Pasting a string of variables into a function is not working - r

I was looking at this question: Find how many times duplicated rows repeat in R data frame, which provides the following code:
library(plyr)
ddply(df,.(a,b),nrow)
However, I have a dataset with many variables, so I can't type them out like a,b in this case. I've tried using names(data) with the paste function, but it doesn't seem to work. I tried this:
var_names=paste(names(data),collapse=",")
ddply(data,.(paste(a)),nrow)
It instead gives this output:
However, if I manually type them out, I get the proper output:
What do I need to do differently here?


Instead of paste and evaluating, make use of count from dplyr, which can take multiple columns with across and select-helpers - everything()
library(dplyr)
df %>%
count(across(everything()))
A reproducible example with mtcars dataset
data(mtcars)
df <- mtcars %>%
select(vs:carb)
count(df, across(everything()))
vs am gear carb n
1 0 0 3 2 4
2 0 0 3 3 3
3 0 0 3 4 5
4 0 1 4 4 2
5 0 1 5 2 1
6 0 1 5 4 1
7 0 1 5 6 1
8 0 1 5 8 1
9 1 0 3 1 3
10 1 0 4 2 2
11 1 0 4 4 2
12 1 1 4 1 4
13 1 1 4 2 2
14 1 1 5 2 1
Also, in ddply, we can just pass a vector of column names i.e. no need to create a single string
library(plyr)
ddply(df, names(df), nrow)
vs am gear carb V1
1 0 0 3 2 4
2 0 0 3 3 3
3 0 0 3 4 5
4 0 1 4 4 2
5 0 1 5 2 1
6 0 1 5 4 1
7 0 1 5 6 1
8 0 1 5 8 1
9 1 0 3 1 3
10 1 0 4 2 2
11 1 0 4 4 2
12 1 1 4 1 4
13 1 1 4 2 2
14 1 1 5 2 1
Or if we are creating a single string from names, also paste the whole expression and then evaluate (which is not recommended as there are standard ways of dealing this)
eval(parse(text = paste('ddply(df, .(', toString(names(df)), '), nrow)')))
vs am gear carb V1
1 0 0 3 2 4
2 0 0 3 3 3
3 0 0 3 4 5
4 0 1 4 4 2
5 0 1 5 2 1
6 0 1 5 4 1
7 0 1 5 6 1
8 0 1 5 8 1
9 1 0 3 1 3
10 1 0 4 2 2
11 1 0 4 4 2
12 1 1 4 1 4
13 1 1 4 2 2
14 1 1 5 2 1


You can use aggregate by grouping all the columns and counting it's length.
aggregate(1:nrow(df)~., df, length)

Related

Reshape wide data to long with multiple variables in R (dplyr) [duplicate]

This question already has an answer here:
How to use Pivot_longer to reshape from wide-type data to long-type data with multiple variables
(1 answer)
Closed 2 years ago.
I have a dataset of adolescents over 3 waves. I need to reshape the data from wide to long, but I haven't been able to figure out how to use pivot_longer (I've checked other questions, but maybe I missed one?). Below is sample data:
HAVE DATA:
id c1sports c2sports c3sports c1smoker c2smoker c3smoker c1drinker c2drinker c3drinker
1 1 1 1 1 1 4 1 5 2
2 1 1 1 5 1 3 4 1 4
3 1 0 0 1 1 5 2 3 2
4 0 0 0 1 3 3 4 2 3
5 0 0 0 2 1 2 1 5 3
6 0 0 0 4 1 4 4 3 1
7 1 0 1 2 2 3 1 4 1
8 0 1 1 4 4 1 4 5 4
9 1 1 1 3 2 2 3 4 2
10 0 1 0 2 5 5 4 2 3
WANT DATA:
id wave sports smoker drinker
1 1 1 1 1
1 2 1 1 5
1 3 1 4 2
2 1 1 5 4
2 2 1 1 1
2 3 1 3 4
3 1 1 1 2
3 2 0 1 3
3 3 0 5 2
4 1 0 1 4
4 2 0 3 2
4 3 0 3 3
5 1 0 2 1
5 2 0 1 5
5 3 0 2 3
6 1 0 4 4
6 2 0 1 3
6 3 0 4 1
7 1 1 2 1
7 2 0 2 4
7 3 1 3 1
8 1 0 4 4
8 2 1 4 5
8 3 1 1 4
9 1 1 3 3
9 2 1 2 4
9 3 1 2 2
10 1 0 2 4
10 2 1 2 2
10 3 0 5 3
So far the only think that I've been able to run is:
long_dat <- wide_dat %>%
pivot_longer(., cols = c1sports:c3drinker)
But this doesn't get me separate columns for sports, smoker, drinker.

You could use names_pattern argument in pivot_longer.
tidyr::pivot_longer(df,
cols = -id,
names_to = c('wave', '.value'),
names_pattern = 'c(\\d+)(.*)')
# id wave sports smoker drinker
# <int> <chr> <int> <int> <int>
# 1 1 1 1 1 1
# 2 1 2 1 1 5
# 3 1 3 1 4 2
# 4 2 1 1 5 4
# 5 2 2 1 1 1
# 6 2 3 1 3 4
# 7 3 1 1 1 2
# 8 3 2 0 1 3
# 9 3 3 0 5 2
#10 4 1 0 1 4
# … with 20 more rows

File entire vector by a certain appearance of another vector

I have the following data:
players<-rep(1:3,each=3)
trial<-rep(1:3)
choice<-c(1,0,0,0,0,0,0,1,0)
gamematrix<-data.frame(cbind(players,trial,choice))
players trial choice
1 1 1 1
2 1 2 0
3 1 3 0
4 2 1 0
5 2 2 0
6 2 3 0
7 3 1 0
8 3 2 1
9 3 3 0
Now I want to create a new vector:
for each participant who have at least one choice of "1", to get the value "3" and "0" otherwise:
players trial choice win
1 1 1 1 3
2 1 2 0 3
3 1 3 0 3
4 2 1 0 0
5 2 2 0 0
6 2 3 0 0
7 3 1 0 3
8 3 2 1 3
9 3 3 0 3
In the simple example above, player "1", had "1" in the first trial, while player 3 in the second trial, thus for all their choices the value is "3" in the new vector.
Any ideas how to do it? thanks!

A base R option using ave + ifelse
within(
gamematrix,
win <- ave(choice,players,FUN = function(x) ifelse(any(x==1),3,0))
)
giving
players trial choice win
1 1 1 1 3
2 1 2 0 3
3 1 3 0 3
4 2 1 0 0
5 2 2 0 0
6 2 3 0 0
7 3 1 0 3
8 3 2 1 3
9 3 3 0 3
Update
If you criteria is depending on the first two values of choice, you can try
within(
gamematrix,
win <- ave(choice,players,FUN = function(x) ifelse(all(head(x,2)==1),3,0))
)
which gives
players trial choice win
1 1 1 1 0
2 1 2 0 0
3 1 3 0 0
4 2 1 0 0
5 2 2 0 0
6 2 3 0 0
7 3 1 0 0
8 3 2 1 0
9 3 3 0 0

Try this dplyr approach:
library(dplyr)
#Code
gamematrix <- gamematrix %>% group_by(players) %>%
mutate(win=ifelse(length(choice[choice==1])>=1,3,0))
Output:
# A tibble: 9 x 4
# Groups: players [3]
players trial choice win
<dbl> <dbl> <dbl> <dbl>
1 1 1 1 3
2 1 2 0 3
3 1 3 0 3
4 2 1 0 0
5 2 2 0 0
6 2 3 0 0
7 3 1 0 3
8 3 2 1 3
9 3 3 0 3

There is no reason for this data to be a data.frame. Keep it as a numeric matrix. If you do so you can do in one line using only vectorized functions.
cbind(gamematrix, win = (rowSums(gamematrix == 1) > 0) * 3)
for your second case:
I would like it to be only for those players who had "choice=1" in the first N (e.g., first 2 trials)
cbind(gamematrix, win = (rowSums(gamematrix[,c(1,2)] == 1) > 0) * 3)
Vectorized solutions are usually more performant than solutions incorporating a buried loop (e.g. ave).

An option with rowsum from base R
gamematrix$win <- with(gamematrix, 3 * players %in%
names(which(rowsum(choice, players)[,1] > 0)))
gamematrix$win
#[1] 3 3 3 0 0 0 3 3 3

Specify effect size in simulated data

I'm using the simstudy package to create simulated data sets for a power analysis of a mixed effects model (using lmer). What I would like to do is to be able to simulate a specific effect size for each group in the simulated data for the dependent/outcome variable. What I'm trying to do is set an r value for each group in the simulated data set across each time point.
Here's a sample data frame that I generated using simstudy
cid period Group male time
1 0 3 1 0
1 1 3 1 3
1 2 3 1 6
1 3 3 1 9
1 4 3 1 25
2 0 1 1 0
2 1 1 1 3
2 2 1 1 6
2 3 1 1 9
2 4 1 1 25
3 0 1 0 0
3 1 1 0 3
3 2 1 0 6
3 3 1 0 9
3 4 1 0 25
4 0 1 1 0
4 1 1 1 3
4 2 1 1 6
4 3 1 1 9
4 4 1 1 25
5 0 3 0 0
5 1 3 0 3
5 2 3 0 6
5 3 3 0 9
5 4 3 0 25
6 0 1 1 0
6 1 1 1 3
6 2 1 1 6
6 3 1 1 9
6 4 1 1 25
7 0 3 0 0
7 1 3 0 3
7 2 3 0 6
7 3 3 0 9
7 4 3 0 25
8 0 2 1 0
8 1 2 1 3
8 2 2 1 6
8 3 2 1 9
8 4 2 1 25
9 0 3 1 0
9 1 3 1 3
9 2 3 1 6
9 3 3 1 9
9 4 3 1 25
10 0 3 1 0
10 1 3 1 3
10 2 3 1 6
10 3 3 1 9
10 4 3 1 25
Let's assume a variable y with a mean = 25 and an SD = 10. Then assume an r for group 1 = .2, group 2 = .5, group 3 = .8
How would I simulate a variable (y) that has those properties? I was thinking something along the lines of rnorm, but really not having a lot of success.
~Note that simstudy provides a formula module to define the outcome variable - it takes the form of
#define the column variable
def <- defData(def, varname = "small.eff", dist = "normal", formula = 0.3)
#define the values for the column
dtAdd <- defDataAdd(varname = "PRCA.small", dist = "normal", formula =
"25 - (Group + 1) * (period * small.eff)", variance = 10)
But I couldn't figure out how to actually create standardized variables in the formula space that would allow me to set a concrete r value for each group.

Sum rows in a group, starting when a specific value occurs

I want to accumulate the values of a column till the end of the group, though starting the addition when a specific value occurs in another column. I am only interested in the first instance of the specific value within a group. So if that value occurs again within the group, the addition column should continue to add the values. I know this sounds like a rather strange problem, so hopefully the example table makes sense.
The following data frame is what I have now:
> df = data.frame(group = c(1,1,1,1,2,2,2,2,2,3,3,3,4,4,4),numToAdd = c(1,1,3,2,4,2,1,3,2,1,2,1,2,3,2),occurs = c(0,0,1,0,0,1,0,0,0,0,1,1,0,0,0))
> df
group numToAdd occurs
1 1 1 0
2 1 1 0
3 1 3 1
4 1 2 0
5 2 4 0
6 2 2 1
7 2 1 0
8 2 3 0
9 2 2 0
10 3 1 0
11 3 2 1
12 3 1 1
13 4 2 0
14 4 3 0
15 4 2 0
Thus, whenever a 1 occurs within a group, I want a cumulative sum of the values from the column numToAdd, until a new group starts. This would look like the following:
> finalDF = data.frame(group = c(1,1,1,1,2,2,2,2,2,3,3,3,4,4,4),numToAdd = c(1,1,3,2,4,2,1,3,2,1,2,1,2,3,2),occurs = c(0,0,1,0,0,1,0,0,0,0,1,1,0,0,0),added = c(0,0,3,5,0,2,3,6,8,0,2,3,0,0,0))
> finalDF
group numToAdd occurs added
1 1 1 0 0
2 1 1 0 0
3 1 3 1 3
4 1 2 0 5
5 2 4 0 0
6 2 2 1 2
7 2 1 0 3
8 2 3 0 6
9 2 2 0 8
10 3 1 0 0
11 3 2 1 2
12 3 1 1 3
13 4 2 0 0
14 4 3 0 0
15 4 2 0 0
Thus, the added column is 0 until a 1 occurs within the group, then accumulates the values from numToAdd until it moves to a new group, turning the added column back to 0. In group three, a value of 1 is found a second time, yet the cumulated sum continues. Additionally, in group 4, a value of 1 is never found, thus the value within the added column remains 0.
I've played around with dplyr, but can't get it to work. The following solution only outputs the total sum, and not the increasing cumulated number at each row.
library(dplyr)
df =
df %>%
mutate(added=ifelse(occurs == 1,cumsum(numToAdd),0)) %>%
group_by(group)

Try
df %>%
group_by(group) %>%
mutate(added= cumsum(numToAdd*cummax(occurs)))
# group numToAdd occurs added
# 1 1 1 0 0
# 2 1 1 0 0
# 3 1 3 1 3
# 4 1 2 0 5
# 5 2 4 0 0
# 6 2 2 1 2
# 7 2 1 0 3
# 8 2 3 0 6
# 9 2 2 0 8
# 10 3 1 0 0
# 11 3 2 1 2
# 12 3 1 1 3
# 13 4 2 0 0
# 14 4 3 0 0
# 15 4 2 0 0
Or using data.table
library(data.table)#v1.9.5+
i1 <-setDT(df)[, .I[(rleid(occurs) + (occurs>0))>1], group]$V1
df[, added:=0][i1, added:=cumsum(numToAdd), by = group]
Or a similar option as in dplyr
setDT(df)[,added := cumsum(numToAdd * cummax(occurs)) , by = group]

You can use split-apply-combine in base R with something like:
df$added <- unlist(lapply(split(df, df$group), function(x) {
y <- rep(0, nrow(x))
pos <- cumsum(x$occurs) > 0
y[pos] <- cumsum(x$numToAdd[pos])
y
}))
df
# group numToAdd occurs added
# 1 1 1 0 0
# 2 1 1 0 0
# 3 1 3 1 3
# 4 1 2 0 5
# 5 2 4 0 0
# 6 2 2 1 2
# 7 2 1 0 3
# 8 2 3 0 6
# 9 2 2 0 8
# 10 3 1 0 0
# 11 3 2 1 2
# 12 3 1 1 3
# 13 4 2 0 0
# 14 4 3 0 0
# 15 4 2 0 0

To add another base R approach:
df$added <- unlist(lapply(split(df, df$group), function(x) {
c(x[,'occurs'][cumsum(x[,'occurs']) == 0L],
cumsum(x[,'numToAdd'][cumsum(x[,'occurs']) != 0L]))
}))
# group numToAdd occurs added
# 1 1 1 0 0
# 2 1 1 0 0
# 3 1 3 1 3
# 4 1 2 0 5
# 5 2 4 0 0
# 6 2 2 1 2
# 7 2 1 0 3
# 8 2 3 0 6
# 9 2 2 0 8
# 10 3 1 0 0
# 11 3 2 1 2
# 12 3 1 1 3
# 13 4 2 0 0
# 14 4 3 0 0
# 15 4 2 0 0

Another base R:
df$added <- unlist(lapply(split(df,df$group),function(x){
cumsum((cumsum(x$occurs) > 0) * x$numToAdd)
}))

Transforming a two-way table() into pairwise list of counts in R

Starting with some sample two-way frequency table:
a <- c(1,2,3,4,4,3,4,2,2,2)
b <- c(1,2,3,4,1,2,4,3,2,2)
tab <- table(a,b)
> tab
b
a 1 2 3 4
1 1 0 0 0
2 0 3 1 0
3 0 1 1 0
4 1 0 0 2
I need to transform the table into the following format:
goal <- data.frame(a=c(1,2,3,4),b=c(1,2,3,4),count=c(1,3,1,2))
> goal
a b count
1 1 1 1
2 2 2 3
3 3 3 1
4 4 4 2
. . . .
How can I form all pairwise combinations from the two-way table and add the frequency counts in the third column?
Intuition tells me there should be a simple kind of 'reverse' function for table, but I could not find anything on SO or Google.

Naturally, after posting the question I found the right search query for Google...
> as.data.frame(tab)
a b Freq
1 1 1 1
2 2 1 0
3 3 1 0
4 4 1 1
5 1 2 0
6 2 2 3
7 3 2 1
8 4 2 0
9 1 3 0
10 2 3 1
11 3 3 1
12 4 3 0
13 1 4 0
14 2 4 0
15 3 4 0
16 4 4 2

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