I'm working on the output off an online questionnaire and have some trouble handling the data. This is the setups: 200 images have been rated on two 9-point-scales, totaling in 400 combinations. Unfortunately, the data hasn't been in encoded in 400 variables with values ranging from 1 to 9, but for each scale-image combination, 9 binary variables have been encoded, looking like this for two image-scale combinations:
Part. V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18
1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
2 0 0 0 0 0 0 1 0 0
3 0 0 1 0 0 0 0 0 0
As you can see, there are also some N/A values in the data set. That's because of all 400 combinations, each participant only rated a randomised 50. Given the 400 combinations, we have a total of 3600 variables in the data set. I would now like to condense and recode those values in a sense, that R counts the vars in intervals of 9, then recodes the binary 1 for a value of 1 to 9, depending on its position on the scale, and then condenses everything into 400 combination variables. In the end, it should look something like this:
Part. C1 C2
1 3 2
2 7
3 3
I've looked into the reshape package, but couldn't exactly figure out the way to do this.
Any suggestions?
Using apply family functions:
#dummy data
df <- read.table(text = "
Part.,V1,V2,V3,V4,V5,V6,V7,V8,V9,V10,V11,V12,V13,V14,V15,V16,V17,V18
1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0
2,0,0,0,0,0,0,1,0,0,,,,,,,,,
3,,,,,,,,,,0,0,1,0,0,0,0,0,0
", header = TRUE, sep = ",")
# result
# cbind - column bind, put columns side by side
cbind(
# First column is the "Part." column
df[, "Part.", drop = FALSE],
# other columns are coming from below code
# sapply returns matrix, converting it to data.frame so we can use cbind.
as.data.frame(
# get data column index 9 columns each, first 2 to 9, then 10 to 18, etc.
sapply(seq(2, ncol(df), 9), function(i)
# for each 9 columns check at which position it is equal to 1,
# using which() function
apply(df[, i:(i + 8)], 1, function(j) which(j == 1)))
)
)
#output
# Part. V1 V2
# 1 1 3 2
# 2 2 7
# 3 3 3
Here is a solution for a small example. I did it for only 2 possible outcomes. So v1 = 1 for pic 1, v2 = 2 for pic one, v3 = 1 for pic 2 ... . If you have 9 possible outcomes you have to change id <- rep(1:2, each = 2) to id <- rep(1:n, each = 9) where n is the total number of pictures. Also change the 2 in final <- matrix(nrow = nrow(dat), ncol = ncol(dat)/2) to 9.
I hope that helps.
dat <- data.frame(v1 = c(NA,0,1,0), v2 = c(NA,1,0,1), v3 = c(0,1,NA,0), v4 = c(1,0,NA,1))
id <- rep(1:2, each = 2)
final <- matrix(nrow = nrow(dat), ncol = ncol(dat)/2)
for (i in unique(id)){
wdat <- dat[ ,which(id == i)]
for (j in 1:nrow(wdat)){
if(is.na(wdat[j,1] )) {
final[j,i] <- NA
} else {
final[j,i] <- which(wdat[j, ] == 1)
}
}
}
The input and output for my example:
> dat
v1 v2 v3 v4
1 NA NA 0 1
2 0 1 1 0
3 1 0 NA NA
4 0 1 0 1
> final
[,1] [,2]
[1,] NA 2
[2,] 2 1
[3,] 1 NA
[4,] 2 2
Related
I have a data frame where each observation is comprehended in two columns. In this way, columns 1 and 2 represents the individual 1, 3 and 4 the individual 2 and so on.
Basically what I want to do is to add two contigous columns so I have the individual real score.
In this example V1 and V2 represent individual I and V3 and V4 represent individual II. So for the result data frame I will have the half of columns, the same number of rows and each value will be the addition of each value between two contigous colums.
Data
V1 V2 V3 V4
1 0 0 1 1
2 1 0 0 0
3 0 1 1 1
4 0 1 0 1
Desire Output
I II
1 0 2
2 1 0
3 1 2
4 1 1
I tried something like this
a <- data.frame(V1= c(0,1,0,0),V2=c(0,0,1,1),V3=c(1,0,1,0),V4=c(1,0,1,1))
b <- data.frame(NA, nrow = nrow(a), ncol = ncol(data))
for (i in seq(2,ncol(a),by=2)){
for (k in 1:nrow(a)){
b[k,i] <- a[k,i] + a[k,i-1]
}
}
b <- as.data.frame(b)
b <- b[,-c(seq(1,length(b),by=2))]
Is there a way to make it simplier?
We could use split.default to split the data and then do rowSums by looping over the list
sapply(split.default(a, as.integer(gl(ncol(a), 2, ncol(a)))), rowSums)
1 2
[1,] 0 2
[2,] 1 0
[3,] 1 2
[4,] 1 1
You can use vector recycling to select columns and add them.
res <- a[c(TRUE, FALSE)] + a[c(FALSE, TRUE)]
names(res) <- paste0('col', seq_along(res))
res
# col1 col2
#1 0 2
#2 1 0
#3 1 2
#4 1 1
dplyr's approach with row-wise operations (rowwise is a special type of grouping per row)
a <- data.frame(V1= c(0,1,0,0),V2=c(0,0,1,1),V3=c(1,0,1,0),V4=c(1,0,1,1))
library(dplyr)
a%>%
rowwise()%>%
transmute(I=sum(c(V1,V2)),
II=sum(c(V3,V4)))
or alternatively with a built-in row-wise variant of the sum
a %>% transmute(I = rowSums(across(1:2)),
II = rowSums(across(3:4)))
I want to make every element in the dataframe (except fot the ID column) become a 0 if it is any number other than 1.
I have:
ID A B C D E
abc 5 3 1 4 1
def 4 1 3 2 5
I want:
ID A B C D E
abc 0 0 1 0 1
def 0 1 0 0 0
I am having trouble figuring out how to specify for this to be done to do to every entry in every column and row.
Here is my code:
apply(dat.lec, 2 , function(y)
if(!is.na(y)){
if(y==1){y <- 1}
else{y <-0}
}
else {y<- NA}
)
Thank you for your help!
No need for implicit or explicit looping.
# Sample data
set.seed(2016);
df <- as.data.frame(matrix(sample(10, replace = TRUE), nrow = 2));
df <- cbind.data.frame(id = sample(letters, 2), df);
df;
# id V1 V2 V3 V4 V5
#1 k 2 9 5 7 1
#2 g 2 2 2 9 1
# Replace all entries != 1 with 0's
df[, -1][df[, -1] != 1] <- 0;
df;
# id V1 V2 V3 V4 V5
#1 k 0 0 0 0 1
#2 g 0 0 0 0 1
I'm looking to compare values within a dataset
Every row starts with a unique ID followed by a couple binary variables
The data looks like this:
row.name v1 v2 v3 ...
1 0 0 0
2 1 1 1
3 1 0 1
I want to know which values are the same (if equal assign value of 1) and which are different (if not equal assign value of 0) for all unique pairings.
For example in column v1: row1 == 0 and row2 == 1, which should result in an assignment of 0.
So, the output should look like this
id1 id2 v1 v2 v3 ...
1 2 0 0 0 ...
1 3 0 1 0 ...
2 3 1 0 1 ...
I'm looking for an efficient way of doing this for more than 1000 rows...
There's no way to do this without expanding each combination of rows, so with 1000 rows, it is going to take a bit of time. But here is a solution:
dat <- read.table(header=T, text="row.name v1 v2 v3
1 0 0 0
2 1 1 1
3 1 0 1")
Create the index rows:
indices <- t(combn(dat$row.name, 2))
colnames(indices) <- c('id1', 'id2')
Loop through the index rows, and collect the comparisons:
res1 <- t(apply(indices, 1, function(x) as.numeric(dat[x[1],-1] == dat[x[2],-1])))
colnames(res1) <- names(dat[-1])
Put them together:
result <- cbind(indices, res1)
result
## id1 id2 v1 v2 v3
## [1,] 1 2 0 0 0
## [2,] 1 3 0 1 0
## [3,] 2 3 1 0 1
I'm trying to multiply column and get its names.
I have a data frame:
v1 v2 v3 v4 v5
0 1 1 1 1
0 1 1 0 1
1 0 1 1 0
I'm trying to multiplying each column with other, like:
v1v2
v1v3
v1v4
v1v5
and
v2v3
v2v4
v2v5
etc, and
v1v2v3
v1v2v4
v1v2v5
v2v3v4
v2v3v5
4 combination and 5 combination...if there is n column then n combination.
I'm try to use following code in while loop, but it is not working:
i<-1
while(i<=ncol(data)
{
results<-data.frame()
v<-i
results<- t(apply(data,1,function(x) combn(x,v,prod)))
comb <- combn(colnames(data),v)
colnames(results) <- apply(comb,v,function(x) paste(x[1],x[2],sep="*"))
results <- colSums(results)
}
but it is not working.
sample out put..
if n=3
v1v2 v1v3 v2v3
0 0 1
0 0 1
0 1 0
and colsum
v1v2 v1v3 v2v3
0 1 2
then
v1v2=0
v1v3=1
v2v3=2
this one is I'm trying?
Try this:
df <- read.table(text = "v1 v2 v3 v4 v5
0 1 1 1 1
0 1 1 0 1
1 0 1 1 0", skip = 1)
df
ll <- vector(mode = "list", length = ncol(df)-1)
ll <- lapply(2:ncol(df), function(ncols){
tmp <- t(apply(df, 1, function(rows) combn(x = rows, m = ncols, prod)))
if(ncols < ncol(df)){
tmp <- colSums(tmp)
}
else{
tmp <- sum(tmp)
}
names1 <- t(combn(x = colnames(df), m = ncols))
names(tmp) <- apply(names1, 1, function(rows) paste0(rows, collapse = ""))
ll[[ncols]] <- tmp
})
ll
# [[1]]
# V1V2 V1V3 V1V4 V1V5 V2V3 V2V4 V2V5 V3V4 V3V5 V4V5
# 0 1 1 0 2 1 2 2 2 1
#
# [[2]]
# V1V2V3 V1V2V4 V1V2V5 V1V3V4 V1V3V5 V1V4V5 V2V3V4 V2V3V5 V2V4V5 V3V4V5
# 0 0 0 1 0 0 1 2 1 1
#
# [[3]]
# V1V2V3V4 V1V2V3V5 V1V2V4V5 V1V3V4V5 V2V3V4V5
# 0 0 0 0 1
#
# [[4]]
# V1V2V3V4V5
# 0
Edit following comment
The results of the different set of column combinations can then be accessed by indexing (subsetting) the list. E.g. to access the "2 combinations", select the first element of the list, to access the "3rd combination", select the second element of the list, et c.
ll[[1]]
# V1V2 V1V3 V1V4 V1V5 V2V3 V2V4 V2V5 V3V4 V3V5 V4V5
# 0 1 1 0 2 1 2 2 2 1
t has 20 values, c has also 20 values 0, 1. I am interested in t matrix. Here I have a Loop, repeating 5 times. Every time sel give 20 values. I want to store there frequency in t.mat. But how can I get the required results, the resulting table may look like the below table
t <- 1:20
# c <- seq(0:1, 10)
t.mat <- array(dim = c(20, 5))
rep <- 5
for(mm in 1:rep){
sel <- sample(1:20, replace = TRUE)
tt <- t[sel]
# cc <- c[sel]
t.mat[, mm] = tt[1:20] # here the problem lies, I have no clue how
}
The output for the above may be look like below. But t will be of 20 values, I roughly give just six lines:
t v1 v2 v3 v4 v5
1 1 0 1 0 1
2 0 0 2 0 1
3 0 1 1 1 0
4 1 1 0 0 1
5 2 0 2 1 0
6 0 0 0 1 2
I'm guessing a little bit as to what you want, but it's probably this:
do.call(cbind, lapply(1:5, function(i)
tabulate(sample(t, replace = T), nbins = 20)))
sample generates the samples you want, tabulate counts frequencies (with max specified manually as it will not always occur in sample), lapply iterates the procedure 5 times, and finally do.call(cbind, binds it all together by column.