I want to create a new column based on some conditions imposed on several columns. For example, here is an example dataset:
a <- data.frame(x=c(1,0,1,0,0), y=c(0,0,0,0,0), z=c(1,1,0,0,0))
a
x y z
1 1 0 1
2 0 0 1
3 1 0 0
4 0 0 0
5 0 0 0
Specifically, if for any particular row 1 is present, then the new column returns 1. If all are 0, then the new column returns 0. So the dataset with the new column will be
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
My initial thought was to use %in% but couldn't get the result I want. Thank you for your help!
If your data frame consists of binary values, e.g., only 0 and 1, you can try the code below with rowSums
a$w <- +(rowSums(a)>0)
such that
> a
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
We can use rowMaxs from matrixStats
library(matrixStats)
a$w <- rowMaxs(as.matrix(a))
a$w
#[1] 1 1 1 0 0
You can find max of each row :
a$w <- do.call(pmax, a)
a
# x y z w
#1 1 0 1 1
#2 0 0 1 1
#3 1 0 0 1
#4 0 0 0 0
#5 0 0 0 0
which can also be done with apply :
a$w <- apply(a, 1, max)
Related
I'm currently stuck on a part of my code that feels intuitive but I can't figure a way to do it. I have a very big data frame (nrows = 34036, ncol = 43) in which I want to create a continuous sequence of the variables where the value of the row is 1 (without having multiple columns with 1). It consists of only zeros and ones similar to the following:
A B C D
1 0 0 0
0 0 0 1
0 0 0 1
0 0 0 0
0 0 0 0
1 0 1 0
1 0 1 0
0 1 0 0
0 1 0 0
1 0 0 1
I was able to remove the zeroes using:
#find the sum of each row
placeholderData <- transform(placeholderData, sum=rowSums(placeholderData))
placeholderData <- placeholderData[!(placeholderData$sum <= 0),]
And the data frame now looks like:
A B C D sum
1 0 0 0 1
0 0 0 1 1
0 0 0 1 1
1 0 1 0 2
1 0 1 0 2
0 1 0 0 1
0 1 0 0 1
1 0 0 1 2
My main problem comes when there are two or more 1's in a row. To try to solve this, I used the following code to identify the columns that have a sum of 2 or more:
placeholderData$Matches <- lapply(apply(placeholderData == 1, 1, which), names)
Which added the following column to the data frame:
A B C D sum Matches
1 0 0 0 1 A
0 0 0 1 1 D
0 0 0 1 1 D
1 0 1 0 2 c("A","C")
1 0 1 0 2 c("A","C")
0 1 0 0 1 B
0 1 0 0 1 B
1 0 0 1 2 c("A", "D")
I added the Matches column as an approach to solve the problem, but I'm not sure how would I do it without using a lot of logical operators (I don't know what columns have matches or not). What I would like to do is to aggregate the rows that have more than (or equal to) two 1's into a new column, to be able to have a data frame like this:
A B C D AC AD sum Matches
1 0 0 0 0 0 1 A
0 0 0 1 0 0 1 D
0 0 0 1 0 0 1 D
0 0 0 0 1 0 1 c("A","C")
0 0 0 0 1 0 1 c("A","C")
0 1 0 0 0 0 1 B
0 1 0 0 0 0 1 B
0 0 0 0 0 1 1 c("A", "D")
Then, I would be able to use my code as normal (It works just fine when there are no repeated values in rows). I tried searching to find similar questions, but I'm not sure if I was even asking the right question. I was wondering if anyone could provide some help or some ideas that I could try.
Thank you very much!
This seems a lot like making dummy variables, so I would use the model.matrix function commonly used for dummy variables (one-hot encoding):
m = read.table(header = T, text = "A B C D
1 0 0 0
0 0 0 1
0 0 0 1
0 0 0 0
0 0 0 0
1 0 1 0
1 0 1 0
0 1 0 0
0 1 0 0
1 0 0 1")
m = m[rowSums(m) > 0, ]
d = factor(sapply(apply(m == 1, 1, which), function(x) paste(names(m)[x], collapse = "")))
result = data.frame(model.matrix(~ d + 0))
names(result) = levels(d)
# A AC AD B D
# 1 1 0 0 0 0
# 2 0 0 0 0 1
# 3 0 0 0 0 1
# 4 0 1 0 0 0
# 5 0 1 0 0 0
# 6 0 0 0 1 0
# 7 0 0 0 1 0
# 8 0 0 1 0 0
I have a data set with user to user. It doesn't have all users as col and row. For example,
U1 U2 T
1 3 1
1 6 1
2 4 1
3 5 1
u1 and u2 represent users of the dataset. When I create a sparse matrix using following code, (df- keep all data of above dataset as a dataframe)
trustmatrix <- xtabs(T~U1+U2,df,sparse = TRUE)
3 4 5 6
1 1 0 0 1
2 0 1 0 0
3 0 0 1 0
Because this matrix doesn't have all the users in row and columns as below.
1 2 3 4 5 6
1 0 0 1 0 0 1
2 0 0 0 1 0 0
3 0 0 0 0 1 0
4 0 0 0 0 0 0
5 0 0 0 0 0 0
6 0 0 0 0 0 0
If I want to get above matrix after sparse matrix, How can I do so in R?
We can convert the columns to factor with levels as 1 through 6 and then use xtabs
df1[1:2] <- lapply(df1[1:2], factor, levels = 1:6)
as.matrix(xtabs(T~U1+U2,df1,sparse = TRUE))
# U2
#U1 1 2 3 4 5 6
# 1 0 0 1 0 0 1
# 2 0 0 0 1 0 0
# 3 0 0 0 0 1 0
# 4 0 0 0 0 0 0
# 5 0 0 0 0 0 0
# 6 0 0 0 0 0 0
Or another option is to get the expanded index filled with 0s and then use sparseMatrix
library(tidyverse)
library(Matrix)
df2 <- crossing(U1 = 1:6, U2 = 1:6) %>%
left_join(df1) %>%
mutate(T = replace(T, is.na(T), 0))
sparseMatrix(i = df2$U1, j = df2$U2, x = df2$T)
Or use spread
spread(df2, U2, T)
Let's say I have 3 vectors (strings of 10):
X <- c(1,1,0,1,0, 1,1, 0, NA,NA)
H <- c(0,0,1,0,NA,1,NA,1, 1, 1 )
I <- c(0,0,0,0,0, 1,NA,NA,NA,1 )
Data.frame Y contains 10 columns and 6 rows:
1 2 3 4 5 6 7 8 9 10
0 1 0 0 1 1 1 0 1 0
1 1 1 0 1 0 1 0 0 0
0 0 0 0 1 0 0 1 0 1
1 0 1 1 0 1 1 1 0 0
0 0 0 0 0 0 1 0 0 0
1 1 0 1 0 0 0 0 1 1
I'd like to use vector X, H en I to make column selections in data.frame Y, using "1's" and "0's" in the vector as selection criterium .
So the results for vector X using the '1' as selection criterium should be:
X <- c(1,1,0,1,0, 1,1, 0, NA,NA)
1 2 4 6 7
0 1 0 1 1
1 1 0 0 1
0 0 0 0 0
1 0 1 1 1
0 0 0 0 1
1 1 1 0 0
For vector H using the '1' as selection criterium:
H <- c(0,0,1,0,NA,1,NA,1, 1, 1 )
3 6 8 9 10
0 1 0 1 0
1 0 0 0 0
0 0 1 0 1
1 1 1 0 0
0 0 0 0 0
0 0 0 1 1
For vector I using the '1' as selection criterium:
I <- c(0,0,0,0,0, 1,NA,NA,NA,1 )
6 10
1 0
0 0
0 1
1 0
0 0
0 1
For convenience and speed I'd like to use a loop. It might be something like this:
all.ones <- lapply[,function(x) x %in% 1]
In the outcome (all.ones), the result for each vector should stay separate. For example:
X 1,2,4,6,7
H 3,6,8,9,10
I 6,10
The standard way of doing this is using the %in% operator:
Y[, X %in% 1]
To do this for multiple vectors (assuming you want an AND operation):
mylist = list(X, H, I, D, E, K)
Y[, Reduce(`&`, lapply(mylist, function(x) x %in% 1))]
The problem is the NA, use which to get round it. Consider the following:
x <- c(1,0,1,NA)
x[x==1]
[1] 1 1 NA
x[which(x==1)]
[1] 1 1
How about this?
idx <- which(X==1)
Y[,idx]
EDIT: For six vectors, do
idx <- which(X==1 & H==1 & I==1 & D==1 & E==1 & K==1)
Y[,idx]
Replace & with | if you want all columns of Y where at least one of the lists has a 1.
I'm trying to generate the following matrix, based on a multinomial framework. For example, if I had three columns, I'd get:
0 0 0
1 0 0
0 1 0
0 0 1
1 1 0
1 0 1
0 1 1
1 1 1
But, I want many more columns. I know I can use expand.grid, like:
u <- list(0:1)
expand.grid(rep(u,3))
But, it returns what I want in the wrong order:
0 0 0
1 0 0
0 1 0
1 1 0
0 0 1
1 0 1
0 1 1
1 1 1
Any ideas? Thanks.
You can reorder your rows to match your expected output:
u <- list(0:1)
g <- expand.grid(rep(u,3))
g <- g[order(rowSums(g)), ]
I have a long format unbalanced longitudinal data. I would like to exclude all the cases that do not contain complete information. By that I mean all cases that do not repeat 8 times. Someone can help me finding a solution?
Below an example: I have three subjects {A, B, and C}. I have 8 information for A and B, but only 2 for C. How can I delete rows in which C is present based on the information it has less than 8 repeated measurements?
temp = scan()
A 1 1 1 0
A 1 1 0 1
A 1 0 0 0
A 1 1 1 1
A 0 1 0 0
A 1 1 1 0
A 1 1 0 1
A 1 0 0 0
B 1 1 1 0
B 1 1 0 1
B 1 0 0 0
B 1 1 1 1
B 0 1 0 0
B 1 1 1 0
B 1 1 0 1
B 1 0 0 0
C 1 1 1 1
C 0 1 0 0
Any help?
Assuming your variable names are V1, V2... and so on, here's one approach:
temp[temp$V1 %in% names(which(table(temp$V1) == 8)), ]
The table(temp$V1) == 8 matches the values in the V1 column that have exactly 8 cases. The names(which(... part creates a basic character vector that we can match using %in%.
And another:
temp[ave(as.character(temp$V1), temp$V1, FUN = length) == "8", ]
Here's another approach:
temp <- read.table(text="
A 1 1 1 0
A 1 1 0 1
A 1 0 0 0
A 1 1 1 1
A 0 1 0 0
A 1 1 1 0
A 1 1 0 1
A 1 0 0 0
B 1 1 1 0
B 1 1 0 1
B 1 0 0 0
B 1 1 1 1
B 0 1 0 0
B 1 1 1 0
B 1 1 0 1
B 1 0 0 0
C 1 1 1 1
C 0 1 0 0", header=FALSE)
do.call(rbind,
Filter(function(subgroup) nrow(subgroup) == 8,
split(temp, temp[[1]])))
split breaks the data.frame up by its first column, then Filter drops the subgroups that don't have 8 rows. Finally, do.call(rbind, ...) collapses the remaining subgroups back into a single data.frame.
If the first column of temp is character (rather than factor, which you can verify with str(temp)) and the rows are ordered by subgroup, you could also do:
with(rle(temp[[1]]), temp[rep(lengths==8, times=lengths), ])