I have a dataset with multiple variables. Each question has the actual survey answer and three other characteristics. So there are four variables for each question. I want to specify if Q135_L ==1 , leave Q135_RT as it is, otherwise code it as NA. I can do that with an ifelse statement.
df$Q135_RT <- ifelse(df$Q135_L == 1, df$Q22_RT, NA)
However, I have hundreds of variables and the names are not related. For example, in the picture we can see Q135, SG1_1 and so on. How can I specify for the whole dataset if a variable ends at _L, then for the same variable ending at _RT should remain as it is, otherwise the variable ending at _RT should be coded as NA.
I tried this but it only returns NAs
ifelse(grepl("//b_L" ==1, df), "//b_RT" , NA)
If I understand your problem correctly, you have a data frame of which the columns represent survey question variables. Each column contains two identifiers, namely: a survey question number (134, 135, etc) and a variable letter (L, R, etc). Because you provide no reproducible example, I tried to make a simplified example of your data frame:
set.seed(5)
DF <- data.frame(array(sample(1:4, 24, replace = TRUE), c(4,6)))
colnames(DF) <- c("Q134_L","Q135_L", "Q134_R", "Q135_R", "Q_L1", "Q134_S")
DF
# Q134_L Q135_L Q134_R Q135_R Q_L1 Q134_S
# 1 2 3 2 3 1 1
# 2 3 1 3 2 4 4
# 3 1 1 3 2 4 3
# 4 3 1 3 3 2 1
What you want is that if Q135_L == 1, leave Q135_RT as it is, otherwise code it as NA. Here is a function that implements this recoding logic:
recode <- function(yourdf, questnums) {
for (k in 1:length(questnums)) {
charnum <- as.character(questnums)
col_end_L_k <- yourdf[grepl("_L\\b", colnames(yourdf)) &
grepl(charnum[k], colnames(yourdf))]
col_end_R_k <- yourdf[grepl("_RT\\b", colnames(yourdf)) &
grepl(charnum[k], colnames(yourdf))]
row_is_1 <- which(col_end_L_k == 1)
col_end_R_k[-row_is_1, ] <- NA
yourdf[, colnames(col_end_R_k)] <- col_end_R_k
}
return(yourdf)
}
This function takes a data frame and a vector of question numbers, and then returns the data frame that has been recoded.
What this function does:
Selecting each question number using for.
Using grepl to identify any column that contains the selected number and contains _L at the end of the column name.
Similar with above but for _RT at the end of the column name.
Using which to identify the location of rows in the _L column that contain 1.
Keeping the values of the _RT column, which has the same question number with the corresponding _L column, in those rows, and change values on other rows to NA.
The result:
recode(DF, 134:135)
# Q134_L Q135_L Q134_RT Q135_RT Q_L1 Q134_S
# 1 2 3 NA NA 1 1
# 2 3 1 NA 2 4 4
# 3 1 1 3 2 4 3
# 4 3 1 NA 3 2 1
Note that the Q_L1 column is not affected because _L in this column is not located on the end of the column name.
As for how to define questnums, the question numbers, you just need to create a numeric vector. Examples:
Your questnums are 1 to 200. Then use 1:200 or seq(200), so recode(DF, 1:200).
Your questnums are 1, 3, 134, 135. Then, use recode(DF, c(1, 3, 134, 135)).
You can also assign the question numbers to an object first, such as n = c(25, 135, 145) and the use it : recode(DF, n)
Related
I want to make a table called Count_Table and in it, Id like to count the number of 0s, 1s, and 5s when column "num" == 1,2,3,4, etc.
For example, the code below will count the 0s,1s,and 5s in column "num" when "num == "1". This is great but i need to do this 34 more times since "num" goes from 1-35.
Count_Table <- table(SASS_data[num == "1"]$Visited5)
I am new to R and I don't know how to add 1 to the "num" and loop it until 35 so that the Count_Table includes the counts of 0,1,5 for all nums that exist (1-35). I am sorry if this is confusing and thank you for your help.
lapply will generate a list of tables that span the columns of a dataframe. E.g.,
tablist <- lapply(mtcars, table)
If your dataframe contains columns you want to exclude, can do that by restricting the dataframe. E.g.,
tablist2 <- lapply(mtcars[, c(2, 4, 7)], table)
Answer
Table works on multiple dimensions. Just put both num and Visited5 as arguments. This also works if not all unique values of Visited5 are present in every level of num, those cells will simply be set to 0.
Example
SASS_data <- data.frame(
num = rep(1:5, each = 5),
Visited5 = sample(1:3, 25, r = T)
)
table(SASS_data$num, SASS_data$Visited5)
# 1 2 3
# 1 2 1 2
# 2 1 3 1
# 3 1 1 3
# 4 2 0 3
# 5 2 2 1
I need to pull records from a first data set (called df1 here) based on a combination of specific dates, ID#s, event start time, and event end time that match with a second data set (df2). Everything works fine when there is just 1 date, ID, and event start and end time, but some of the matching records between the data sets contain multiple IDs, dates, or times, and I can't get the records from df1 to subset properly in those cases. I ultimately want to put this in a FOR loop or independent function since I have a rather large dataset. Here's what I've got so far:
I started just by matching the dates between the two data sets as follows:
match_dates <- as.character(intersect(df1$Date, df2$Date))
Then I selected the records in df2 based on the first matching date, also keeping the other columns so I have the other ID and time information I need:
records <- df2[which(df2$Date == match_dates[1]), ]
The date, ID, start, and end time from records are then:
[1] "01-04-2009" "599091" "12:00" "17:21"
Finally I subset df1 for before and after the event based on the date, ID, and times in records and combined them into a new data frame called final to get at the data contained in df1 that I ultimately need.
before <- subset(df1, NUM==records$ID & Date==records$Date & Time<records$Start)
after <- subset(df1, NUM==records$ID & Date==records$Date & Time>records$End)
final <- rbind(before, after)
Here's the real problem - some of the matching dates have more than 1 corresponding row in df2, and return multiple IDs or times. Here is what an example of multiple records looks like:
records <- df2[which(df2$Date == match_dates[25]), ]
> records$ID
[1] 507646 680845 680845
> records$Date
[1] "04-02-2009" "04-02-2009" "04-02-2009"
> records$Start
[1] "09:43" "05:37" "11:59"
> records$End
[1] "05:19" "11:29" "16:47"
When I try to subset df1 based on this I get an error:
before <- subset(df1, NUM==records$ID & Date==records$Date & Time<records$Start)
Warning messages:
1: In NUM == records$ID :
longer object length is not a multiple of shorter object length
2: In Date == records$Date :
longer object length is not a multiple of shorter object length
3: In Time < records$Start :
longer object length is not a multiple of shorter object length
Trying to do it manually for each ID-date-time combination would be way to tedious. I have 9 years worth of data, all with multiple matching dates for a given year between the data sets, so ideally I would like to set this up as a FOR loop, or a function with a FOR loop in it, but I can't get past this. Thanks in advance for any tips!
If you're asking what I think you are the filter() function from the dplyr package combined with the match function does what you're looking for.
> df1 <- data.frame(A = c(rep(1,4),rep(2,4),rep(3,4)), B = c(rep(1:4,3)))
> df1
A B
1 1 1
2 1 2
3 1 3
4 1 4
5 2 1
6 2 2
7 2 3
8 2 4
9 3 1
10 3 2
11 3 3
12 3 4
> df2 <- data.frame(A = c(1,2), B = c(3,4))
> df2
A B
1 1 3
2 2 4
> filter(df1, A %in% df2$A, B %in% df2$B)
A B
1 1 3
2 1 4
3 2 3
4 2 4
I have Valence Category for word stimuli in my psychology experiment.
1 = Negative, 2 = Neutral, 3 = Positive
I need to sort the thousands of stimuli with a pseudo-randomised condition.
Val_Category cannot have more than 2 of the same valence stimuli in a row i.e. no more than 2x negative stimuli in a row.
for example - 2, 2, 2 = not acceptable
2, 2, 1 = ok
I can't sequence the data i.e. decide the whole experiment will be 1,3,2,3,1,3,2,3,2,2,1 because I'm not allowed to have a pattern.
I tried various packages like dylpr, sample, order, sort and nothing so far solves the problem.
I think there's a thousand ways to do this, none of which are probably very pretty. I wrote a small function that takes care of the ordering. It's a bit hacky, but it appeared to work for what I tried.
To explain what I did, the function works as follows:
Take the vector of valences and samples from it.
If sequences are found that are larger than the desired length, then, (for each such sequence), take the last value of that sequence at places it "somewhere else".
Check if the problem is solved. If so, return the reordered vector. If not, then go back to 2.
# some vector of valences
val <- rep(1:3,each=50)
pseudoRandomize <- function(x, n){
# take an initial sample
out <- sample(val)
# check if the sample is "bad" (containing sequences longer than n)
bad.seq <- any(rle(out)$lengths > n)
# length of the whole sample
l0 <- length(out)
while(bad.seq){
# get lengths of all subsequences
l1 <- rle(out)$lengths
# find the bad ones
ind <- l1 > n
# take the last value of each bad sequence, and...
for(i in cumsum(l1)[ind]){
# take it out of the original sample
tmp <- out[-i]
# pick new position at random
pos <- sample(2:(l0-2),1)
# put the value back into the sample at the new position
out <- c(tmp[1:(pos-1)],out[i],tmp[pos:(l0-1)])
}
# check if bad sequences (still) exist
# if TRUE, then 'while' continues; if FALSE, then it doesn't
bad.seq <- any(rle(out)$lengths > n)
}
# return the reordered sequence
out
}
Example:
The function may be used on a vector with or without names. If the vector was named, then these names will still be present on the pseudo-randomized vector.
# simple unnamed vector
val <- rep(1:3,each=5)
pseudoRandomize(val, 2)
# gives:
# [1] 1 3 2 1 2 3 3 2 1 2 1 3 3 1 2
# when names assigned to the vector
names(val) <- 1:length(val)
pseudoRandomize(val, 2)
# gives (first row shows the names):
# 1 13 9 7 3 11 15 8 10 5 12 14 6 4 2
# 1 3 2 2 1 3 3 2 2 1 3 3 2 1 1
This property can be used for randomizing a whole data frame. To achieve that, the "valence" vector is taken out of the data frame, and names are assigned to it either by row index (1:nrow(dat)) or by row names (rownames(dat)).
# reorder a data.frame using a named vector
dat <- data.frame(val=rep(1:3,each=5), stim=rep(letters[1:5],3))
val <- dat$val
names(val) <- 1:nrow(dat)
new.val <- pseudoRandomize(val, 2)
new.dat <- dat[as.integer(names(new.val)),]
# gives:
# val stim
# 5 1 e
# 2 1 b
# 9 2 d
# 6 2 a
# 3 1 c
# 15 3 e
# ...
I believe this loop will set the Valence Category's appropriately. I've called the valence categories treat.
#Generate example data
s1 = data.frame(id=c(1:10),treat=NA)
#Setting the first two rows
s1[1,"treat"] <- sample(1:3,1)
s1[2,"treat"] <- sample(1:3,1)
#Looping through the remainder of the rows
for (i in 3:length(s1$id))
{
s1[i,"treat"] <- sample(1:3,1)
#Check if the treat value is equal to the previous two values.
if (s1[i,"treat"]==s1[i-1,"treat"] & s1[i-1,"treat"]==s1[i-2,"treat"])
#If so draw one of the values not equal to that value
{
a = 1:3
remove <- s1[i,"treat"]
a=a[!a==remove]
s1[i,"treat"] <- sample(a,1)
}
}
This solution is not particularly elegant. There may be a much faster way to accomplish this by sorting several columns or something.
I have large data-frame consists of two columns. I want to calculate the average of the second column values for each subset of the first column. The subset of the first column is based on a specified granularity. For example, for the following data-frame, df, I want to calculate the average of df$B values for each subset of df$A with an increment(granularity) of 1 for each subset. The results should be in two new columns.
A B expected results newA newB
0.22096 1 0 1.142857
0.33489 1 1 2
0.33655 1 2 4
0.43953 1
0.64933 2
0.86668 1
0.96932 1
1.09342 2
1.58314 2
1.88481 2
2.07654 4
2.34652 3
2.79777 5
This is a simple example, I'm not sure how to loop over the whole data-frame and perform the calculation i.e. the average of the df$B.
tried below to subset, but couldn't figure how to append the results and create final results:
Tried something like :
increment<-1
mx<-max(df$A)
i<-0
newDF<-data.frame()
while(i < mx){
tmp<-subset(df, (A >i & A< (i+increment)))
i<-i+granualrity
}
Not sure about the logic. But I'm sure there is a short way to do the required calculation. Any thoughts?
I would use findInterval for the subset selection (In your example a simple ceiling for each A value should be sufficient, too. But if your increment is different from 1 you need findInterval.) and tapply to calculate the mean:
df <- read.table(textConnection("
A B
0.22096 1
0.33489 1
0.33655 1
0.43953 1
0.64933 2
0.86668 1
0.96932 1
1.09342 2
1.58314 2
1.88481 2
2.07654 4
2.34652 3
2.79777 5"), header=TRUE)
## sort data.frame by column A (needed for findInterval)
df <- df[order(df$A), ]
## define granuality
subsets <- seq(1, max(ceiling(df$A)), by=1) # change the "by" argument for different increments
df$subset <- findInterval(df$A, subsets)
tapply(df$B, df$subset, mean)
# 0 1 2
#1.142857 2.000000 4.000000
I'm pretty confused on how to go about this. Say I have two columns in a dataframe. One column a numerical series in order (x), the other specifying some value from the first, or -1 (y). These are results from a matching experiment, where the goal is to see if multiple photos are taken of the same individual. In the example below, there 10 photos, but 6 are unique individuals. In the y column, the corresponding x is reported if there is a match. y is -1 for no match (might as well be NAs). If there is more than 2 photos per individual, the match # will be the most recent record (photo 1, 5 and 7 are the same individual below). The group is the time period the photo was take (no matches within a group!). Hopefully I've got this example right:
x <- c(1,2,3,4,5,6,7,8,9,10)
y <- c(-1,-1,-1,-1,1,-1,1,-1,2,4)
group <- c(1,1,1,2,2,2,3,3,3,3)
DF <- data.frame(x,y,group)
I would like to create a new variable to name the unique individuals, and have a final dataset with a single row per individual (i.e. only have 6 rows instead of 10), that also includes the group information. I.e. if an individual is in all three groups, there could be a value of "111" or if just in the first and last group it would be "101". Any tips?
Thanks for asking about the resulting dataset. I realized my group explanation was bad based on the actual numbers I gave, so I changed the results slightly. Bonus would also be nice to have, but not critical.
name <- c(1,2,3,4,6,8)
group_history <- as.character(c('111','101','100','011','010','001'))
bonus <- as.character(c('1,5,7','2,9','3','4,10','6','8'))
results_I_want <- data.frame(name,group_history,bonus)
My word, more mistakes fixed above...
Using the (updated) example you gave
x <- c(1,2,3,4,5,6,7,8,9,10)
y <- c(-1,-1,-1,-1,1,-1,1,-1,3,4)
group <- c(1,1,1,2,2,2,3,3,3,3)
DF <- data.frame(x,y,group)
Use the x and y to create a mapping from higher numbers to lower numbers that are the same person. Note that names is a string, despite it be a string of digits.
bottom.df <- DF[DF$y==-1,]
mapdown.df <- DF[DF$y!=-1,]
mapdown <- c(mapdown.df$y, bottom.df$x)
names(mapdown) <- c(mapdown.df$x, bottom.df$x)
We don't know how many times it might take to get everything down to the lowest number, so have to use a while loop.
oldx <- DF$x
newx <- mapdown[as.character(oldx)]
while(any(oldx != newx)) {
oldx = newx
newx = mapdown[as.character(oldx)]
}
The result is the group it belongs to, names by the lowest number of that set.
DF$id <- unname(newx)
Getting the group membership is harder. Using reshape2 to convert this into wide format (one column per group) where the column is "1" if there was something in that one and "0" if not.
library("reshape2")
wide <- dcast(DF, id~group, value.var="id",
fun.aggregate=function(x){if(length(x)>0){"1"}else{"0"}})
Finally, paste these "0"/"1" memberships together to get the grouping variable you described.
wide$grouping = apply(wide[,-1], 1, paste, collapse="")
The result:
> wide
id 1 2 3 grouping
1 1 1 1 1 111
2 2 1 0 0 100
3 3 1 0 1 101
4 4 0 1 1 011
5 6 0 1 0 010
6 8 0 0 1 001
No "bonus" yet.
EDIT:
To get the bonus information, it helps to redo the mapping to keep everything. If you have a lot of cases, this could be slow.
Replace the oldx/newx part with:
iterx <- matrix(DF$x, ncol=1)
iterx <- cbind(iterx, mapdown[as.character(iterx[,1])])
while(any(iterx[,ncol(iterx)]!=iterx[,ncol(iterx)-1])) {
iterx <- cbind(iterx, mapdown[as.character(iterx[,ncol(iterx)])])
}
DF$id <- iterx[,ncol(iterx)]
To generate the bonus data, then you can use
bonus <- tapply(iterx[,1], iterx[,ncol(iterx)], paste, collapse=",")
wide$bonus <- bonus[as.character(wide$id)]
Which gives:
> wide
id 1 2 3 grouping bonus
1 1 1 1 1 111 1,5,7
2 2 1 0 0 100 2
3 3 1 0 1 101 3,9
4 4 0 1 1 011 4,10
5 6 0 1 0 010 6
6 8 0 0 1 001 8
Note this isn't same as your example output, but I don't think your example output is right (how can you have a grouping_history of "000"?)
EDIT:
Now it agrees.
Another solution for bonus variable
f_bonus <- function(data=df){
data_a <- subset(data,y== -1,select=x)
data_a$pos <- seq(nrow(data_a))
data_b <- subset(df,y!= -1,select=c(x,y))
data_b$pos <- match(data_b$y, data_a$x)
data_t <- rbind(data_a,data_b[-2])
data_t <- with(data_t,tapply(x,pos,paste,sep="",collapse=","))
return(data_t)
}