Expand a vector and count instances - r

I have a data.frame
orig.DF<-data.frame(V1=c("A", "B", "C"), V2=c(3,2,4))
and I have to expand it so that it takes the following form
A 1
A 2
A 3
B 1
B 2
C 1
C 2
C 3
C 4
I tried taaply and ave but I can't get it to count to 1:x and repeat the V1 accordingly

df <- data.frame(V1 = c("A", "B", "C"), V2 = c(3, 2, 4))
data.frame(x = rep(df$V1, df$V2), y = sequence(df$V2))
x y
1 A 1
2 A 2
3 A 3
4 B 1
5 B 2
6 C 1
7 C 2
8 C 3
9 C 4

Here is one approach:
do.call(
rbind,
apply(orig.DF, 1, function(row) expand.grid(row["V1"], 1:row["V2"]))
)

Related

R-Converting Incidence matrix(csv file) to edge list format

I am studying social network analysis and will be using Ucinet to draw network graphs. For this, I have to convert the csv file to an edge list format. Converting the adjacency matrix to the edge list was successful. However, it is difficult to convert an incidence matrix to the edge list format.
The csv file('some.csv') I have, with a incidence matrix like this:
A B C D
a 1 0 3 1
b 0 0 0 2
c 3 2 0 1
The code that converted the adjacency matrix to the edge list was as follows:
x<-read.csv("C:/.../something.csv", header=T, row.names=1)
net<-as.network(x, matrix.type='adjacency', ignore.eval=FALSE, names.eval='dd', loops=FALSE)
el<-edgelist(net, attrname='dd')
write.csv(el, file='C:/.../result.csv')
Now It only succeedded in loading the file. I tried to follow the above method, but I get an error.
y<-read.csv("C:/.../some.csv", header=T, row.names=1)
net2<-network(y, matrix.type='incidence', ignore.eval=FALSE, names.eval='co', loops=FALSE)
Error in network.incidence(x, g, ignore.eval, names.eval, na.rm, edge.check) :
Supplied incidence matrix has empty head/tail lists. (Did you get the directedness right?)
I want to see the result in this way:
a A 1
a C 3
a D 1
b D 2
c A 3
c B 2
c D 1
I tried to put the values as the error said, but I could not get the result i wanted.
Thank you for any assistance with this.
Here's your data:
inc_mat <- matrix(
c(1, 0, 3, 1,
0, 0, 0, 2,
3, 2, 0, 1),
nrow = 3, ncol = 4, byrow = TRUE
)
rownames(inc_mat) <- letters[1:3]
colnames(inc_mat) <- LETTERS[1:4]
inc_mat
#> A B C D
#> a 1 0 3 1
#> b 0 0 0 2
#> c 3 2 0 1
Here's a generalized function that does the trick:
as_edgelist.weighted_incidence_matrix <- function(x, drop_rownames = TRUE) {
melted <- do.call(cbind, lapply(list(row(x), col(x), x), as.vector)) # 3 col matrix of row index, col index, and `x`'s values
filtered <- melted[melted[, 3] != 0, ] # drop rows where column 3 is 0
# data frame where first 2 columns are...
df <- data.frame(mode1 = rownames(x)[filtered[, 1]], # `x`'s rownames, indexed by first column in `filtered``
mode2 = colnames(x)[filtered[, 2]], # `x`'s colnames, indexed by the second column in `filtered`
weight = filtered[, 3], # the third column in `filtered`
stringsAsFactors = FALSE)
out <- df[order(df$mode1), ] # sort by first column
if (!drop_rownames) {
return(out)
}
`rownames<-`(out, NULL)
}
Take it for a spin:
el <- as_edgelist.weighted_incidence_matrix(inc_mat)
el
#> mode1 mode2 weight
#> 1 a A 1
#> 2 a C 3
#> 3 a D 1
#> 4 b D 2
#> 5 c A 3
#> 6 c B 2
#> 7 c D 1
Here are the results you wanted:
control_df <- data.frame(
mode1 = c("a", "a", "a", "b", "c", "c", "c"),
mode2 = c("A", "C", "D", "D", "A", "B", "D"),
weight = c(1, 3, 1, 2, 3, 2, 1),
stringsAsFactors = FALSE
)
control_df
#> mode1 mode2 weight
#> 1 a A 1
#> 2 a C 3
#> 3 a D 1
#> 4 b D 2
#> 5 c A 3
#> 6 c B 2
#> 7 c D 1
Do they match?
identical(control_df, el)
#> [1] TRUE
This might not be the most efficient way, but it produces expected result:
y <- matrix( c(1,0,3,0,0,2,3,0,0,1,2,1), nrow=3)
colnames(y) <- c("e.A","e.B","e.C","e.D")
dt <- data.frame(rnames=c("a","b","c"))
dt <- cbind(dt, y)
# rnames e.A e.B e.C e.D
#1 a 1 0 3 1
#2 b 0 0 0 2
#3 c 3 2 0 1
# use reshape () function to convert dataframe into the long format
M <- reshape(dt, direction="long", idvar = "rnames", varying = c("e.A","e.B","e.C","e.D"))
M <- M[M$e >0,]
M
# rnames time e
# a.A a A 1
# c.A c A 3
# c.B c B 2
# a.C a C 3
# a.D a D 1
# b.D b D 2
# c.D c D 1
# If M needs to be sorted by the column rnames:
M[order(M$rnames), ]
# rnames time e
# a.A a A 1
# a.C a C 3
# a.D a D 1
# b.D b D 2
# c.A c A 3
# c.B c B 2
# c.D c D 1

Convert a matrix to columns

Assuming I have a matrix looks like below, the values up or down the diagonal are the same. In other words, [,1] x [2,] and [,2] x [1,] both are 2 in the matrix.
> m = cbind(c(1,2,3),c(2,4,5),c(3,5,6))
> m
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 4 5
[3,] 3 5 6
Then I have real name for 1, 2, and 3 as well.
>Real_name
A B C # A represents 1, B represents 2, and C represents 3.
If I would like to convert the matrix to 3 columns containing corresponding real name for each pair, and the pair must be unique, A x B is the same as B x A, so we keep A x B only. How can I achieve it using R?
A A 1
A B 2
A C 3
B B 4
B C 5
C C 6
The following is straightforward:
m <- cbind(c(1,2,3), c(2,4,5), c(3,5,6))
## read `?lower.tri` and try `v <- lower.tri(m, diag = TRUE)` to see what `v` is
## read `?which` and try `which(v, arr.ind = TRUE)` to see what it gives
ij <- which(lower.tri(m, diag = TRUE), arr.ind = TRUE)
Real_name <- LETTERS[1:3]
data.frame(row = Real_name[ij[, 1]], col = Real_name[ij[, 2]], val = c(m[ij]))
# row col val
#1 A A 1
#2 B A 2
#3 C A 3
#4 B B 4
#5 C B 5
#6 C C 6
colnames(m) <- c("A", "B", "C")
rownames(m) <- c("A", "B", "C")
m[lower.tri(m)] = NA # replace lower triangular elements with NA
data.table::melt(m, na.rm = TRUE) # melt and remove NA
# Var1 Var2 value
#1 A A 1
#4 A B 2
#5 B B 4
#7 A C 3
#8 B C 5
#9 C C 6
Or you can do it in a single line: melt(replace(m, lower.tri(m), NA), na.rm = TRUE)
This will also work:
g <- expand.grid(1:ncol(m), 1:ncol(m))
g <- g[g[,2]>=g[,1],]
cbind.data.frame(sapply(g, function(x) Real_name[x]), Val=m[as.matrix(g)])
Var1 Var2 Val
1 A A 1
2 A B 2
3 B B 4
4 A C 3
5 B C 5
6 C C 6

cbind warnings : row names were found from a short variable and have been discarded

I have below line of code for cbind, but I am getting a warning message everytime.
Though the code still functions as it should be, is there any way to resolve the warning?
dateset = subset(all_data[,c("VAR1","VAR2","VAR3","VAR4","VAR5","RATE1","RATE2","RATE3")])
dateset = cbind(dateset[c(1,2,3,4,5)],stack(dateset[,-c(1,2,3,4,5)]))
Warnings :
Warning message:
In data.frame(..., check.names = FALSE) :
row names were found from a short variable and have been discarded
Thanks in advance!
I'm guessing your data.frame has row.names:
A <- data.frame(a = c("A", "B", "C"),
b = c(1, 2, 3),
c = c(4, 5, 6),
row.names=c("A", "B", "C"))
cbind(A[1], stack(A[-1]))
# a values ind
# 1 A 1 b
# 2 B 2 b
# 3 C 3 b
# 4 A 4 c
# 5 B 5 c
# 6 C 6 c
# Warning message:
# In data.frame(..., check.names = FALSE) :
# row names were found from a short variable and have been discarded
What's happening here is that since you can't by default have duplicated row.names in a data.frame and since you don't tell R at any point to duplicate the row.names when recycling the first column to the same number of rows of the stacked column, R just discards the row.names.
Compare with a similar data.frame, but one without row.names:
B <- data.frame(a = c("A", "B", "C"),
b = c(1, 2, 3),
c = c(4, 5, 6))
cbind(B[1], stack(B[-1]))
# a values ind
# 1 A 1 b
# 2 B 2 b
# 3 C 3 b
# 4 A 4 c
# 5 B 5 c
# 6 C 6 c
Alternatively, you can set row.names = NULL in your cbind statement:
cbind(A[1], stack(A[-1]), row.names = NULL)
# a values ind
# 1 A 1 b
# 2 B 2 b
# 3 C 3 b
# 4 A 4 c
# 5 B 5 c
# 6 C 6 c
If your original row.names are important, you can also add them back in with:
cbind(rn = rownames(A), A[1], stack(A[-1]), row.names = NULL)
# rn a values ind
# 1 A A 1 b
# 2 B B 2 b
# 3 C C 3 b
# 4 A A 4 c
# 5 B B 5 c
# 6 C C 6 c

I have multiple dataframes under one name and I need to create a new column in each one by combining two of the other columns? [duplicate]

I have several csv files all named with dates and for all of them I want to create a new column in each file that contains data from two other columns placed together. Then, I want to combine them into one big dataframe and choose only two of those columns to keep. Here's an example:
Say I have two dataframes:
a b c a b c
x 1 2 3 x 3 2 1
y 2 3 1 y 2 1 3
Then I want to create a new column d in each of them:
a b c d a b c d
x 1 2 3 13 x 3 2 1 31
y 2 3 1 21 y 2 1 3 23
Then I want to combine them like this:
a b c d
x 1 2 3 13
y 2 3 1 21
x 3 2 1 31
y 2 1 3 23
Then keep two of the columns a and d and delete the other two columns b and c:
a d
x 1 13
y 2 21
x 3 31
y 2 23
Here is my current code (It doesn't work when I try to combine two of the columns or when I try to only keep two of the columns):
f <- list.files(pattern="201\\d{5}\\.csv") # reading in all the files
mydata <- sapply(f, read.csv, simplify=FALSE) # assigning them to a dataframe
do.call(rbind,mydata) # combining all of those dataframes into one
mydata$Data <- paste(mydata$LAST_UPDATE_DT,mydata$px_last) # combining two of the columns into a new column named "Data"
c('X','Data') %in% names(mydata) # keeping two of the columns while deleting the rest
The object mydata is a list of data frames. You can change the data frames in the list with lapply:
lapply(mydata, function(x) "[<-"(x, "c", value = paste0(x$a, x$b)))
file1 <- "a b
x 2 3"
file2 <- "a b
x 3 1"
mydata <- lapply(c(file1, file2), function(x) read.table(text = x, header =TRUE))
lapply(mydata, function(x) "[<-"(x, "c", value = paste0(x$a, x$b)))
# [[1]]
# a b c
# x 2 3 23
#
# [[2]]
# a b c
# x 3 1 31
You can use rbind (data1,data2)[,c(1,3)] for that. I assume that you can create col d in each dataframe which is a basic thing.
data1<-structure(list(a = 1:2, b = 2:3, c = c(3L, 1L), d = c(13L, 21L
)), .Names = c("a", "b", "c", "d"), row.names = c("x", "y"), class = "data.frame")
> data1
a b c d
x 1 2 3 13
y 2 3 1 21
data2<-structure(list(a = c(3L, 2L), b = c(2L, 1L), c = c(1L, 3L), d = c(31L,
23L)), .Names = c("a", "b", "c", "d"), row.names = c("x", "y"
), class = "data.frame")
> data2
a b c d
x 3 2 1 31
y 2 1 3 23
data3<-rbind(data1,data2)
> data3
a b c d
x 1 2 3 13
y 2 3 1 21
x1 3 2 1 31
y1 2 1 3 23
finaldata<-data3[,c("a","d")]
> finaldata
a d
x 1 13
y 2 21
x1 3 31
y1 2 23

change data.frame column into rows in R

A <- c(1,6)
B <- c(2,7)
C <- c(3,8)
D <- c(4,9)
E <- c(5,0)
df <- data.frame(A,B,C,D,E)
df
A B C D E
1 1 2 3 4 5
2 6 7 8 9 0
I would like to have this:
df
1 2
A 1 6
B 2 7
C 3 8
D 4 9
E 5 0
If your dataframe is truly in that format, then all of your vectors will be character vectors. Or, you basically have a character matrix and you could do this:
data.frame(t(df))
It would be better, though, to just define it the way you want it from the get-go
df <- data.frame(c('A','B','C','D','E'),
c(1, 2, 3, 4, 5),
c(6, 7, 8, 9, 0))
You could also do this
df <- data.frame(LETTERS[1:5], 1:5, c(6:9, 0))
If you wanted to give the columns names, you could do this
df <- data.frame(L = LETTERS[1:5], N1 = 1:5, N2 = c(6:9, 0))
Sometimes, if I use read.DIF of Excel data the data gets transposed. Is that how you got the original data in? If so, you can call
read.DIF(filename, transpose = T)
to get the data in the correct orientation.
I really recommend data.table approach without manual steps becauce they are error-prone
A <- c(1,6)
B <- c(2,7)
C <- c(3,8)
D <- c(4,9)
E <- c(5,0)
df <- data.frame(A,B,C,D,E)
df
library('data.table')
dat.m <- melt(as.data.table(df, keep.rownames = "Vars"), id.vars = "Vars") # https://stackoverflow.com/a/44128640/54964
dat.m
Output
A B C D E
1 1 2 3 4 5
2 6 7 8 9 0
Vars variable value
1: 1 A 1
2: 2 A 6
3: 1 B 2
4: 2 B 7
5: 1 C 3
6: 2 C 8
7: 1 D 4
8: 2 D 9
9: 1 E 5
10: 2 E 0
R: 3.4.0 (backports)
OS: Debian 8.7

Resources