looks like it will take about a day and a half just to run. Is there something I'm missing here? Any suggestions would be appreciated.
My current code does the following: pos, change, and fraction each receive a vector of the above values split use strsplit. I then loop through the entire database, getting the ith value from those three vectors, and creating new columns with the values I want.
Once the database has been formatted, I should be able to easily check a large number of samples by chromosome number, base, reference allele, alternate allele, etc.
pos <- strsplit(total.esp$NCBI.Base, ":")
change <- strsplit(total.esp$Alleles, ">")
fraction <- strsplit(total.esp$'MAFinPercent(EA/AA/All)', "/")
for (i in 1:length(pos)){
current <- pos[[i]]
mutation <- change[[i]]
af <- fraction[[i]]
total.esp$chrom[i] <- current[1]
total.esp$base[i] <- current [2]
total.esp$ref[i] <- mutation[1]
total.esp$alt[i] <- mutation[2]
total.esp$af[i] <- af[3]
}
Thanks!
Here is a data.table solution. We convert the 'data.frame' to 'data.table' (setDT(df1)), loop over the Subset of Data.table (.SD) with lapply, use tstrsplit and split the columns by specifying the split character, unlist the output with recursive=FALSE.
library(data.table)#v1.9.6+
setDT(df1)[, unlist(lapply(.SD, tstrsplit,
split='[>:/]', type.convert=TRUE), recursive=FALSE)]
# Alleles1 Alleles2 NCBI.Base1 NCBI.Base2 MAFinPercent1 MAFinPercent2
#1: C G 10 130448 0.02 0.03
#2: A C 2 40483 0.05 0.03
#3: A G 5 30821291 0.02 0.04
# MAFinPercent3
#1: 0.20
#2: 0.04
#3: 0.03
NOTE: I assumed that there are only 3 columns in the dataset. If there are more columns, and want to do the split only for the 3 columns, we can specify the .SDcols= 1:3 i.e. column index or the actual column names, assign (:=) the output to new columns and subset the columns that are only needed in the output.
data
df1 <- data.frame(Alleles =c('C>G', 'A>C', 'A>G'),
NCBI.Base=c('10:130448', '2:40483', '5:30821291'),
MAFinPercent= c('.02/.03/.20', '.05/.03/.04', '.02/.04/.03'),
stringsAsFactors=FALSE)
You can use tidyr, dplyr and separate:
library(tidyr)
library(dplyr)
total.esp %>% separate(Alleles, c("ref", "alt"), sep=">") %>%
separate(NCBI.Base, c("chrom", "base"), sep=":") %>%
separate(MAFinPercent.EA.AA.All., c("af1", "af2", "af3"), sep="/") %>%
select(-af1, -af2, af = af3)
You'll need to be careful about that last MAFinPercent.EA.AA.All. - you have a horrible column name so may have to rename it/quote it depending on how exactly r has it (this is also a good reason to include at least some data in your question, such as the output of dput(head(total.esp))).
data used to check:
total.esp <- data.frame(Alleles= rep("C>G", 50), NCBI.Base = rep("10:130448", 50), 'MAFinPercent(EA/AA/All)'= rep(".02/.03/.20", 50))
Because we now have a tidyr/dplyr solution, a data.table solution and a base solution, let's benchmark them. First, data from #akrun, 300,000 rows in total:
df1 <- data.frame(Alleles =rep(c('C>G', 'A>C', 'A>G'), 100000),
NCBI.Base=rep(c('10:130448', '2:40483', '5:30821291'), 100000),
MAFinPercent= rep(c('.02/.03/.20', '.05/.03/.04', '.02/.04/.03'), 100000),
stringsAsFactors=FALSE)
Now, the benchmark:
microbenchmark::microbenchmark(
tidyr = {df1 %>% separate(Alleles, c("ref", "alt"), sep=">") %>%
separate(NCBI.Base, c("chrom", "base"), sep=":") %>%
separate(MAFinPercent, c("af1", "af2", "af3"), sep="/") %>%
select(-af1, -af2, af = af3)},
data.table = {setDT(df1)[, unlist(lapply(.SD, tstrsplit,
split='[>:/]', type.convert=TRUE), recursive=FALSE)]},
base = {pos <- strsplit(df1$NCBI.Base, ":");
change <- strsplit(df1$Alleles, ">");
fraction <- strsplit(df1$MAFinPercent, "/");
data.frame( chrom =sapply( pos, "[", 1),
base = sapply( pos, "[", 2),
ref = sapply( change, "[", 1),
alt = sapply(change, "[", 2),
af = sapply( fraction, "[", 3)
)}
)
Unit: seconds
expr min lq mean median uq max neval
tidyr 1.295970 1.398792 1.514862 1.470185 1.629978 1.889703 100
data.table 2.140007 2.209656 2.315608 2.249883 2.481336 2.666345 100
base 2.718375 3.079861 3.183766 3.154202 3.221133 3.791544 100
tidyr is the winner
Try this (after retaining your first three lines of code):
total.esp <- data.frame( chrom =sapply( pos, "[", 1),
base = sapply( pos, "[", 2),
ref = sapply( change, "[", 1),
alt = sapply(change, "[", 2),
af = sapply( af, "[", 3)
)
I cannot imagine this taking more than a couple of minutes. (I do work with R objects of similar size.)
I have the following dataframe:
df = data.frame(id=c("A","A","A","A","B","B","B","B","C","C","C","C","D","D","D","D"),
sub=rep(c(1:4),4),
acc1=runif(16,0,3),
acc2=runif(16,0,3),
acc3=runif(16,0,3),
acc4=runif(16,0,3))
What I want is to obtain the mean rows for each ID, which is to say I want to obtain the mean acc1, acc2, acc3 and acc4 for each level A, B, C and D by averaging the values for each sub (4 levels for each id), which would give something like this in the end (with the NAs replaced by the means I want of course):
dfavg = data.frame(id=c("A","B","C","D"),meanacc1=NA,meanacc2=NA,meanacc3=NA,meanacc4=NA)
Thanks in advance!
Try:
You can use any of the specialized packages dplyr or data.table or using base R. Because you have a lot of columns that starts with acc to get the mean of, I choose dplyr. Here, the idea is to first group the variable by id and then use summarise_each to get the mean of each column by id that starts_with acc
library(dplyr)
df1 <- df %>%
group_by(id) %>%
summarise_each(funs(mean=mean(., na.rm=TRUE)), starts_with("acc")) %>%
rename(meanacc1=acc1, meanacc2=acc2, meanacc3=acc3, meanacc4=acc4) #this works but it requires more typing.
I would rename using paste
# colnames(df1)[-1] <- paste0("mean", colnames(df1)[-1])
gives the result
# id meanacc1 meanacc2 meanacc3 meanacc4
#1 A 1.7061929 2.401601 2.057538 1.643627
#2 B 1.7172095 1.405389 2.132378 1.769410
#3 C 1.4424233 1.737187 1.998414 1.137112
#4 D 0.5468509 1.281781 1.790294 1.429353
Or using data.table
library(data.table)
nm1 <- paste0("acc", 1:4) #names of columns to do the `means`
dt1 <- setDT(df)[, lapply(.SD, mean, na.rm=TRUE), by=id, .SDcols=nm1]
Here.SD implies Subset of Data.table, .SDcols are the columns to which we apply the mean operation.
setnames(dt1, 2:5, paste0("mean", nm1)) #change the names of the concerned columns in the result
dt1
(This must have been asked at least 20 times.) The `aggregate function applies the same function (given as the third argument) to all the columns of its first argument within groups defined by its second argument:
aggregate(df[-(1:2)], df[1],mean)
If you want to append the letters "mean" to the column names:
names(df2) <- paste0("mean", names(df2)
If you had wanted to do the column selection automatically then grep or grepl would work:
aggregate(df[ grepl("acc", names(df) )], df[1], mean)
Here are a couple of other base R options:
split + vapply (since we know vapply would simplify to a matrix whenever possible)
t(vapply(split(df[-c(1, 2)], df[, 1]), colMeans, numeric(4L)))
by (with a do.call(rbind, ...) to get the final structure)
do.call(rbind, by(data = df[-c(1, 2)], INDICES = df[[1]], FUN = colMeans))
Both will give you something like this as your result:
# acc1 acc2 acc3 acc4
# A 1.337496 2.091926 1.978835 1.799669
# B 1.287303 1.447884 1.297933 1.312325
# C 1.870008 1.145385 1.768011 1.252027
# D 1.682446 1.413716 1.582506 1.274925
The sample data used here was (with set.seed, for reproducibility):
set.seed(1)
df = data.frame(id = rep(LETTERS[1:4], 4),
sub = rep(c(1:4), 4),
acc1 = runif(16, 0, 3),
acc2 = runif(16, 0, 3),
acc3 = runif(16, 0, 3),
acc4 = runif(16, 0, 3))
Scaling up to 1M rows, these both perform quite well (though obviously not as fast as "dplyr" or "data.table").
You can do this in base package itself using this:
a <- list();
for (i in 1:nlevels(df$id))
{
a[[i]] = colMeans(subset(df, id==levels(df$id)[i])[,c(3,4,5,6)]) ##select columns of df of which you want to compute the means. In your example, 3, 4, 5 and 6 are the columns
}
meanDF <- cbind(data.frame(levels(df$id)), data.frame(matrix(unlist(a), nrow=4, ncol=4, byrow=T)))
colnames(meanDF) = c("id", "meanacc1", "meanacc2", "meanacc3", "meanacc4")
meanDF
id meanacc1 meanacc2 meanacc3 meanacc4
A 1.464635 1.645898 1.7461862 1.026917
B 1.807555 1.097313 1.7135346 1.517892
C 1.350708 1.922609 0.8068907 1.607274
D 1.458911 0.726527 2.4643733 2.141865