I just have a simple question, I really appreciate everyones input, you have been a great help to my project. I have an additional question about data frames in R.
I have data frame that looks similar to something like this:
C <- c("","","","","","","","A","B","D","A","B","D","A","B","D")
D <- c(NA,NA,NA,2,NA,NA,1,1,4,2,2,5,2,1,4,2)
G <- list(C=C,D=D)
T <- as.data.frame(G)
T
C D
1 NA
2 NA
3 NA
4 2
5 NA
6 NA
7 1
8 A 1
9 B 4
10 D 2
11 A 2
12 B 5
13 D 2
14 A 1
15 B 4
16 D 2
I would like to be able to condense all the repeat characters into one, and look similar to this:
J B C E
1 2 1
2 A 1 2 1
3 B 4 5 4
4 D 2 2 2
So of course, the data is all the same, it is just that it is condensed and new columns are formed to hold the data. I am sure there is an easy way to do it, but from the books I have looked through, I haven't seen anything for this!
EDIT I edited the example because it wasn't working with the answers so far. I wonder if the NA's, blanks, and unevenness from the blanks are contributing??
here´s a reshape solution:
require(reshape)
cast(T, C ~ ., function(x) x)
Changed T to df to avoid a bad habit. Returns a list, which my not be what you want but you can convert from there.
C <- c("A","B","D","A","B","D","A","B","D")
D <- c(1,4,2,2,5,2,1,4,2)
my.df <- data.frame(id=C,val=D)
ret <- function(x) x
by.df <- by(my.df$val,INDICES=my.df$id,ret)
This seems to get the results you are looking for. I'm assuming it's OK to remove the NA values since that matches the desired output you show.
T <- na.omit(T)
T$ind <- ave(1:nrow(T), T$C, FUN = seq_along)
reshape(T, direction = "wide", idvar = "C", timevar = "ind")
# C D.1 D.2 D.3
# 4 2 1 NA
# 8 A 1 2 1
# 9 B 4 5 4
# 10 D 2 2 2
library(reshape2)
dcast(T, C ~ ind, value.var = "D", fill = "")
# C 1 2 3
# 1 2 1
# 2 A 1 2 1
# 3 B 4 5 4
# 4 D 2 2 2
Related
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 have a long data list similar to the following one:
set.seed(9)
part_number<-sample(1:5,5,replace=TRUE)
Type<-sample( c("A","B","C"),5, replace=TRUE)
rank<-sample(1:20,5,replace=TRUE)
data<-data.frame(cbind(part_number,Type,rank))
data
part_number Type rank
1 2 A 3
2 1 B 1
3 2 B 18
4 2 C 7
5 3 C 10
I want to rearrange the data in the following way:
part_number A B C
1 1
2 3 18 7
3 10
I think I need to use the reshape library. But I am not sure.
libary(tidyr)
data %>% spread(Type,rank)
# part_number A B C
# 1 1 <NA> 1 <NA>
# 2 2 3 18 7
# 3 3 <NA> <NA> 10
You would go about doing the following:
data <- reshape(data, idvar = "part_number", timevar = "Type", direction = "wide")
data
To format it exactly as you asked, I would add in,
library(tidyverse)
data %>%
arrange(part_number) %>%
dplyr::select(part_number, A = rank.A, B = rank.B, C = rank.C)
If you however had a lot more columns to rename, I would use the gsub function to rename by pattern. In addition, since now the row names are messy,
rownames(data) <- c()
Let me know if this doesn't work or this wasn't what you had in mind.
I have some data in long form that looks like this:
dat1 = data.frame(
id = rep(LETTERS[1:2], each=4),
value = 1:8
)
In table form:
id value
A 1
A 2
A 3
A 4
B 5
B 6
B 7
B 8
And I want it to be in short form and look like this:
dat1 = data.frame(A = 1:4, B = 5:8)
In table form:
A B
1 5
2 6
3 7
4 8
Now I could solve this by looping with cbind() and stuff, but I want to use some kind of reshape/melt function as these are the best way to do this kind of thing I think.
However, from spending >30 minutes trying to get melt() and reshape() to work, reading answers on SO, it seems that these functions requires the id.var to be set. Now, it is plainly redundant for this kind of thing, so how do I do what I want to do without having to resort to some kind of looping?
I'm pretty sure this has been answered before. Anyway, unstack is convenient in this particular case with equal group size:
unstack(dat1, form = value ~ id)
# A B
# 1 1 5
# 2 2 6
# 3 3 7
# 4 4 8
Solution below works when there are different numbers of As and Bs. For equal counts, unstack works great and with less code (Henrik's answer).
# create more general data (unbalanced 'id')
each <- c(4,2,3)
dat1 = data.frame(
id = unlist(mapply(rep, x = LETTERS[1:length(each)], each = each)),
value = 1:sum(each),
row.names = 1:sum(each) # to reproduce original row.names
)
tab <- table(dat1$id)
dat1$timevar <- unlist(sapply(tab, seq))
library(reshape2)
dcast(dat1, timevar ~ id )[-1]
initial data:
id value
1 A 1
2 A 2
3 A 3
4 A 4
5 B 5
6 B 6
7 C 7
8 C 8
9 C 9
result:
A B C
1 1 5 7
2 2 6 8
3 3 NA 9
4 4 NA NA
Here's a base R approach to consider. It uses the lengths function, which I believe was introduced in R 3.2.
x <- split(dat1$value, dat1$id)
as.data.frame(lapply(x, function(y) `length<-`(y, max(lengths(x)))))
# A B C
# 1 1 5 7
# 2 2 6 8
# 3 3 NA 9
# 4 4 NA NA
I am dealing with a dataset that is in wide format, as in
> data=read.csv("http://www.kuleuven.be/bio/ento/temp/data.csv")
> data
factor1 factor2 count_1 count_2 count_3
1 a a 1 2 0
2 a b 3 0 0
3 b a 1 2 3
4 b b 2 2 0
5 c a 3 4 0
6 c b 1 1 0
where factor1 and factor2 are different factors which I would like to take along (in fact I have more than 2, but that shouldn't matter), and count_1 to count_3 are counts of aggressive interactions on an ordinal scale (3>2>1). I would now like to convert this dataset to long format, to get something like
factor1 factor2 aggression
1 a a 1
2 a a 2
3 a a 2
4 a b 1
5 a b 1
6 a b 1
7 b a 1
8 b a 2
9 b a 2
10 b a 3
11 b a 3
12 b a 3
13 b b 1
14 b b 1
15 b b 2
16 b b 2
17 c a 1
18 c a 1
19 c a 1
20 c a 2
21 c a 2
22 c a 2
23 c a 2
24 c b 1
25 c b 2
Would anyone happen to know how to do this without using for...to loops, e.g. using package reshape2? (I realize it should work using melt, but I just haven't been able to figure out the right syntax yet)
Edit: For those of you that would also happen to need this kind of functionality, here is Ananda's answer below wrapped into a little function:
widetolong.ordinal<-function(data,factors,responses,responsename) {
library(reshape2)
data$ID=1:nrow(data) # add an ID to preserve row order
dL=melt(data, id.vars=c("ID", factors)) # `melt` the data
dL=dL[order(dL$ID), ] # sort the molten data
dL[,responsename]=match(dL$variable,responses) # convert reponses to ordinal scores
dL[,responsename]=factor(dL[,responsename],ordered=T)
dL=dL[dL$value != 0, ] # drop rows where `value == 0`
out=dL[rep(rownames(dL), dL$value), c(factors, responsename)] # use `rep` to "expand" `data.frame` & drop unwanted columns
rownames(out) <- NULL
return(out)
}
# example
data <- read.csv("http://www.kuleuven.be/bio/ento/temp/data.csv")
widetolong.ordinal(data,c("factor1","factor2"),c("count_1","count_2","count_3"),"aggression")
melt from "reshape2" will only get you part of the way through this problem. To go the rest of the way, you just need to use rep from base R:
data <- read.csv("http://www.kuleuven.be/bio/ento/temp/data.csv")
library(reshape2)
## Add an ID if the row order is importantt o you
data$ID <- 1:nrow(data)
## `melt` the data
dL <- melt(data, id.vars=c("ID", "factor1", "factor2"))
## Sort the molten data, if necessary
dL <- dL[order(dL$ID), ]
## Extract the numeric portion of the "variable" variable
dL$aggression <- gsub("count_", "", dL$variable)
## Drop rows where `value == 0`
dL <- dL[dL$value != 0, ]
## Use `rep` to "expand" your `data.frame`.
## Drop any unwanted columns at this point.
out <- dL[rep(rownames(dL), dL$value), c("factor1", "factor2", "aggression")]
This is what the output finally looks like. If you want to remove the funny row names, just use rownames(out) <- NULL.
out
# factor1 factor2 aggression
# 1 a a 1
# 7 a a 2
# 7.1 a a 2
# 2 a b 1
# 2.1 a b 1
# 2.2 a b 1
# 3 b a 1
# 9 b a 2
# 9.1 b a 2
# 15 b a 3
# 15.1 b a 3
# 15.2 b a 3
# 4 b b 1
# 4.1 b b 1
# 10 b b 2
# 10.1 b b 2
# 5 c a 1
# 5.1 c a 1
# 5.2 c a 1
# 11 c a 2
# 11.1 c a 2
# 11.2 c a 2
# 11.3 c a 2
# 6 c b 1
# 12 c b 2
Hoping there's a simple answer here but I can't find it anywhere.
I have a numeric matrix with row names and column names:
# 1 2 3 4
# a 6 7 8 9
# b 8 7 5 7
# c 8 5 4 1
# d 1 6 3 2
I want to melt the matrix to a long format, with the values in one column and matrix row and column names in one column each. The result could be a data.table or data.frame like this:
# col row value
# 1 a 6
# 1 b 8
# 1 c 8
# 1 d 1
# 2 a 7
# 2 c 5
# 2 d 6
...
Any tips appreciated.
Use melt from reshape2:
library(reshape2)
#Fake data
x <- matrix(1:12, ncol = 3)
colnames(x) <- letters[1:3]
rownames(x) <- 1:4
x.m <- melt(x)
x.m
Var1 Var2 value
1 1 a 1
2 2 a 2
3 3 a 3
4 4 a 4
...
The as.table and as.data.frame functions together will do this:
> m <- matrix( sample(1:12), nrow=4 )
> dimnames(m) <- list( One=letters[1:4], Two=LETTERS[1:3] )
> as.data.frame( as.table(m) )
One Two Freq
1 a A 7
2 b A 2
3 c A 1
4 d A 5
5 a B 9
6 b B 6
7 c B 8
8 d B 10
9 a C 11
10 b C 12
11 c C 3
12 d C 4
Assuming 'm' is your matrix...
data.frame(col = rep(colnames(m), each = nrow(m)),
row = rep(rownames(m), ncol(m)),
value = as.vector(m))
This executes extremely fast on a large matrix and also shows you a bit about how a matrix is made, how to access things in it, and how to construct your own vectors.
A modification that doesn't require you to know anything about the storage structure, and that easily extends to high dimensional arrays if you use the dimnames, and slice.index functions:
data.frame(row=rownames(m)[as.vector(row(m))],
col=colnames(m)[as.vector(col(m))],
value=as.vector(m))